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What changes for linear algebra over a finite field?
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This question asks which standard results from linear algebra over a field no longer hold when we generalize the algebraic structure of the scalars to be an arbitrary division ring.
My question is similar but considers a less drastic generalization. In elementary courses on linear algebra, the underlying field is virtually always assumed to be either the real or the complex numbers. (Maybe once in a blue moon, the rationals.) As such, all my intuition is for infinite fields. Moreover, I know that fields of characteristic 2 are especially problematic.
Which theorems from linear algebra no longer hold when we go from an infinite field to a finite field of characteristic greater than 2?
Which further theorems break down (nontrivially) when we go from characteristic greater than 2 to characteristic 2?
linear-algebra finite-fields intuition
asked Jul 29 at 0:08
tparker
1,679629
 |Â
show 11 more comments
up vote
9
down vote
favorite
This question asks which standard results from linear algebra over a field no longer hold when we generalize the algebraic structure of the scalars to be an arbitrary division ring.
My question is similar but considers a less drastic generalization. In elementary courses on linear algebra, the underlying field is virtually always assumed to be either the real or the complex numbers. (Maybe once in a blue moon, the rationals.) As such, all my intuition is for infinite fields. Moreover, I know that fields of characteristic 2 are especially problematic.
Which theorems from linear algebra no longer hold when we go from an infinite field to a finite field of characteristic greater than 2?
Which further theorems break down (nontrivially) when we go from characteristic greater than 2 to characteristic 2?
linear-algebra finite-fields intuition
asked Jul 29 at 0:08
tparker
1,679629
2
My experience is that almost everything works exactly the same way until you hit eigenvalues.
– Randall
Jul 29 at 0:14
2
One loses the notion of an inner product. Thus the "geometric aspects", like the Cauhy Schwarz inequality, no longer make sense.
– caffeinemachine
Jul 29 at 0:38
1
@caffeinemachine But one doesn't have the notion of an inner product in a general vector space anyway - only an inner product space. Anyway, what's stopping you from using the standard inner product $(u, v) = sum_i u_i v_i$ for an arbitrary field (or even ring), at least for finite-dimensional vector spaces?
– tparker
Jul 29 at 0:49
1
@CaveJohnson Sure, but you enlarge to the complexes for algebraic-closure reasons. This isn't so nice in prime characteristic.
– Randall
Jul 29 at 1:12
2
@tparker One can use the bilinear form that you have defined. I wouldn't call it an inner product. Because there is no meaning of positivity ($xcdot xgeq$ 0 for all $x$ and $xcdot x=0$ if and only if $x=0$) if the field is not real or complex (and hence no Cauchy-Schwarz). You are right in saying that there is no notion of an inner product on a plain simple (real or complex) vector space. But one can put an inner product forcefully and deduce purely linear algebraic results which do not have anything to do with any extra structure on the vector space.
– caffeinemachine
Jul 29 at 3:58
 |Â
show 11 more comments
up vote
9
down vote
favorite
up vote
9
down vote
favorite
This question asks which standard results from linear algebra over a field no longer hold when we generalize the algebraic structure of the scalars to be an arbitrary division ring.
My question is similar but considers a less drastic generalization. In elementary courses on linear algebra, the underlying field is virtually always assumed to be either the real or the complex numbers. (Maybe once in a blue moon, the rationals.) As such, all my intuition is for infinite fields. Moreover, I know that fields of characteristic 2 are especially problematic.
Which theorems from linear algebra no longer hold when we go from an infinite field to a finite field of characteristic greater than 2?
Which further theorems break down (nontrivially) when we go from characteristic greater than 2 to characteristic 2?
linear-algebra finite-fields intuition
asked Jul 29 at 0:08
tparker
1,679629
This question asks which standard results from linear algebra over a field no longer hold when we generalize the algebraic structure of the scalars to be an arbitrary division ring.
My question is similar but considers a less drastic generalization. In elementary courses on linear algebra, the underlying field is virtually always assumed to be either the real or the complex numbers. (Maybe once in a blue moon, the rationals.) As such, all my intuition is for infinite fields. Moreover, I know that fields of characteristic 2 are especially problematic.
Which theorems from linear algebra no longer hold when we go from an infinite field to a finite field of characteristic greater than 2?
Which further theorems break down (nontrivially) when we go from characteristic greater than 2 to characteristic 2?
linear-algebra finite-fields intuition
asked Jul 29 at 0:08
tparker
1,679629
asked Jul 29 at 0:08
tparker
1,679629
asked Jul 29 at 0:08
tparker
1,679629
asked Jul 29 at 0:08
tparker
1,679629
1,679629
2
My experience is that almost everything works exactly the same way until you hit eigenvalues.
– Randall
Jul 29 at 0:14
2
One loses the notion of an inner product. Thus the "geometric aspects", like the Cauhy Schwarz inequality, no longer make sense.
– caffeinemachine
Jul 29 at 0:38
1
@caffeinemachine But one doesn't have the notion of an inner product in a general vector space anyway - only an inner product space. Anyway, what's stopping you from using the standard inner product $(u, v) = sum_i u_i v_i$ for an arbitrary field (or even ring), at least for finite-dimensional vector spaces?
– tparker
Jul 29 at 0:49
1
@CaveJohnson Sure, but you enlarge to the complexes for algebraic-closure reasons. This isn't so nice in prime characteristic.
– Randall
Jul 29 at 1:12
2
@tparker One can use the bilinear form that you have defined. I wouldn't call it an inner product. Because there is no meaning of positivity ($xcdot xgeq$ 0 for all $x$ and $xcdot x=0$ if and only if $x=0$) if the field is not real or complex (and hence no Cauchy-Schwarz). You are right in saying that there is no notion of an inner product on a plain simple (real or complex) vector space. But one can put an inner product forcefully and deduce purely linear algebraic results which do not have anything to do with any extra structure on the vector space.
– caffeinemachine
Jul 29 at 3:58
 |Â
show 11 more comments
2
My experience is that almost everything works exactly the same way until you hit eigenvalues.
– Randall
Jul 29 at 0:14
2
One loses the notion of an inner product. Thus the "geometric aspects", like the Cauhy Schwarz inequality, no longer make sense.
– caffeinemachine
Jul 29 at 0:38
1
@caffeinemachine But one doesn't have the notion of an inner product in a general vector space anyway - only an inner product space. Anyway, what's stopping you from using the standard inner product $(u, v) = sum_i u_i v_i$ for an arbitrary field (or even ring), at least for finite-dimensional vector spaces?
– tparker
Jul 29 at 0:49
1
@CaveJohnson Sure, but you enlarge to the complexes for algebraic-closure reasons. This isn't so nice in prime characteristic.
– Randall
Jul 29 at 1:12
2
@tparker One can use the bilinear form that you have defined. I wouldn't call it an inner product. Because there is no meaning of positivity ($xcdot xgeq$ 0 for all $x$ and $xcdot x=0$ if and only if $x=0$) if the field is not real or complex (and hence no Cauchy-Schwarz). You are right in saying that there is no notion of an inner product on a plain simple (real or complex) vector space. But one can put an inner product forcefully and deduce purely linear algebraic results which do not have anything to do with any extra structure on the vector space.
– caffeinemachine
Jul 29 at 3:58
2
2
My experience is that almost everything works exactly the same way until you hit eigenvalues.
– Randall
Jul 29 at 0:14
My experience is that almost everything works exactly the same way until you hit eigenvalues.
– Randall
Jul 29 at 0:14
2
2
One loses the notion of an inner product. Thus the "geometric aspects", like the Cauhy Schwarz inequality, no longer make sense.
– caffeinemachine
Jul 29 at 0:38
One loses the notion of an inner product. Thus the "geometric aspects", like the Cauhy Schwarz inequality, no longer make sense.
– caffeinemachine
Jul 29 at 0:38
1
1
@caffeinemachine But one doesn't have the notion of an inner product in a general vector space anyway - only an inner product space. Anyway, what's stopping you from using the standard inner product $(u, v) = sum_i u_i v_i$ for an arbitrary field (or even ring), at least for finite-dimensional vector spaces?
– tparker
Jul 29 at 0:49
@caffeinemachine But one doesn't have the notion of an inner product in a general vector space anyway - only an inner product space. Anyway, what's stopping you from using the standard inner product $(u, v) = sum_i u_i v_i$ for an arbitrary field (or even ring), at least for finite-dimensional vector spaces?
– tparker
Jul 29 at 0:49
1
1
@CaveJohnson Sure, but you enlarge to the complexes for algebraic-closure reasons. This isn't so nice in prime characteristic.
– Randall
Jul 29 at 1:12
@CaveJohnson Sure, but you enlarge to the complexes for algebraic-closure reasons. This isn't so nice in prime characteristic.
– Randall
Jul 29 at 1:12
2
2
@tparker One can use the bilinear form that you have defined. I wouldn't call it an inner product. Because there is no meaning of positivity ($xcdot xgeq$ 0 for all $x$ and $xcdot x=0$ if and only if $x=0$) if the field is not real or complex (and hence no Cauchy-Schwarz). You are right in saying that there is no notion of an inner product on a plain simple (real or complex) vector space. But one can put an inner product forcefully and deduce purely linear algebraic results which do not have anything to do with any extra structure on the vector space.
– caffeinemachine
Jul 29 at 3:58
@tparker One can use the bilinear form that you have defined. I wouldn't call it an inner product. Because there is no meaning of positivity ($xcdot xgeq$ 0 for all $x$ and $xcdot x=0$ if and only if $x=0$) if the field is not real or complex (and hence no Cauchy-Schwarz). You are right in saying that there is no notion of an inner product on a plain simple (real or complex) vector space. But one can put an inner product forcefully and deduce purely linear algebraic results which do not have anything to do with any extra structure on the vector space.
– caffeinemachine
Jul 29 at 3:58
 |Â
show 11 more comments
2 Answers
2
active
oldest
votes
up vote
9
down vote
accepted
This is a rather approximative overview of what generalizations can be explored in an early course of linear algebra.
The short answer is that all that does not use the fact that $Bbb R$ is ordered, $Bbb C$ has a norm, or that $Bbb C=Bbb R[i]=overlineBbb R$ carries on identically to all fields and it can, in principle and in point of fact, be taught directly as "linear algebra", instead of "$Bbb R$-or-$Bbb C$ linear algebra". More specifically
All the things that are genuinely linear, like basis, matricial representations for finite dimensional spaces, dual and bi-dual, Gaussian elimination, determinants, Rouché-Capelli theorem carry on verbatim or with very obvious adjustments.
The results around Jordan normal form stay unchanged for algebraically closed fields. Phenomena like "real Jordan normal form", though, use heavily the fact that $dim_Bbb RBbb C=2$, and need to be heavily amended to be generalized to other extensions (which are almost always of infinite degree) in an interesting way.
The "theory of real inner products on finitely dimensional spaces" is generalized by the theory of quadratic forms, and it is interesting even as part of an early course. It studies the symmetric and bilinear maps $phi:k^ntimes k^nto k$. There is a generalized notion of orthogonality, of adjoint, of degenerate quadratic forms, of orthogonal maps (sometimes called isometries). The main differences revolve around the fact that:
$Bbb R$ is ordered, and so there is a notion of sign and positive definiteness that can be used to control/distinguish a lot of things. For a general field, the only thing that can be controlled is the presence of vectors $v$ such that $phi(v,v)=0$ (isotropic vectors) and/or such that $phi(v,w)=0$ for all $w$ (orthogonal to the whole space). This reflects on terminology and choice of "canonical forms". If you want to quickly gage the flavour of it, have a look at these results by Witt.
Fields of characteristic $2$, and $Bbb F_2$ especially, need (if any) a separate treatment. The issue is that, in fields where $1+1ne0$, there is a bijective correspondence between symmetric bilinear maps and homogeneous polynomial functions of degree $2$ - i.e., maps $q:k^nto k$ that can be written as $q(v)=sum_i,jq_ijv_iv_j$ for some constants $q_ij$. This correspondence is established by calling $Q_phi(v)=phi(v,v)$, and $Phi_q(v,w)=fracq(v+w)-q(v)-q(w)2$. It's straight-forward to verify that $Phi_q_phi=phi$ and $Q_Phi_q=q$. You can't divide by $2$ when $1+1=0$, and it turns out that the map $phimapsto Q_phi$ is not injective in characteristic $2$.
However, you may want to look into an actual textbook for further detail on (3); "Introduction To Quadratic Forms Over Fields" by Lam (or its earlier, more famous version "Algebraic Theory of Quadratic Forms") is something that you may find in your local library. It doesn't quite cover what happens in characteristic $2$, though.
answered Jul 29 at 2:14
Saucy O'Path
2,393217
1
I would say that the theory of real/complex inner products is generalized by the theory of bilinear/sesquilinear forms (specifically by reflexive forms for which $B(x,y)=0$ if and only if $B(y,x)=0$), not by quadratic forms. Quadratic forms are related but not always exactly equivalent to these forms, and can be defined on vector spaces over any field.
– Morgan Rodgers
Jul 29 at 4:25
1
(In particular, a quadratic form is a map from $V to mathbbF$, so cannot on its own be used to define orthogonality, while inner producst/bilinear/sesquilinear forms are all maps from $V times V to mathbbF$.)
– Morgan Rodgers
Jul 29 at 4:27
@MorganRodgers Can't you use the polarization identity to convert any quadratic form (except over a field of characteristic 2) into a sesquilinear or symmetric bilinear form?
– tparker
Jul 29 at 16:15
@tparker As long as you are not over characteristic 2 you can obtain a symmetric bilinear form from a quadratic form. But there are the alternating bilinear forms, as well as sesquilinear forms associated with a nonidentity field automorphisms, which do not arise in this way. These general bilinear/sesquilinear forms more closely match up to the concept of an inner product than does the notion of a quadratic form.
– Morgan Rodgers
Jul 29 at 16:28
@MorganRodgers I wouldn't say "more closely", since symmetric bilinear forms are a special case of bilinear forms. I'd rather say that which direction is to be investigated first is probably a matter of taste. Introducing tensor, symmetric and alternatig algebras is certainly a valid option for a first course in linear algebra.
– Saucy O'Path
Jul 29 at 16:44
 |Â
show 1 more comment
up vote
1
down vote
The problems that arise over arbitrary fields mostly have nothing to do with linear algebra itself, more with the applications.
You have to realize that linear algebra arose as a conglomerate of many different concepts and applications: solving linear equations, linear transformations between vector spaces, general matrix theory, matrix groups and rings, geometric problems, engineering applications.
All of the algebra essentially only depends on the fact that you are working over a field. But when you are working with an arbitrary field, you often don't have a notion of distance, angles, slopes, etc. So
- anything that involves having something be greater than/less something else could create a problem (as mentioned in another answer, inner products can give problems because asking that $x cdot x geq 0$ for all $x$ can become meaningless)
- anything involving length requires a bit of consideration or redefinition (what does it mean to "normalize" a vector if you don't have a way to measure its length? How do you measure the distance between a vector and a subspace for a least-squares problem?)
Many things can be redefined however.
- Distance metrics such as the Hamming distance can be imposed (as well as others).
- Sometimes "normalizing" a vector can just mean scaling so the first (or last) nonzero entry is a 1.
- Orthogonality can be defined in terms of an arbitrary bilinear or sesquilinear form $B: Vtimes V to mathbbF$ (usually we require that $B$ is reflexive, that is that $B(x,y) = 0$ if and only if $B(y,x) = 0$ so that orthogonality is a symmetric relation).
Making these redefinitions will require verifying and reproving that you have analogous properties that you have in the real and complex case. Often you do have similarities, but often with some subtle differences (eg you can often have nonzero vectors that are orthogonal to themselves).
answered Jul 29 at 16:45
Morgan Rodgers
9,06321338
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
This is a rather approximative overview of what generalizations can be explored in an early course of linear algebra.
The short answer is that all that does not use the fact that $Bbb R$ is ordered, $Bbb C$ has a norm, or that $Bbb C=Bbb R[i]=overlineBbb R$ carries on identically to all fields and it can, in principle and in point of fact, be taught directly as "linear algebra", instead of "$Bbb R$-or-$Bbb C$ linear algebra". More specifically
All the things that are genuinely linear, like basis, matricial representations for finite dimensional spaces, dual and bi-dual, Gaussian elimination, determinants, Rouché-Capelli theorem carry on verbatim or with very obvious adjustments.
The results around Jordan normal form stay unchanged for algebraically closed fields. Phenomena like "real Jordan normal form", though, use heavily the fact that $dim_Bbb RBbb C=2$, and need to be heavily amended to be generalized to other extensions (which are almost always of infinite degree) in an interesting way.
The "theory of real inner products on finitely dimensional spaces" is generalized by the theory of quadratic forms, and it is interesting even as part of an early course. It studies the symmetric and bilinear maps $phi:k^ntimes k^nto k$. There is a generalized notion of orthogonality, of adjoint, of degenerate quadratic forms, of orthogonal maps (sometimes called isometries). The main differences revolve around the fact that:
$Bbb R$ is ordered, and so there is a notion of sign and positive definiteness that can be used to control/distinguish a lot of things. For a general field, the only thing that can be controlled is the presence of vectors $v$ such that $phi(v,v)=0$ (isotropic vectors) and/or such that $phi(v,w)=0$ for all $w$ (orthogonal to the whole space). This reflects on terminology and choice of "canonical forms". If you want to quickly gage the flavour of it, have a look at these results by Witt.
Fields of characteristic $2$, and $Bbb F_2$ especially, need (if any) a separate treatment. The issue is that, in fields where $1+1ne0$, there is a bijective correspondence between symmetric bilinear maps and homogeneous polynomial functions of degree $2$ - i.e., maps $q:k^nto k$ that can be written as $q(v)=sum_i,jq_ijv_iv_j$ for some constants $q_ij$. This correspondence is established by calling $Q_phi(v)=phi(v,v)$, and $Phi_q(v,w)=fracq(v+w)-q(v)-q(w)2$. It's straight-forward to verify that $Phi_q_phi=phi$ and $Q_Phi_q=q$. You can't divide by $2$ when $1+1=0$, and it turns out that the map $phimapsto Q_phi$ is not injective in characteristic $2$.
However, you may want to look into an actual textbook for further detail on (3); "Introduction To Quadratic Forms Over Fields" by Lam (or its earlier, more famous version "Algebraic Theory of Quadratic Forms") is something that you may find in your local library. It doesn't quite cover what happens in characteristic $2$, though.
answered Jul 29 at 2:14
Saucy O'Path
2,393217
1
I would say that the theory of real/complex inner products is generalized by the theory of bilinear/sesquilinear forms (specifically by reflexive forms for which $B(x,y)=0$ if and only if $B(y,x)=0$), not by quadratic forms. Quadratic forms are related but not always exactly equivalent to these forms, and can be defined on vector spaces over any field.
– Morgan Rodgers
Jul 29 at 4:25
1
(In particular, a quadratic form is a map from $V to mathbbF$, so cannot on its own be used to define orthogonality, while inner producst/bilinear/sesquilinear forms are all maps from $V times V to mathbbF$.)
– Morgan Rodgers
Jul 29 at 4:27
@MorganRodgers Can't you use the polarization identity to convert any quadratic form (except over a field of characteristic 2) into a sesquilinear or symmetric bilinear form?
– tparker
Jul 29 at 16:15
@tparker As long as you are not over characteristic 2 you can obtain a symmetric bilinear form from a quadratic form. But there are the alternating bilinear forms, as well as sesquilinear forms associated with a nonidentity field automorphisms, which do not arise in this way. These general bilinear/sesquilinear forms more closely match up to the concept of an inner product than does the notion of a quadratic form.
– Morgan Rodgers
Jul 29 at 16:28
@MorganRodgers I wouldn't say "more closely", since symmetric bilinear forms are a special case of bilinear forms. I'd rather say that which direction is to be investigated first is probably a matter of taste. Introducing tensor, symmetric and alternatig algebras is certainly a valid option for a first course in linear algebra.
– Saucy O'Path
Jul 29 at 16:44
 |Â
show 1 more comment
up vote
9
down vote
accepted
This is a rather approximative overview of what generalizations can be explored in an early course of linear algebra.
The short answer is that all that does not use the fact that $Bbb R$ is ordered, $Bbb C$ has a norm, or that $Bbb C=Bbb R[i]=overlineBbb R$ carries on identically to all fields and it can, in principle and in point of fact, be taught directly as "linear algebra", instead of "$Bbb R$-or-$Bbb C$ linear algebra". More specifically
All the things that are genuinely linear, like basis, matricial representations for finite dimensional spaces, dual and bi-dual, Gaussian elimination, determinants, Rouché-Capelli theorem carry on verbatim or with very obvious adjustments.
The results around Jordan normal form stay unchanged for algebraically closed fields. Phenomena like "real Jordan normal form", though, use heavily the fact that $dim_Bbb RBbb C=2$, and need to be heavily amended to be generalized to other extensions (which are almost always of infinite degree) in an interesting way.
The "theory of real inner products on finitely dimensional spaces" is generalized by the theory of quadratic forms, and it is interesting even as part of an early course. It studies the symmetric and bilinear maps $phi:k^ntimes k^nto k$. There is a generalized notion of orthogonality, of adjoint, of degenerate quadratic forms, of orthogonal maps (sometimes called isometries). The main differences revolve around the fact that:
$Bbb R$ is ordered, and so there is a notion of sign and positive definiteness that can be used to control/distinguish a lot of things. For a general field, the only thing that can be controlled is the presence of vectors $v$ such that $phi(v,v)=0$ (isotropic vectors) and/or such that $phi(v,w)=0$ for all $w$ (orthogonal to the whole space). This reflects on terminology and choice of "canonical forms". If you want to quickly gage the flavour of it, have a look at these results by Witt.
Fields of characteristic $2$, and $Bbb F_2$ especially, need (if any) a separate treatment. The issue is that, in fields where $1+1ne0$, there is a bijective correspondence between symmetric bilinear maps and homogeneous polynomial functions of degree $2$ - i.e., maps $q:k^nto k$ that can be written as $q(v)=sum_i,jq_ijv_iv_j$ for some constants $q_ij$. This correspondence is established by calling $Q_phi(v)=phi(v,v)$, and $Phi_q(v,w)=fracq(v+w)-q(v)-q(w)2$. It's straight-forward to verify that $Phi_q_phi=phi$ and $Q_Phi_q=q$. You can't divide by $2$ when $1+1=0$, and it turns out that the map $phimapsto Q_phi$ is not injective in characteristic $2$.
However, you may want to look into an actual textbook for further detail on (3); "Introduction To Quadratic Forms Over Fields" by Lam (or its earlier, more famous version "Algebraic Theory of Quadratic Forms") is something that you may find in your local library. It doesn't quite cover what happens in characteristic $2$, though.
answered Jul 29 at 2:14
Saucy O'Path
2,393217
1
I would say that the theory of real/complex inner products is generalized by the theory of bilinear/sesquilinear forms (specifically by reflexive forms for which $B(x,y)=0$ if and only if $B(y,x)=0$), not by quadratic forms. Quadratic forms are related but not always exactly equivalent to these forms, and can be defined on vector spaces over any field.
– Morgan Rodgers
Jul 29 at 4:25
1
(In particular, a quadratic form is a map from $V to mathbbF$, so cannot on its own be used to define orthogonality, while inner producst/bilinear/sesquilinear forms are all maps from $V times V to mathbbF$.)
– Morgan Rodgers
Jul 29 at 4:27
@MorganRodgers Can't you use the polarization identity to convert any quadratic form (except over a field of characteristic 2) into a sesquilinear or symmetric bilinear form?
– tparker
Jul 29 at 16:15
@tparker As long as you are not over characteristic 2 you can obtain a symmetric bilinear form from a quadratic form. But there are the alternating bilinear forms, as well as sesquilinear forms associated with a nonidentity field automorphisms, which do not arise in this way. These general bilinear/sesquilinear forms more closely match up to the concept of an inner product than does the notion of a quadratic form.
– Morgan Rodgers
Jul 29 at 16:28
@MorganRodgers I wouldn't say "more closely", since symmetric bilinear forms are a special case of bilinear forms. I'd rather say that which direction is to be investigated first is probably a matter of taste. Introducing tensor, symmetric and alternatig algebras is certainly a valid option for a first course in linear algebra.
– Saucy O'Path
Jul 29 at 16:44
 |Â
show 1 more comment
up vote
9
down vote
accepted
up vote
9
down vote
accepted
This is a rather approximative overview of what generalizations can be explored in an early course of linear algebra.
The short answer is that all that does not use the fact that $Bbb R$ is ordered, $Bbb C$ has a norm, or that $Bbb C=Bbb R[i]=overlineBbb R$ carries on identically to all fields and it can, in principle and in point of fact, be taught directly as "linear algebra", instead of "$Bbb R$-or-$Bbb C$ linear algebra". More specifically
All the things that are genuinely linear, like basis, matricial representations for finite dimensional spaces, dual and bi-dual, Gaussian elimination, determinants, Rouché-Capelli theorem carry on verbatim or with very obvious adjustments.
The results around Jordan normal form stay unchanged for algebraically closed fields. Phenomena like "real Jordan normal form", though, use heavily the fact that $dim_Bbb RBbb C=2$, and need to be heavily amended to be generalized to other extensions (which are almost always of infinite degree) in an interesting way.
The "theory of real inner products on finitely dimensional spaces" is generalized by the theory of quadratic forms, and it is interesting even as part of an early course. It studies the symmetric and bilinear maps $phi:k^ntimes k^nto k$. There is a generalized notion of orthogonality, of adjoint, of degenerate quadratic forms, of orthogonal maps (sometimes called isometries). The main differences revolve around the fact that:
$Bbb R$ is ordered, and so there is a notion of sign and positive definiteness that can be used to control/distinguish a lot of things. For a general field, the only thing that can be controlled is the presence of vectors $v$ such that $phi(v,v)=0$ (isotropic vectors) and/or such that $phi(v,w)=0$ for all $w$ (orthogonal to the whole space). This reflects on terminology and choice of "canonical forms". If you want to quickly gage the flavour of it, have a look at these results by Witt.
Fields of characteristic $2$, and $Bbb F_2$ especially, need (if any) a separate treatment. The issue is that, in fields where $1+1ne0$, there is a bijective correspondence between symmetric bilinear maps and homogeneous polynomial functions of degree $2$ - i.e., maps $q:k^nto k$ that can be written as $q(v)=sum_i,jq_ijv_iv_j$ for some constants $q_ij$. This correspondence is established by calling $Q_phi(v)=phi(v,v)$, and $Phi_q(v,w)=fracq(v+w)-q(v)-q(w)2$. It's straight-forward to verify that $Phi_q_phi=phi$ and $Q_Phi_q=q$. You can't divide by $2$ when $1+1=0$, and it turns out that the map $phimapsto Q_phi$ is not injective in characteristic $2$.
However, you may want to look into an actual textbook for further detail on (3); "Introduction To Quadratic Forms Over Fields" by Lam (or its earlier, more famous version "Algebraic Theory of Quadratic Forms") is something that you may find in your local library. It doesn't quite cover what happens in characteristic $2$, though.
answered Jul 29 at 2:14
Saucy O'Path
2,393217
This is a rather approximative overview of what generalizations can be explored in an early course of linear algebra.
The short answer is that all that does not use the fact that $Bbb R$ is ordered, $Bbb C$ has a norm, or that $Bbb C=Bbb R[i]=overlineBbb R$ carries on identically to all fields and it can, in principle and in point of fact, be taught directly as "linear algebra", instead of "$Bbb R$-or-$Bbb C$ linear algebra". More specifically
All the things that are genuinely linear, like basis, matricial representations for finite dimensional spaces, dual and bi-dual, Gaussian elimination, determinants, Rouché-Capelli theorem carry on verbatim or with very obvious adjustments.
The results around Jordan normal form stay unchanged for algebraically closed fields. Phenomena like "real Jordan normal form", though, use heavily the fact that $dim_Bbb RBbb C=2$, and need to be heavily amended to be generalized to other extensions (which are almost always of infinite degree) in an interesting way.
The "theory of real inner products on finitely dimensional spaces" is generalized by the theory of quadratic forms, and it is interesting even as part of an early course. It studies the symmetric and bilinear maps $phi:k^ntimes k^nto k$. There is a generalized notion of orthogonality, of adjoint, of degenerate quadratic forms, of orthogonal maps (sometimes called isometries). The main differences revolve around the fact that:
$Bbb R$ is ordered, and so there is a notion of sign and positive definiteness that can be used to control/distinguish a lot of things. For a general field, the only thing that can be controlled is the presence of vectors $v$ such that $phi(v,v)=0$ (isotropic vectors) and/or such that $phi(v,w)=0$ for all $w$ (orthogonal to the whole space). This reflects on terminology and choice of "canonical forms". If you want to quickly gage the flavour of it, have a look at these results by Witt.
Fields of characteristic $2$, and $Bbb F_2$ especially, need (if any) a separate treatment. The issue is that, in fields where $1+1ne0$, there is a bijective correspondence between symmetric bilinear maps and homogeneous polynomial functions of degree $2$ - i.e., maps $q:k^nto k$ that can be written as $q(v)=sum_i,jq_ijv_iv_j$ for some constants $q_ij$. This correspondence is established by calling $Q_phi(v)=phi(v,v)$, and $Phi_q(v,w)=fracq(v+w)-q(v)-q(w)2$. It's straight-forward to verify that $Phi_q_phi=phi$ and $Q_Phi_q=q$. You can't divide by $2$ when $1+1=0$, and it turns out that the map $phimapsto Q_phi$ is not injective in characteristic $2$.
However, you may want to look into an actual textbook for further detail on (3); "Introduction To Quadratic Forms Over Fields" by Lam (or its earlier, more famous version "Algebraic Theory of Quadratic Forms") is something that you may find in your local library. It doesn't quite cover what happens in characteristic $2$, though.
answered Jul 29 at 2:14
Saucy O'Path
2,393217
edited Jul 30 at 21:12
answered Jul 29 at 2:14
Saucy O'Path
2,393217
answered Jul 29 at 2:14
Saucy O'Path
2,393217
answered Jul 29 at 2:14
Saucy O'Path
2,393217
2,393217
1
I would say that the theory of real/complex inner products is generalized by the theory of bilinear/sesquilinear forms (specifically by reflexive forms for which $B(x,y)=0$ if and only if $B(y,x)=0$), not by quadratic forms. Quadratic forms are related but not always exactly equivalent to these forms, and can be defined on vector spaces over any field.
– Morgan Rodgers
Jul 29 at 4:25
1
(In particular, a quadratic form is a map from $V to mathbbF$, so cannot on its own be used to define orthogonality, while inner producst/bilinear/sesquilinear forms are all maps from $V times V to mathbbF$.)
– Morgan Rodgers
Jul 29 at 4:27
@MorganRodgers Can't you use the polarization identity to convert any quadratic form (except over a field of characteristic 2) into a sesquilinear or symmetric bilinear form?
– tparker
Jul 29 at 16:15
@tparker As long as you are not over characteristic 2 you can obtain a symmetric bilinear form from a quadratic form. But there are the alternating bilinear forms, as well as sesquilinear forms associated with a nonidentity field automorphisms, which do not arise in this way. These general bilinear/sesquilinear forms more closely match up to the concept of an inner product than does the notion of a quadratic form.
– Morgan Rodgers
Jul 29 at 16:28
@MorganRodgers I wouldn't say "more closely", since symmetric bilinear forms are a special case of bilinear forms. I'd rather say that which direction is to be investigated first is probably a matter of taste. Introducing tensor, symmetric and alternatig algebras is certainly a valid option for a first course in linear algebra.
– Saucy O'Path
Jul 29 at 16:44
 |Â
show 1 more comment
1
I would say that the theory of real/complex inner products is generalized by the theory of bilinear/sesquilinear forms (specifically by reflexive forms for which $B(x,y)=0$ if and only if $B(y,x)=0$), not by quadratic forms. Quadratic forms are related but not always exactly equivalent to these forms, and can be defined on vector spaces over any field.
– Morgan Rodgers
Jul 29 at 4:25
1
(In particular, a quadratic form is a map from $V to mathbbF$, so cannot on its own be used to define orthogonality, while inner producst/bilinear/sesquilinear forms are all maps from $V times V to mathbbF$.)
– Morgan Rodgers
Jul 29 at 4:27
@MorganRodgers Can't you use the polarization identity to convert any quadratic form (except over a field of characteristic 2) into a sesquilinear or symmetric bilinear form?
– tparker
Jul 29 at 16:15
@tparker As long as you are not over characteristic 2 you can obtain a symmetric bilinear form from a quadratic form. But there are the alternating bilinear forms, as well as sesquilinear forms associated with a nonidentity field automorphisms, which do not arise in this way. These general bilinear/sesquilinear forms more closely match up to the concept of an inner product than does the notion of a quadratic form.
– Morgan Rodgers
Jul 29 at 16:28
@MorganRodgers I wouldn't say "more closely", since symmetric bilinear forms are a special case of bilinear forms. I'd rather say that which direction is to be investigated first is probably a matter of taste. Introducing tensor, symmetric and alternatig algebras is certainly a valid option for a first course in linear algebra.
– Saucy O'Path
Jul 29 at 16:44
1
1
I would say that the theory of real/complex inner products is generalized by the theory of bilinear/sesquilinear forms (specifically by reflexive forms for which $B(x,y)=0$ if and only if $B(y,x)=0$), not by quadratic forms. Quadratic forms are related but not always exactly equivalent to these forms, and can be defined on vector spaces over any field.
– Morgan Rodgers
Jul 29 at 4:25
I would say that the theory of real/complex inner products is generalized by the theory of bilinear/sesquilinear forms (specifically by reflexive forms for which $B(x,y)=0$ if and only if $B(y,x)=0$), not by quadratic forms. Quadratic forms are related but not always exactly equivalent to these forms, and can be defined on vector spaces over any field.
– Morgan Rodgers
Jul 29 at 4:25
1
1
(In particular, a quadratic form is a map from $V to mathbbF$, so cannot on its own be used to define orthogonality, while inner producst/bilinear/sesquilinear forms are all maps from $V times V to mathbbF$.)
– Morgan Rodgers
Jul 29 at 4:27
(In particular, a quadratic form is a map from $V to mathbbF$, so cannot on its own be used to define orthogonality, while inner producst/bilinear/sesquilinear forms are all maps from $V times V to mathbbF$.)
– Morgan Rodgers
Jul 29 at 4:27
@MorganRodgers Can't you use the polarization identity to convert any quadratic form (except over a field of characteristic 2) into a sesquilinear or symmetric bilinear form?
– tparker
Jul 29 at 16:15
@MorganRodgers Can't you use the polarization identity to convert any quadratic form (except over a field of characteristic 2) into a sesquilinear or symmetric bilinear form?
– tparker
Jul 29 at 16:15
@tparker As long as you are not over characteristic 2 you can obtain a symmetric bilinear form from a quadratic form. But there are the alternating bilinear forms, as well as sesquilinear forms associated with a nonidentity field automorphisms, which do not arise in this way. These general bilinear/sesquilinear forms more closely match up to the concept of an inner product than does the notion of a quadratic form.
– Morgan Rodgers
Jul 29 at 16:28
@tparker As long as you are not over characteristic 2 you can obtain a symmetric bilinear form from a quadratic form. But there are the alternating bilinear forms, as well as sesquilinear forms associated with a nonidentity field automorphisms, which do not arise in this way. These general bilinear/sesquilinear forms more closely match up to the concept of an inner product than does the notion of a quadratic form.
– Morgan Rodgers
Jul 29 at 16:28
@MorganRodgers I wouldn't say "more closely", since symmetric bilinear forms are a special case of bilinear forms. I'd rather say that which direction is to be investigated first is probably a matter of taste. Introducing tensor, symmetric and alternatig algebras is certainly a valid option for a first course in linear algebra.
– Saucy O'Path
Jul 29 at 16:44
@MorganRodgers I wouldn't say "more closely", since symmetric bilinear forms are a special case of bilinear forms. I'd rather say that which direction is to be investigated first is probably a matter of taste. Introducing tensor, symmetric and alternatig algebras is certainly a valid option for a first course in linear algebra.
– Saucy O'Path
Jul 29 at 16:44
 |Â
show 1 more comment
up vote
1
down vote
The problems that arise over arbitrary fields mostly have nothing to do with linear algebra itself, more with the applications.
You have to realize that linear algebra arose as a conglomerate of many different concepts and applications: solving linear equations, linear transformations between vector spaces, general matrix theory, matrix groups and rings, geometric problems, engineering applications.
All of the algebra essentially only depends on the fact that you are working over a field. But when you are working with an arbitrary field, you often don't have a notion of distance, angles, slopes, etc. So
- anything that involves having something be greater than/less something else could create a problem (as mentioned in another answer, inner products can give problems because asking that $x cdot x geq 0$ for all $x$ can become meaningless)
- anything involving length requires a bit of consideration or redefinition (what does it mean to "normalize" a vector if you don't have a way to measure its length? How do you measure the distance between a vector and a subspace for a least-squares problem?)
Many things can be redefined however.
- Distance metrics such as the Hamming distance can be imposed (as well as others).
- Sometimes "normalizing" a vector can just mean scaling so the first (or last) nonzero entry is a 1.
- Orthogonality can be defined in terms of an arbitrary bilinear or sesquilinear form $B: Vtimes V to mathbbF$ (usually we require that $B$ is reflexive, that is that $B(x,y) = 0$ if and only if $B(y,x) = 0$ so that orthogonality is a symmetric relation).
Making these redefinitions will require verifying and reproving that you have analogous properties that you have in the real and complex case. Often you do have similarities, but often with some subtle differences (eg you can often have nonzero vectors that are orthogonal to themselves).
answered Jul 29 at 16:45
Morgan Rodgers
9,06321338
add a comment |Â
up vote
1
down vote
The problems that arise over arbitrary fields mostly have nothing to do with linear algebra itself, more with the applications.
You have to realize that linear algebra arose as a conglomerate of many different concepts and applications: solving linear equations, linear transformations between vector spaces, general matrix theory, matrix groups and rings, geometric problems, engineering applications.
All of the algebra essentially only depends on the fact that you are working over a field. But when you are working with an arbitrary field, you often don't have a notion of distance, angles, slopes, etc. So
- anything that involves having something be greater than/less something else could create a problem (as mentioned in another answer, inner products can give problems because asking that $x cdot x geq 0$ for all $x$ can become meaningless)
- anything involving length requires a bit of consideration or redefinition (what does it mean to "normalize" a vector if you don't have a way to measure its length? How do you measure the distance between a vector and a subspace for a least-squares problem?)
Many things can be redefined however.
- Distance metrics such as the Hamming distance can be imposed (as well as others).
- Sometimes "normalizing" a vector can just mean scaling so the first (or last) nonzero entry is a 1.
- Orthogonality can be defined in terms of an arbitrary bilinear or sesquilinear form $B: Vtimes V to mathbbF$ (usually we require that $B$ is reflexive, that is that $B(x,y) = 0$ if and only if $B(y,x) = 0$ so that orthogonality is a symmetric relation).
Making these redefinitions will require verifying and reproving that you have analogous properties that you have in the real and complex case. Often you do have similarities, but often with some subtle differences (eg you can often have nonzero vectors that are orthogonal to themselves).
answered Jul 29 at 16:45
Morgan Rodgers
9,06321338
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The problems that arise over arbitrary fields mostly have nothing to do with linear algebra itself, more with the applications.
You have to realize that linear algebra arose as a conglomerate of many different concepts and applications: solving linear equations, linear transformations between vector spaces, general matrix theory, matrix groups and rings, geometric problems, engineering applications.
All of the algebra essentially only depends on the fact that you are working over a field. But when you are working with an arbitrary field, you often don't have a notion of distance, angles, slopes, etc. So
- anything that involves having something be greater than/less something else could create a problem (as mentioned in another answer, inner products can give problems because asking that $x cdot x geq 0$ for all $x$ can become meaningless)
- anything involving length requires a bit of consideration or redefinition (what does it mean to "normalize" a vector if you don't have a way to measure its length? How do you measure the distance between a vector and a subspace for a least-squares problem?)
Many things can be redefined however.
- Distance metrics such as the Hamming distance can be imposed (as well as others).
- Sometimes "normalizing" a vector can just mean scaling so the first (or last) nonzero entry is a 1.
- Orthogonality can be defined in terms of an arbitrary bilinear or sesquilinear form $B: Vtimes V to mathbbF$ (usually we require that $B$ is reflexive, that is that $B(x,y) = 0$ if and only if $B(y,x) = 0$ so that orthogonality is a symmetric relation).
Making these redefinitions will require verifying and reproving that you have analogous properties that you have in the real and complex case. Often you do have similarities, but often with some subtle differences (eg you can often have nonzero vectors that are orthogonal to themselves).
answered Jul 29 at 16:45
Morgan Rodgers
9,06321338
The problems that arise over arbitrary fields mostly have nothing to do with linear algebra itself, more with the applications.
You have to realize that linear algebra arose as a conglomerate of many different concepts and applications: solving linear equations, linear transformations between vector spaces, general matrix theory, matrix groups and rings, geometric problems, engineering applications.
All of the algebra essentially only depends on the fact that you are working over a field. But when you are working with an arbitrary field, you often don't have a notion of distance, angles, slopes, etc. So
- anything that involves having something be greater than/less something else could create a problem (as mentioned in another answer, inner products can give problems because asking that $x cdot x geq 0$ for all $x$ can become meaningless)
- anything involving length requires a bit of consideration or redefinition (what does it mean to "normalize" a vector if you don't have a way to measure its length? How do you measure the distance between a vector and a subspace for a least-squares problem?)
Many things can be redefined however.
- Distance metrics such as the Hamming distance can be imposed (as well as others).
- Sometimes "normalizing" a vector can just mean scaling so the first (or last) nonzero entry is a 1.
- Orthogonality can be defined in terms of an arbitrary bilinear or sesquilinear form $B: Vtimes V to mathbbF$ (usually we require that $B$ is reflexive, that is that $B(x,y) = 0$ if and only if $B(y,x) = 0$ so that orthogonality is a symmetric relation).
Making these redefinitions will require verifying and reproving that you have analogous properties that you have in the real and complex case. Often you do have similarities, but often with some subtle differences (eg you can often have nonzero vectors that are orthogonal to themselves).
answered Jul 29 at 16:45
Morgan Rodgers
9,06321338
answered Jul 29 at 16:45
Morgan Rodgers
9,06321338
answered Jul 29 at 16:45
Morgan Rodgers
9,06321338
answered Jul 29 at 16:45
Morgan Rodgers
9,06321338
9,06321338
add a comment |Â
add a comment |Â
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2
My experience is that almost everything works exactly the same way until you hit eigenvalues.
– Randall
Jul 29 at 0:14
2
One loses the notion of an inner product. Thus the "geometric aspects", like the Cauhy Schwarz inequality, no longer make sense.
– caffeinemachine
Jul 29 at 0:38
1
@caffeinemachine But one doesn't have the notion of an inner product in a general vector space anyway - only an inner product space. Anyway, what's stopping you from using the standard inner product $(u, v) = sum_i u_i v_i$ for an arbitrary field (or even ring), at least for finite-dimensional vector spaces?
– tparker
Jul 29 at 0:49
1
@CaveJohnson Sure, but you enlarge to the complexes for algebraic-closure reasons. This isn't so nice in prime characteristic.
– Randall
Jul 29 at 1:12
2
@tparker One can use the bilinear form that you have defined. I wouldn't call it an inner product. Because there is no meaning of positivity ($xcdot xgeq$ 0 for all $x$ and $xcdot x=0$ if and only if $x=0$) if the field is not real or complex (and hence no Cauchy-Schwarz). You are right in saying that there is no notion of an inner product on a plain simple (real or complex) vector space. But one can put an inner product forcefully and deduce purely linear algebraic results which do not have anything to do with any extra structure on the vector space.
– caffeinemachine
Jul 29 at 3:58