Mixing
Subspaces in $mathbb R^n times n$
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If our space is $mathbbR^n times n$, then can we definitely conclude that
$n times n$ invertible matrices belong to its subspace
the set of all $ntimes n$ matrices of the form $k*mathcalI$, where $mathcalI$ is the identity matrix, and $k$ is any real number also belongs to its subspace.
What kind of example can I give to show the following?
- the set of matrices similar to $A$ are not necessarily subspaces of $mathbbR^ntimes n$. We're assuming that $A$ is a fixed matrix
- the set of all matrices that commute with $A$ do not fall into the subspace of $mathbbR^ntimes n$. We're assuming that $A$ is a fixed matrix as well.
Are these assumptions correct?
linear-algebra matrices vector-spaces linear-transformations
edited Jul 22 at 12:11
Rodrigo de Azevedo
12.6k41751
asked Jul 22 at 10:40
PERTURBATIONFLOW
396
add a comment |Â
up vote
0
down vote
favorite
If our space is $mathbbR^n times n$, then can we definitely conclude that
$n times n$ invertible matrices belong to its subspace
the set of all $ntimes n$ matrices of the form $k*mathcalI$, where $mathcalI$ is the identity matrix, and $k$ is any real number also belongs to its subspace.
What kind of example can I give to show the following?
- the set of matrices similar to $A$ are not necessarily subspaces of $mathbbR^ntimes n$. We're assuming that $A$ is a fixed matrix
- the set of all matrices that commute with $A$ do not fall into the subspace of $mathbbR^ntimes n$. We're assuming that $A$ is a fixed matrix as well.
Are these assumptions correct?
linear-algebra matrices vector-spaces linear-transformations
edited Jul 22 at 12:11
Rodrigo de Azevedo
12.6k41751
asked Jul 22 at 10:40
PERTURBATIONFLOW
396
3
Are You asking whether the invertible matrices form a subspace of $mathbbR^ntimes n$? If so consider an invertible matrix $A$ and $-A$ and their sum, etc. However it´s not clear what You´re asking.
– Peter Melech
Jul 22 at 10:50
@PeterMelech The question is identifying subspaces that exist in the space Rnxn. The choices ares: invertible matrices of size nxn/the set of all matrices that are similar to A where A is a fixed matrix/the set of all matrices that commute with A where A is a fixed matrix/the set of all nxn matrices of the form KI where K is any real number. I know that the 1st and 4th can be conclusively defined as subspaces in Rxn, but what about the second and the third
– PERTURBATIONFLOW
Jul 22 at 10:57
" I know that the 1st and 4th can be conclusively defined as subspaces". How do You "know" that?? Consider e.g. the example I gave in my first comment.
– Peter Melech
Jul 22 at 11:35
E.g. ad 3:Let $V_A=T^-1AT:Tin GL_n(mathbbR)$. If $Aneq 0$ then $0notin V_A$
– Peter Melech
Jul 22 at 12:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If our space is $mathbbR^n times n$, then can we definitely conclude that
$n times n$ invertible matrices belong to its subspace
the set of all $ntimes n$ matrices of the form $k*mathcalI$, where $mathcalI$ is the identity matrix, and $k$ is any real number also belongs to its subspace.
What kind of example can I give to show the following?
- the set of matrices similar to $A$ are not necessarily subspaces of $mathbbR^ntimes n$. We're assuming that $A$ is a fixed matrix
- the set of all matrices that commute with $A$ do not fall into the subspace of $mathbbR^ntimes n$. We're assuming that $A$ is a fixed matrix as well.
Are these assumptions correct?
linear-algebra matrices vector-spaces linear-transformations
edited Jul 22 at 12:11
Rodrigo de Azevedo
12.6k41751
asked Jul 22 at 10:40
PERTURBATIONFLOW
396
If our space is $mathbbR^n times n$, then can we definitely conclude that
$n times n$ invertible matrices belong to its subspace
the set of all $ntimes n$ matrices of the form $k*mathcalI$, where $mathcalI$ is the identity matrix, and $k$ is any real number also belongs to its subspace.
What kind of example can I give to show the following?
- the set of matrices similar to $A$ are not necessarily subspaces of $mathbbR^ntimes n$. We're assuming that $A$ is a fixed matrix
- the set of all matrices that commute with $A$ do not fall into the subspace of $mathbbR^ntimes n$. We're assuming that $A$ is a fixed matrix as well.
Are these assumptions correct?
linear-algebra matrices vector-spaces linear-transformations
edited Jul 22 at 12:11
Rodrigo de Azevedo
12.6k41751
asked Jul 22 at 10:40
PERTURBATIONFLOW
396
edited Jul 22 at 12:11
Rodrigo de Azevedo
12.6k41751
edited Jul 22 at 12:11
Rodrigo de Azevedo
12.6k41751
edited Jul 22 at 12:11
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Jul 22 at 10:40
PERTURBATIONFLOW
396
asked Jul 22 at 10:40
PERTURBATIONFLOW
396
asked Jul 22 at 10:40
PERTURBATIONFLOW
396
396
3
Are You asking whether the invertible matrices form a subspace of $mathbbR^ntimes n$? If so consider an invertible matrix $A$ and $-A$ and their sum, etc. However it´s not clear what You´re asking.
– Peter Melech
Jul 22 at 10:50
@PeterMelech The question is identifying subspaces that exist in the space Rnxn. The choices ares: invertible matrices of size nxn/the set of all matrices that are similar to A where A is a fixed matrix/the set of all matrices that commute with A where A is a fixed matrix/the set of all nxn matrices of the form KI where K is any real number. I know that the 1st and 4th can be conclusively defined as subspaces in Rxn, but what about the second and the third
– PERTURBATIONFLOW
Jul 22 at 10:57
" I know that the 1st and 4th can be conclusively defined as subspaces". How do You "know" that?? Consider e.g. the example I gave in my first comment.
– Peter Melech
Jul 22 at 11:35
E.g. ad 3:Let $V_A=T^-1AT:Tin GL_n(mathbbR)$. If $Aneq 0$ then $0notin V_A$
– Peter Melech
Jul 22 at 12:24
add a comment |Â
3
Are You asking whether the invertible matrices form a subspace of $mathbbR^ntimes n$? If so consider an invertible matrix $A$ and $-A$ and their sum, etc. However it´s not clear what You´re asking.
– Peter Melech
Jul 22 at 10:50
@PeterMelech The question is identifying subspaces that exist in the space Rnxn. The choices ares: invertible matrices of size nxn/the set of all matrices that are similar to A where A is a fixed matrix/the set of all matrices that commute with A where A is a fixed matrix/the set of all nxn matrices of the form KI where K is any real number. I know that the 1st and 4th can be conclusively defined as subspaces in Rxn, but what about the second and the third
– PERTURBATIONFLOW
Jul 22 at 10:57
" I know that the 1st and 4th can be conclusively defined as subspaces". How do You "know" that?? Consider e.g. the example I gave in my first comment.
– Peter Melech
Jul 22 at 11:35
E.g. ad 3:Let $V_A=T^-1AT:Tin GL_n(mathbbR)$. If $Aneq 0$ then $0notin V_A$
– Peter Melech
Jul 22 at 12:24
3
3
Are You asking whether the invertible matrices form a subspace of $mathbbR^ntimes n$? If so consider an invertible matrix $A$ and $-A$ and their sum, etc. However it´s not clear what You´re asking.
– Peter Melech
Jul 22 at 10:50
Are You asking whether the invertible matrices form a subspace of $mathbbR^ntimes n$? If so consider an invertible matrix $A$ and $-A$ and their sum, etc. However it´s not clear what You´re asking.
– Peter Melech
Jul 22 at 10:50
@PeterMelech The question is identifying subspaces that exist in the space Rnxn. The choices ares: invertible matrices of size nxn/the set of all matrices that are similar to A where A is a fixed matrix/the set of all matrices that commute with A where A is a fixed matrix/the set of all nxn matrices of the form KI where K is any real number. I know that the 1st and 4th can be conclusively defined as subspaces in Rxn, but what about the second and the third
– PERTURBATIONFLOW
Jul 22 at 10:57
@PeterMelech The question is identifying subspaces that exist in the space Rnxn. The choices ares: invertible matrices of size nxn/the set of all matrices that are similar to A where A is a fixed matrix/the set of all matrices that commute with A where A is a fixed matrix/the set of all nxn matrices of the form KI where K is any real number. I know that the 1st and 4th can be conclusively defined as subspaces in Rxn, but what about the second and the third
– PERTURBATIONFLOW
Jul 22 at 10:57
" I know that the 1st and 4th can be conclusively defined as subspaces". How do You "know" that?? Consider e.g. the example I gave in my first comment.
– Peter Melech
Jul 22 at 11:35
" I know that the 1st and 4th can be conclusively defined as subspaces". How do You "know" that?? Consider e.g. the example I gave in my first comment.
– Peter Melech
Jul 22 at 11:35
E.g. ad 3:Let $V_A=T^-1AT:Tin GL_n(mathbbR)$. If $Aneq 0$ then $0notin V_A$
– Peter Melech
Jul 22 at 12:24
E.g. ad 3:Let $V_A=T^-1AT:Tin GL_n(mathbbR)$. If $Aneq 0$ then $0notin V_A$
– Peter Melech
Jul 22 at 12:24
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Let's start by the definition
If $V$ is a vector space on a field $K$ and $W$ is a subset of $V$, then $W$ is a subspace if
- The zero vector is in $W$
- $W$ is closed under addition and multiplication by a scalar in $K$
Let us see now if the sets that you gave us are indeed subspaces o $mathbbR^ntimes n$:
- The set of all invertible $ntimes n$ matrices: being invertible means that $det(A)neq 0$. This is enough to say that this cannot be a subspace because the zero vector is the zero matrix which clearly has zero determinant and so it's not an invertible matrix
$ntimes n$ matrices of the form $kmathcalI$: this indeed is a subspace because it has the zero vector (just set $k=0$) and is closed under addition $$kmathcalI+wmathcalI= (k+w)mathcalI = nmathcalI$$ and it's closed under scalar multiplication which is clear even from the definition $kmathcalI$
All matrices similar to a matrix $A$: this is trickier because clearly if $A$ is the null matrix than all the matrix similar to $A$ is just the zero matrix. In this case this is a subspace because it has zero matrix, adding zero matrices together get's you zero matrices, and multiplying by a scalar the zero matrix gives the zero matrix. In all other cases this is not a subspace because of the lack of zero vector.
All matrices that commute with $A$: if two matrices commute then $$[A,B]= AB-BA = 0$$ let's see if this is a subspace: the zero matrix commutes with every matrix, in fact $$[A,0] = A0-0A = 0-0=0$$ For every matrix $B$ that commute with $A$, the matrix $kB$ $$[A, kB] = AkB-kBA = kAB-kBA = k[A,B] = 0$$ commutes as well with $A$. If two matrices $B$ and $C$ commutes with $A$, then $$beginalign&[A,B+C] = A(B+C)-(B+C)A = \ &AB + AC - BA - CA = \ &(AB-BA)+(AC-CA) = \ &[A,B]+[A,C] = 0+0=0endalign$$ their sum commute. So this is a subspace.
answered Jul 22 at 12:48
Davide Morgante
1,798220
Hi Davide! I hope you are doing well. I am struggling a bit in my linear algebra homework, and have posted some questions that people are not responding to. I was wondering if you can take a look at these questions and clarify them to me since your responses have been consistently the best. 1) math.stackexchange.com/questions/2871861/… 2) math.stackexchange.com/questions/2871059/… I have already posted my line of thinking. If it isn't any inconvenience to you. Thank you sir.
– PERTURBATIONFLOW
Aug 4 at 11:07
Hi! I've answered to the second question, on the first I think you already got a clear answer!
– Davide Morgante
Aug 4 at 12:45
add a comment |Â
up vote
-1
down vote
I think you should rephrase your question.
You have a given dimension space of â„Ân×n, and wish to identify which of the following are subspaces?
- n×n invertible matrices belong to its subspace
- the set of all n×n matrices of the form k∗I, where I is the identity matrix, and k is any real number also belongs to its subspace
- the set of matrices similar to A. We're assuming that A is a fixed matrix
- the set of all matrices that commute with A. We're assuming that A is a fixed matrix as well.
Your previous question was a bit jumbled to understand. You're making the assumption that 1 and 4 hold true, but 2 and 3 don't. You need to justify these claims first.
answered Jul 22 at 11:53
FireMeUP
456
This isn´t really an answer to the question and equally unclear.
– Peter Melech
Jul 22 at 11:58
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let's start by the definition
If $V$ is a vector space on a field $K$ and $W$ is a subset of $V$, then $W$ is a subspace if
- The zero vector is in $W$
- $W$ is closed under addition and multiplication by a scalar in $K$
Let us see now if the sets that you gave us are indeed subspaces o $mathbbR^ntimes n$:
- The set of all invertible $ntimes n$ matrices: being invertible means that $det(A)neq 0$. This is enough to say that this cannot be a subspace because the zero vector is the zero matrix which clearly has zero determinant and so it's not an invertible matrix
$ntimes n$ matrices of the form $kmathcalI$: this indeed is a subspace because it has the zero vector (just set $k=0$) and is closed under addition $$kmathcalI+wmathcalI= (k+w)mathcalI = nmathcalI$$ and it's closed under scalar multiplication which is clear even from the definition $kmathcalI$
All matrices similar to a matrix $A$: this is trickier because clearly if $A$ is the null matrix than all the matrix similar to $A$ is just the zero matrix. In this case this is a subspace because it has zero matrix, adding zero matrices together get's you zero matrices, and multiplying by a scalar the zero matrix gives the zero matrix. In all other cases this is not a subspace because of the lack of zero vector.
All matrices that commute with $A$: if two matrices commute then $$[A,B]= AB-BA = 0$$ let's see if this is a subspace: the zero matrix commutes with every matrix, in fact $$[A,0] = A0-0A = 0-0=0$$ For every matrix $B$ that commute with $A$, the matrix $kB$ $$[A, kB] = AkB-kBA = kAB-kBA = k[A,B] = 0$$ commutes as well with $A$. If two matrices $B$ and $C$ commutes with $A$, then $$beginalign&[A,B+C] = A(B+C)-(B+C)A = \ &AB + AC - BA - CA = \ &(AB-BA)+(AC-CA) = \ &[A,B]+[A,C] = 0+0=0endalign$$ their sum commute. So this is a subspace.
answered Jul 22 at 12:48
Davide Morgante
1,798220
Hi Davide! I hope you are doing well. I am struggling a bit in my linear algebra homework, and have posted some questions that people are not responding to. I was wondering if you can take a look at these questions and clarify them to me since your responses have been consistently the best. 1) math.stackexchange.com/questions/2871861/… 2) math.stackexchange.com/questions/2871059/… I have already posted my line of thinking. If it isn't any inconvenience to you. Thank you sir.
– PERTURBATIONFLOW
Aug 4 at 11:07
Hi! I've answered to the second question, on the first I think you already got a clear answer!
– Davide Morgante
Aug 4 at 12:45
add a comment |Â
up vote
1
down vote
accepted
Let's start by the definition
If $V$ is a vector space on a field $K$ and $W$ is a subset of $V$, then $W$ is a subspace if
- The zero vector is in $W$
- $W$ is closed under addition and multiplication by a scalar in $K$
Let us see now if the sets that you gave us are indeed subspaces o $mathbbR^ntimes n$:
- The set of all invertible $ntimes n$ matrices: being invertible means that $det(A)neq 0$. This is enough to say that this cannot be a subspace because the zero vector is the zero matrix which clearly has zero determinant and so it's not an invertible matrix
$ntimes n$ matrices of the form $kmathcalI$: this indeed is a subspace because it has the zero vector (just set $k=0$) and is closed under addition $$kmathcalI+wmathcalI= (k+w)mathcalI = nmathcalI$$ and it's closed under scalar multiplication which is clear even from the definition $kmathcalI$
All matrices similar to a matrix $A$: this is trickier because clearly if $A$ is the null matrix than all the matrix similar to $A$ is just the zero matrix. In this case this is a subspace because it has zero matrix, adding zero matrices together get's you zero matrices, and multiplying by a scalar the zero matrix gives the zero matrix. In all other cases this is not a subspace because of the lack of zero vector.
All matrices that commute with $A$: if two matrices commute then $$[A,B]= AB-BA = 0$$ let's see if this is a subspace: the zero matrix commutes with every matrix, in fact $$[A,0] = A0-0A = 0-0=0$$ For every matrix $B$ that commute with $A$, the matrix $kB$ $$[A, kB] = AkB-kBA = kAB-kBA = k[A,B] = 0$$ commutes as well with $A$. If two matrices $B$ and $C$ commutes with $A$, then $$beginalign&[A,B+C] = A(B+C)-(B+C)A = \ &AB + AC - BA - CA = \ &(AB-BA)+(AC-CA) = \ &[A,B]+[A,C] = 0+0=0endalign$$ their sum commute. So this is a subspace.
answered Jul 22 at 12:48
Davide Morgante
1,798220
Hi Davide! I hope you are doing well. I am struggling a bit in my linear algebra homework, and have posted some questions that people are not responding to. I was wondering if you can take a look at these questions and clarify them to me since your responses have been consistently the best. 1) math.stackexchange.com/questions/2871861/… 2) math.stackexchange.com/questions/2871059/… I have already posted my line of thinking. If it isn't any inconvenience to you. Thank you sir.
– PERTURBATIONFLOW
Aug 4 at 11:07
Hi! I've answered to the second question, on the first I think you already got a clear answer!
– Davide Morgante
Aug 4 at 12:45
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let's start by the definition
If $V$ is a vector space on a field $K$ and $W$ is a subset of $V$, then $W$ is a subspace if
- The zero vector is in $W$
- $W$ is closed under addition and multiplication by a scalar in $K$
Let us see now if the sets that you gave us are indeed subspaces o $mathbbR^ntimes n$:
- The set of all invertible $ntimes n$ matrices: being invertible means that $det(A)neq 0$. This is enough to say that this cannot be a subspace because the zero vector is the zero matrix which clearly has zero determinant and so it's not an invertible matrix
$ntimes n$ matrices of the form $kmathcalI$: this indeed is a subspace because it has the zero vector (just set $k=0$) and is closed under addition $$kmathcalI+wmathcalI= (k+w)mathcalI = nmathcalI$$ and it's closed under scalar multiplication which is clear even from the definition $kmathcalI$
All matrices similar to a matrix $A$: this is trickier because clearly if $A$ is the null matrix than all the matrix similar to $A$ is just the zero matrix. In this case this is a subspace because it has zero matrix, adding zero matrices together get's you zero matrices, and multiplying by a scalar the zero matrix gives the zero matrix. In all other cases this is not a subspace because of the lack of zero vector.
All matrices that commute with $A$: if two matrices commute then $$[A,B]= AB-BA = 0$$ let's see if this is a subspace: the zero matrix commutes with every matrix, in fact $$[A,0] = A0-0A = 0-0=0$$ For every matrix $B$ that commute with $A$, the matrix $kB$ $$[A, kB] = AkB-kBA = kAB-kBA = k[A,B] = 0$$ commutes as well with $A$. If two matrices $B$ and $C$ commutes with $A$, then $$beginalign&[A,B+C] = A(B+C)-(B+C)A = \ &AB + AC - BA - CA = \ &(AB-BA)+(AC-CA) = \ &[A,B]+[A,C] = 0+0=0endalign$$ their sum commute. So this is a subspace.
answered Jul 22 at 12:48
Davide Morgante
1,798220
Let's start by the definition
If $V$ is a vector space on a field $K$ and $W$ is a subset of $V$, then $W$ is a subspace if
- The zero vector is in $W$
- $W$ is closed under addition and multiplication by a scalar in $K$
Let us see now if the sets that you gave us are indeed subspaces o $mathbbR^ntimes n$:
- The set of all invertible $ntimes n$ matrices: being invertible means that $det(A)neq 0$. This is enough to say that this cannot be a subspace because the zero vector is the zero matrix which clearly has zero determinant and so it's not an invertible matrix
$ntimes n$ matrices of the form $kmathcalI$: this indeed is a subspace because it has the zero vector (just set $k=0$) and is closed under addition $$kmathcalI+wmathcalI= (k+w)mathcalI = nmathcalI$$ and it's closed under scalar multiplication which is clear even from the definition $kmathcalI$
All matrices similar to a matrix $A$: this is trickier because clearly if $A$ is the null matrix than all the matrix similar to $A$ is just the zero matrix. In this case this is a subspace because it has zero matrix, adding zero matrices together get's you zero matrices, and multiplying by a scalar the zero matrix gives the zero matrix. In all other cases this is not a subspace because of the lack of zero vector.
All matrices that commute with $A$: if two matrices commute then $$[A,B]= AB-BA = 0$$ let's see if this is a subspace: the zero matrix commutes with every matrix, in fact $$[A,0] = A0-0A = 0-0=0$$ For every matrix $B$ that commute with $A$, the matrix $kB$ $$[A, kB] = AkB-kBA = kAB-kBA = k[A,B] = 0$$ commutes as well with $A$. If two matrices $B$ and $C$ commutes with $A$, then $$beginalign&[A,B+C] = A(B+C)-(B+C)A = \ &AB + AC - BA - CA = \ &(AB-BA)+(AC-CA) = \ &[A,B]+[A,C] = 0+0=0endalign$$ their sum commute. So this is a subspace.
answered Jul 22 at 12:48
Davide Morgante
1,798220
answered Jul 22 at 12:48
Davide Morgante
1,798220
answered Jul 22 at 12:48
Davide Morgante
1,798220
answered Jul 22 at 12:48
Davide Morgante
1,798220
1,798220
Hi Davide! I hope you are doing well. I am struggling a bit in my linear algebra homework, and have posted some questions that people are not responding to. I was wondering if you can take a look at these questions and clarify them to me since your responses have been consistently the best. 1) math.stackexchange.com/questions/2871861/… 2) math.stackexchange.com/questions/2871059/… I have already posted my line of thinking. If it isn't any inconvenience to you. Thank you sir.
– PERTURBATIONFLOW
Aug 4 at 11:07
Hi! I've answered to the second question, on the first I think you already got a clear answer!
– Davide Morgante
Aug 4 at 12:45
add a comment |Â
Hi Davide! I hope you are doing well. I am struggling a bit in my linear algebra homework, and have posted some questions that people are not responding to. I was wondering if you can take a look at these questions and clarify them to me since your responses have been consistently the best. 1) math.stackexchange.com/questions/2871861/… 2) math.stackexchange.com/questions/2871059/… I have already posted my line of thinking. If it isn't any inconvenience to you. Thank you sir.
– PERTURBATIONFLOW
Aug 4 at 11:07
Hi! I've answered to the second question, on the first I think you already got a clear answer!
– Davide Morgante
Aug 4 at 12:45
Hi Davide! I hope you are doing well. I am struggling a bit in my linear algebra homework, and have posted some questions that people are not responding to. I was wondering if you can take a look at these questions and clarify them to me since your responses have been consistently the best. 1) math.stackexchange.com/questions/2871861/… 2) math.stackexchange.com/questions/2871059/… I have already posted my line of thinking. If it isn't any inconvenience to you. Thank you sir.
– PERTURBATIONFLOW
Aug 4 at 11:07
Hi Davide! I hope you are doing well. I am struggling a bit in my linear algebra homework, and have posted some questions that people are not responding to. I was wondering if you can take a look at these questions and clarify them to me since your responses have been consistently the best. 1) math.stackexchange.com/questions/2871861/… 2) math.stackexchange.com/questions/2871059/… I have already posted my line of thinking. If it isn't any inconvenience to you. Thank you sir.
– PERTURBATIONFLOW
Aug 4 at 11:07
Hi! I've answered to the second question, on the first I think you already got a clear answer!
– Davide Morgante
Aug 4 at 12:45
Hi! I've answered to the second question, on the first I think you already got a clear answer!
– Davide Morgante
Aug 4 at 12:45
add a comment |Â
up vote
-1
down vote
I think you should rephrase your question.
You have a given dimension space of â„Ân×n, and wish to identify which of the following are subspaces?
- n×n invertible matrices belong to its subspace
- the set of all n×n matrices of the form k∗I, where I is the identity matrix, and k is any real number also belongs to its subspace
- the set of matrices similar to A. We're assuming that A is a fixed matrix
- the set of all matrices that commute with A. We're assuming that A is a fixed matrix as well.
Your previous question was a bit jumbled to understand. You're making the assumption that 1 and 4 hold true, but 2 and 3 don't. You need to justify these claims first.
answered Jul 22 at 11:53
FireMeUP
456
This isn´t really an answer to the question and equally unclear.
– Peter Melech
Jul 22 at 11:58
add a comment |Â
up vote
-1
down vote
I think you should rephrase your question.
You have a given dimension space of â„Ân×n, and wish to identify which of the following are subspaces?
- n×n invertible matrices belong to its subspace
- the set of all n×n matrices of the form k∗I, where I is the identity matrix, and k is any real number also belongs to its subspace
- the set of matrices similar to A. We're assuming that A is a fixed matrix
- the set of all matrices that commute with A. We're assuming that A is a fixed matrix as well.
Your previous question was a bit jumbled to understand. You're making the assumption that 1 and 4 hold true, but 2 and 3 don't. You need to justify these claims first.
answered Jul 22 at 11:53
FireMeUP
456
This isn´t really an answer to the question and equally unclear.
– Peter Melech
Jul 22 at 11:58
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
I think you should rephrase your question.
You have a given dimension space of â„Ân×n, and wish to identify which of the following are subspaces?
- n×n invertible matrices belong to its subspace
- the set of all n×n matrices of the form k∗I, where I is the identity matrix, and k is any real number also belongs to its subspace
- the set of matrices similar to A. We're assuming that A is a fixed matrix
- the set of all matrices that commute with A. We're assuming that A is a fixed matrix as well.
Your previous question was a bit jumbled to understand. You're making the assumption that 1 and 4 hold true, but 2 and 3 don't. You need to justify these claims first.
answered Jul 22 at 11:53
FireMeUP
456
I think you should rephrase your question.
You have a given dimension space of â„Ân×n, and wish to identify which of the following are subspaces?
- n×n invertible matrices belong to its subspace
- the set of all n×n matrices of the form k∗I, where I is the identity matrix, and k is any real number also belongs to its subspace
- the set of matrices similar to A. We're assuming that A is a fixed matrix
- the set of all matrices that commute with A. We're assuming that A is a fixed matrix as well.
Your previous question was a bit jumbled to understand. You're making the assumption that 1 and 4 hold true, but 2 and 3 don't. You need to justify these claims first.
answered Jul 22 at 11:53
FireMeUP
456
answered Jul 22 at 11:53
FireMeUP
456
answered Jul 22 at 11:53
FireMeUP
456
answered Jul 22 at 11:53
FireMeUP
456
456
This isn´t really an answer to the question and equally unclear.
– Peter Melech
Jul 22 at 11:58
add a comment |Â
This isn´t really an answer to the question and equally unclear.
– Peter Melech
Jul 22 at 11:58
This isn´t really an answer to the question and equally unclear.
– Peter Melech
Jul 22 at 11:58
This isn´t really an answer to the question and equally unclear.
– Peter Melech
Jul 22 at 11:58
add a comment |Â
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3
Are You asking whether the invertible matrices form a subspace of $mathbbR^ntimes n$? If so consider an invertible matrix $A$ and $-A$ and their sum, etc. However it´s not clear what You´re asking.
– Peter Melech
Jul 22 at 10:50
@PeterMelech The question is identifying subspaces that exist in the space Rnxn. The choices ares: invertible matrices of size nxn/the set of all matrices that are similar to A where A is a fixed matrix/the set of all matrices that commute with A where A is a fixed matrix/the set of all nxn matrices of the form KI where K is any real number. I know that the 1st and 4th can be conclusively defined as subspaces in Rxn, but what about the second and the third
– PERTURBATIONFLOW
Jul 22 at 10:57
" I know that the 1st and 4th can be conclusively defined as subspaces". How do You "know" that?? Consider e.g. the example I gave in my first comment.
– Peter Melech
Jul 22 at 11:35
E.g. ad 3:Let $V_A=T^-1AT:Tin GL_n(mathbbR)$. If $Aneq 0$ then $0notin V_A$
– Peter Melech
Jul 22 at 12:24