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Why is $T^* -barlambdaI$ not one-to-one?
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From Friedberg's Linear Algebra
Let $T$ be a linear operator on a finite-dimensional inner product space $V$. If $T$ has an eigenvector, then so doess $T$*.
Proof. Suppose that $v$ is an eigenvector of $T$ with corresponding eigenvalue $lambda$. Then for any $x in V$,
$0 = langle 0,x rangle = langle(T - lambda I)(v),x rangle = langle v,(T-lambda I)^*(x)rangle = langle v, (T^*-barlambda I)(x)rangle$,
and hence $v$ is orthogonal to the range of $T^* -barlambdaI$. So $T^* -barlambdaI$ is not onto and hence is not one-to-one. Thus $T^* -barlambdaI$ has a nonzero null space, and any nonzero vector in this null space is an eigenvector of $T*$ with correspoinding eigenvalue $barlambda.$
I'm unable to see why $T^* -barlambdaI$ is not one-to-one, and why $T^* -barlambdaI$ has a nonzero null space.
linear-algebra
asked Jul 22 at 16:44
K.M
480312
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From Friedberg's Linear Algebra
Let $T$ be a linear operator on a finite-dimensional inner product space $V$. If $T$ has an eigenvector, then so doess $T$*.
Proof. Suppose that $v$ is an eigenvector of $T$ with corresponding eigenvalue $lambda$. Then for any $x in V$,
$0 = langle 0,x rangle = langle(T - lambda I)(v),x rangle = langle v,(T-lambda I)^*(x)rangle = langle v, (T^*-barlambda I)(x)rangle$,
and hence $v$ is orthogonal to the range of $T^* -barlambdaI$. So $T^* -barlambdaI$ is not onto and hence is not one-to-one. Thus $T^* -barlambdaI$ has a nonzero null space, and any nonzero vector in this null space is an eigenvector of $T*$ with correspoinding eigenvalue $barlambda.$
I'm unable to see why $T^* -barlambdaI$ is not one-to-one, and why $T^* -barlambdaI$ has a nonzero null space.
linear-algebra
asked Jul 22 at 16:44
K.M
480312
1
If the range of $T* -barlambdaI$ was all of $V$, then how can $v$ be orthogonal to all of it? Unless $v$ is zero, but it cannot be zero because it is an eigenvector of something, and zero vectors are not eigenvectors of anything.
– Marcus Aurelius
Jul 22 at 16:49
I thought that would mean that it's not onto, but I still can't see why it's not one-to-one.
– K.M
Jul 22 at 17:14
This is just the dimension formula. Dimension of kernel + dimension of image = dimension of V (or more generally, the domain).
– Marcus Aurelius
Jul 22 at 17:15
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
From Friedberg's Linear Algebra
Let $T$ be a linear operator on a finite-dimensional inner product space $V$. If $T$ has an eigenvector, then so doess $T$*.
Proof. Suppose that $v$ is an eigenvector of $T$ with corresponding eigenvalue $lambda$. Then for any $x in V$,
$0 = langle 0,x rangle = langle(T - lambda I)(v),x rangle = langle v,(T-lambda I)^*(x)rangle = langle v, (T^*-barlambda I)(x)rangle$,
and hence $v$ is orthogonal to the range of $T^* -barlambdaI$. So $T^* -barlambdaI$ is not onto and hence is not one-to-one. Thus $T^* -barlambdaI$ has a nonzero null space, and any nonzero vector in this null space is an eigenvector of $T*$ with correspoinding eigenvalue $barlambda.$
I'm unable to see why $T^* -barlambdaI$ is not one-to-one, and why $T^* -barlambdaI$ has a nonzero null space.
linear-algebra
asked Jul 22 at 16:44
K.M
480312
From Friedberg's Linear Algebra
Let $T$ be a linear operator on a finite-dimensional inner product space $V$. If $T$ has an eigenvector, then so doess $T$*.
Proof. Suppose that $v$ is an eigenvector of $T$ with corresponding eigenvalue $lambda$. Then for any $x in V$,
$0 = langle 0,x rangle = langle(T - lambda I)(v),x rangle = langle v,(T-lambda I)^*(x)rangle = langle v, (T^*-barlambda I)(x)rangle$,
and hence $v$ is orthogonal to the range of $T^* -barlambdaI$. So $T^* -barlambdaI$ is not onto and hence is not one-to-one. Thus $T^* -barlambdaI$ has a nonzero null space, and any nonzero vector in this null space is an eigenvector of $T*$ with correspoinding eigenvalue $barlambda.$
I'm unable to see why $T^* -barlambdaI$ is not one-to-one, and why $T^* -barlambdaI$ has a nonzero null space.
linear-algebra
asked Jul 22 at 16:44
K.M
480312
asked Jul 22 at 16:44
K.M
480312
asked Jul 22 at 16:44
K.M
480312
asked Jul 22 at 16:44
K.M
480312
480312
1
If the range of $T* -barlambdaI$ was all of $V$, then how can $v$ be orthogonal to all of it? Unless $v$ is zero, but it cannot be zero because it is an eigenvector of something, and zero vectors are not eigenvectors of anything.
– Marcus Aurelius
Jul 22 at 16:49
I thought that would mean that it's not onto, but I still can't see why it's not one-to-one.
– K.M
Jul 22 at 17:14
This is just the dimension formula. Dimension of kernel + dimension of image = dimension of V (or more generally, the domain).
– Marcus Aurelius
Jul 22 at 17:15
add a comment |Â
1
If the range of $T* -barlambdaI$ was all of $V$, then how can $v$ be orthogonal to all of it? Unless $v$ is zero, but it cannot be zero because it is an eigenvector of something, and zero vectors are not eigenvectors of anything.
– Marcus Aurelius
Jul 22 at 16:49
I thought that would mean that it's not onto, but I still can't see why it's not one-to-one.
– K.M
Jul 22 at 17:14
This is just the dimension formula. Dimension of kernel + dimension of image = dimension of V (or more generally, the domain).
– Marcus Aurelius
Jul 22 at 17:15
1
1
If the range of $T* -barlambdaI$ was all of $V$, then how can $v$ be orthogonal to all of it? Unless $v$ is zero, but it cannot be zero because it is an eigenvector of something, and zero vectors are not eigenvectors of anything.
– Marcus Aurelius
Jul 22 at 16:49
If the range of $T* -barlambdaI$ was all of $V$, then how can $v$ be orthogonal to all of it? Unless $v$ is zero, but it cannot be zero because it is an eigenvector of something, and zero vectors are not eigenvectors of anything.
– Marcus Aurelius
Jul 22 at 16:49
I thought that would mean that it's not onto, but I still can't see why it's not one-to-one.
– K.M
Jul 22 at 17:14
I thought that would mean that it's not onto, but I still can't see why it's not one-to-one.
– K.M
Jul 22 at 17:14
This is just the dimension formula. Dimension of kernel + dimension of image = dimension of V (or more generally, the domain).
– Marcus Aurelius
Jul 22 at 17:15
This is just the dimension formula. Dimension of kernel + dimension of image = dimension of V (or more generally, the domain).
– Marcus Aurelius
Jul 22 at 17:15
add a comment |Â
2 Answers
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Let $lambda$ be a non-zero eigen value of $T$, i.e. $exists$ non-zero $x in V$ such that
$$Tx = lambda x$$
$$(T-lambda I)x = 0$$
Now let $y in V$ be non-zero.
$$langle Tx,y rangle = langle x,T^*y rangle $$
$$langle Tx,y rangle =langle lambda x,y rangle = langle x,overlinelambda y rangle$$
And therefore,
$$langle x,T^*y rangle = langle x,
overlinelambda y rangle$$.
$$implies langle x, (T^*-overlinelambda I )y rangle = 0$$
Therefore $x in (Rg(T^*-overlinelambda I))^perp$
And we know that $(Rg(T))^perp = N(T)$
Therefore, $x in N(T^*-overlinelambda I)$
i.e.
$$T^*x = overlinelambda x$$
answered Jul 22 at 21:01
Vizag
271111
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Lauds to our colleague Vizag for his elegant demonstration that
$lambda ; textan eigenvalue of ; T Longrightarrow bar lambda ; textan eigenvalue of ; T^dagger; tag 1$
however, his work on this subject leaves unaddressed the title question, that is,
$text"Why is ; T^dagger - bar lambda I ; textnot one-to-one?" tag 2$
I wish to take up this specific topic here, and provide a sort of "classic" answer; specifically, I wish to demonstrate the essential and well-known result,
"A linear map $S:V to V$ from a finite dimensional vector space to itself is one-to-one if and only if it is onto."
Note: in what follows we allow $V$ to be a vector space over any base field $Bbb F$.
Proof: The argument is based upon elementary notions of basis and linear independence.
We first assume $S:V to V$ is onto. Then we let $w_1, w_2, dots w_n$ be a basis for $V$ over the field $Bbb F$, and we see, since $S$ is surjective, that there must be a set of vectors $v_i$, $1 le i le n$, with
$Sv_i = w_i, ; 1 le i le n; tag 3$
I claim the set $ v_i mid 1 le i le n $ is linearly independent over $Bbb F$; for if not, there would exist $alpha_i in Bbb F$, not all zero, with
$displaystyle sum_1^n alpha_i v_i = 0; tag 4$
then
$displaystyle sum_1^n alpha_i w_i = sum_1^n alpha_i Sv_i = S left (sum_1^n alpha_i v_i right ) = S(0) = 0; tag 5$
but this contradicts the linear independence of the $w_i$ unless
$alpha_i = 0, ; 1 le i le n; tag 6$
but condition (6) is precluded by our assumption that not all the $alpha_i = 0$; therefore the $v_i$ are linearly independent over $Bbb F$ and hence form a basis for $V$; then any $x in V$ may be written
$x = displaystyle sum_1^n x_i v_i, ; x_i in Bbb F; tag 7$
now suppose $S$ were not injective. Then we could find $x_1, x_2 in V$ with
$Sx_1 = Sx_2; tag 8$
if, in accord with (7) we set
$x_1 = displaystyle sum_1^n alpha_i v_i, tag 9$
$x_2 = displaystyle sum_1^n beta_i v_i, tag10$
then from (8)-(10),
$displaystyle sum_1^n alpha_i w_i = sum_1^n alpha_i Sv_i = S left (sum_1^n
alpha_i v_i right ) = S left (sum_1^n beta_i v_i right) = sum_1^n beta_i Sv_i = sum_1^n beta_i w_i, tag11$
whence
$displaystyle sum_1^n (alpha_i - beta_i) w_i = 0; tag12$
now the linear independence of the $w_i$ forces
$alpha_i = beta_i, ; 1 le i le n, tag13$
whence again via (9)-(10)
$x_1 = x_2, tag14$
and we see that $S$ is injective.
Going the other way, we now suppose $S$ is injective; and let the set
$v_i mid 1 le i le n $ form a basis for $V$. I claim that the vectors $Sv_1, Sv_2, ldots, Sv_n$ also form a basis; for if not, they must be linearly dependent and we may find $alpha_i in Bbb F$ such that
$displaystyle S left ( sum_1^n alpha_i v_i right ) = sum_1^n alpha_i Sv_i = 0; tag15$
now with $S$ injective this forces
$displaystyle sum_1^n alpha_i v_i = 0, tag16$
impossible by the assumed linear independence of the $v_i$; thus the $Sv_i$ do form a basis and hence any $y in V$ may be written
$y = displaystyle sum_1^n beta_i Sv_i = S left ( sum_1^n beta_i v_i right ); tag17$
thus every $y in V$ lies in the image of $S$ which at last seen to be onto.
End: Proof.
If we apply this result to $T^dagger - bar lambda I$ as in the body of the question, we see that, having shown that $T^dagger - bar lambda I$ is not onto, we may conclude it is also not injective by the preceding basic demonstration; but not injective implies the null space is not $ 0 $, since if $x_1 ne x_2$ but $Sx_1 = Sx_2$, we have $x_1 - x_2 ne 0$ but
$S(x_1 - x_2) = Sx_1 - Sx_2 = 0, tag18$
whence $0 ne x_1 - x_2 in ker S ne 0 $.
answered Jul 24 at 0:28
Robert Lewis
36.9k22255
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2 Answers
2
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let $lambda$ be a non-zero eigen value of $T$, i.e. $exists$ non-zero $x in V$ such that
$$Tx = lambda x$$
$$(T-lambda I)x = 0$$
Now let $y in V$ be non-zero.
$$langle Tx,y rangle = langle x,T^*y rangle $$
$$langle Tx,y rangle =langle lambda x,y rangle = langle x,overlinelambda y rangle$$
And therefore,
$$langle x,T^*y rangle = langle x,
overlinelambda y rangle$$.
$$implies langle x, (T^*-overlinelambda I )y rangle = 0$$
Therefore $x in (Rg(T^*-overlinelambda I))^perp$
And we know that $(Rg(T))^perp = N(T)$
Therefore, $x in N(T^*-overlinelambda I)$
i.e.
$$T^*x = overlinelambda x$$
answered Jul 22 at 21:01
Vizag
271111
add a comment |Â
up vote
2
down vote
Let $lambda$ be a non-zero eigen value of $T$, i.e. $exists$ non-zero $x in V$ such that
$$Tx = lambda x$$
$$(T-lambda I)x = 0$$
Now let $y in V$ be non-zero.
$$langle Tx,y rangle = langle x,T^*y rangle $$
$$langle Tx,y rangle =langle lambda x,y rangle = langle x,overlinelambda y rangle$$
And therefore,
$$langle x,T^*y rangle = langle x,
overlinelambda y rangle$$.
$$implies langle x, (T^*-overlinelambda I )y rangle = 0$$
Therefore $x in (Rg(T^*-overlinelambda I))^perp$
And we know that $(Rg(T))^perp = N(T)$
Therefore, $x in N(T^*-overlinelambda I)$
i.e.
$$T^*x = overlinelambda x$$
answered Jul 22 at 21:01
Vizag
271111
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $lambda$ be a non-zero eigen value of $T$, i.e. $exists$ non-zero $x in V$ such that
$$Tx = lambda x$$
$$(T-lambda I)x = 0$$
Now let $y in V$ be non-zero.
$$langle Tx,y rangle = langle x,T^*y rangle $$
$$langle Tx,y rangle =langle lambda x,y rangle = langle x,overlinelambda y rangle$$
And therefore,
$$langle x,T^*y rangle = langle x,
overlinelambda y rangle$$.
$$implies langle x, (T^*-overlinelambda I )y rangle = 0$$
Therefore $x in (Rg(T^*-overlinelambda I))^perp$
And we know that $(Rg(T))^perp = N(T)$
Therefore, $x in N(T^*-overlinelambda I)$
i.e.
$$T^*x = overlinelambda x$$
answered Jul 22 at 21:01
Vizag
271111
Let $lambda$ be a non-zero eigen value of $T$, i.e. $exists$ non-zero $x in V$ such that
$$Tx = lambda x$$
$$(T-lambda I)x = 0$$
Now let $y in V$ be non-zero.
$$langle Tx,y rangle = langle x,T^*y rangle $$
$$langle Tx,y rangle =langle lambda x,y rangle = langle x,overlinelambda y rangle$$
And therefore,
$$langle x,T^*y rangle = langle x,
overlinelambda y rangle$$.
$$implies langle x, (T^*-overlinelambda I )y rangle = 0$$
Therefore $x in (Rg(T^*-overlinelambda I))^perp$
And we know that $(Rg(T))^perp = N(T)$
Therefore, $x in N(T^*-overlinelambda I)$
i.e.
$$T^*x = overlinelambda x$$
answered Jul 22 at 21:01
Vizag
271111
answered Jul 22 at 21:01
Vizag
271111
answered Jul 22 at 21:01
Vizag
271111
answered Jul 22 at 21:01
Vizag
271111
271111
add a comment |Â
add a comment |Â
up vote
0
down vote
Lauds to our colleague Vizag for his elegant demonstration that
$lambda ; textan eigenvalue of ; T Longrightarrow bar lambda ; textan eigenvalue of ; T^dagger; tag 1$
however, his work on this subject leaves unaddressed the title question, that is,
$text"Why is ; T^dagger - bar lambda I ; textnot one-to-one?" tag 2$
I wish to take up this specific topic here, and provide a sort of "classic" answer; specifically, I wish to demonstrate the essential and well-known result,
"A linear map $S:V to V$ from a finite dimensional vector space to itself is one-to-one if and only if it is onto."
Note: in what follows we allow $V$ to be a vector space over any base field $Bbb F$.
Proof: The argument is based upon elementary notions of basis and linear independence.
We first assume $S:V to V$ is onto. Then we let $w_1, w_2, dots w_n$ be a basis for $V$ over the field $Bbb F$, and we see, since $S$ is surjective, that there must be a set of vectors $v_i$, $1 le i le n$, with
$Sv_i = w_i, ; 1 le i le n; tag 3$
I claim the set $ v_i mid 1 le i le n $ is linearly independent over $Bbb F$; for if not, there would exist $alpha_i in Bbb F$, not all zero, with
$displaystyle sum_1^n alpha_i v_i = 0; tag 4$
then
$displaystyle sum_1^n alpha_i w_i = sum_1^n alpha_i Sv_i = S left (sum_1^n alpha_i v_i right ) = S(0) = 0; tag 5$
but this contradicts the linear independence of the $w_i$ unless
$alpha_i = 0, ; 1 le i le n; tag 6$
but condition (6) is precluded by our assumption that not all the $alpha_i = 0$; therefore the $v_i$ are linearly independent over $Bbb F$ and hence form a basis for $V$; then any $x in V$ may be written
$x = displaystyle sum_1^n x_i v_i, ; x_i in Bbb F; tag 7$
now suppose $S$ were not injective. Then we could find $x_1, x_2 in V$ with
$Sx_1 = Sx_2; tag 8$
if, in accord with (7) we set
$x_1 = displaystyle sum_1^n alpha_i v_i, tag 9$
$x_2 = displaystyle sum_1^n beta_i v_i, tag10$
then from (8)-(10),
$displaystyle sum_1^n alpha_i w_i = sum_1^n alpha_i Sv_i = S left (sum_1^n
alpha_i v_i right ) = S left (sum_1^n beta_i v_i right) = sum_1^n beta_i Sv_i = sum_1^n beta_i w_i, tag11$
whence
$displaystyle sum_1^n (alpha_i - beta_i) w_i = 0; tag12$
now the linear independence of the $w_i$ forces
$alpha_i = beta_i, ; 1 le i le n, tag13$
whence again via (9)-(10)
$x_1 = x_2, tag14$
and we see that $S$ is injective.
Going the other way, we now suppose $S$ is injective; and let the set
$v_i mid 1 le i le n $ form a basis for $V$. I claim that the vectors $Sv_1, Sv_2, ldots, Sv_n$ also form a basis; for if not, they must be linearly dependent and we may find $alpha_i in Bbb F$ such that
$displaystyle S left ( sum_1^n alpha_i v_i right ) = sum_1^n alpha_i Sv_i = 0; tag15$
now with $S$ injective this forces
$displaystyle sum_1^n alpha_i v_i = 0, tag16$
impossible by the assumed linear independence of the $v_i$; thus the $Sv_i$ do form a basis and hence any $y in V$ may be written
$y = displaystyle sum_1^n beta_i Sv_i = S left ( sum_1^n beta_i v_i right ); tag17$
thus every $y in V$ lies in the image of $S$ which at last seen to be onto.
End: Proof.
If we apply this result to $T^dagger - bar lambda I$ as in the body of the question, we see that, having shown that $T^dagger - bar lambda I$ is not onto, we may conclude it is also not injective by the preceding basic demonstration; but not injective implies the null space is not $ 0 $, since if $x_1 ne x_2$ but $Sx_1 = Sx_2$, we have $x_1 - x_2 ne 0$ but
$S(x_1 - x_2) = Sx_1 - Sx_2 = 0, tag18$
whence $0 ne x_1 - x_2 in ker S ne 0 $.
answered Jul 24 at 0:28
Robert Lewis
36.9k22255
add a comment |Â
up vote
0
down vote
Lauds to our colleague Vizag for his elegant demonstration that
$lambda ; textan eigenvalue of ; T Longrightarrow bar lambda ; textan eigenvalue of ; T^dagger; tag 1$
however, his work on this subject leaves unaddressed the title question, that is,
$text"Why is ; T^dagger - bar lambda I ; textnot one-to-one?" tag 2$
I wish to take up this specific topic here, and provide a sort of "classic" answer; specifically, I wish to demonstrate the essential and well-known result,
"A linear map $S:V to V$ from a finite dimensional vector space to itself is one-to-one if and only if it is onto."
Note: in what follows we allow $V$ to be a vector space over any base field $Bbb F$.
Proof: The argument is based upon elementary notions of basis and linear independence.
We first assume $S:V to V$ is onto. Then we let $w_1, w_2, dots w_n$ be a basis for $V$ over the field $Bbb F$, and we see, since $S$ is surjective, that there must be a set of vectors $v_i$, $1 le i le n$, with
$Sv_i = w_i, ; 1 le i le n; tag 3$
I claim the set $ v_i mid 1 le i le n $ is linearly independent over $Bbb F$; for if not, there would exist $alpha_i in Bbb F$, not all zero, with
$displaystyle sum_1^n alpha_i v_i = 0; tag 4$
then
$displaystyle sum_1^n alpha_i w_i = sum_1^n alpha_i Sv_i = S left (sum_1^n alpha_i v_i right ) = S(0) = 0; tag 5$
but this contradicts the linear independence of the $w_i$ unless
$alpha_i = 0, ; 1 le i le n; tag 6$
but condition (6) is precluded by our assumption that not all the $alpha_i = 0$; therefore the $v_i$ are linearly independent over $Bbb F$ and hence form a basis for $V$; then any $x in V$ may be written
$x = displaystyle sum_1^n x_i v_i, ; x_i in Bbb F; tag 7$
now suppose $S$ were not injective. Then we could find $x_1, x_2 in V$ with
$Sx_1 = Sx_2; tag 8$
if, in accord with (7) we set
$x_1 = displaystyle sum_1^n alpha_i v_i, tag 9$
$x_2 = displaystyle sum_1^n beta_i v_i, tag10$
then from (8)-(10),
$displaystyle sum_1^n alpha_i w_i = sum_1^n alpha_i Sv_i = S left (sum_1^n
alpha_i v_i right ) = S left (sum_1^n beta_i v_i right) = sum_1^n beta_i Sv_i = sum_1^n beta_i w_i, tag11$
whence
$displaystyle sum_1^n (alpha_i - beta_i) w_i = 0; tag12$
now the linear independence of the $w_i$ forces
$alpha_i = beta_i, ; 1 le i le n, tag13$
whence again via (9)-(10)
$x_1 = x_2, tag14$
and we see that $S$ is injective.
Going the other way, we now suppose $S$ is injective; and let the set
$v_i mid 1 le i le n $ form a basis for $V$. I claim that the vectors $Sv_1, Sv_2, ldots, Sv_n$ also form a basis; for if not, they must be linearly dependent and we may find $alpha_i in Bbb F$ such that
$displaystyle S left ( sum_1^n alpha_i v_i right ) = sum_1^n alpha_i Sv_i = 0; tag15$
now with $S$ injective this forces
$displaystyle sum_1^n alpha_i v_i = 0, tag16$
impossible by the assumed linear independence of the $v_i$; thus the $Sv_i$ do form a basis and hence any $y in V$ may be written
$y = displaystyle sum_1^n beta_i Sv_i = S left ( sum_1^n beta_i v_i right ); tag17$
thus every $y in V$ lies in the image of $S$ which at last seen to be onto.
End: Proof.
If we apply this result to $T^dagger - bar lambda I$ as in the body of the question, we see that, having shown that $T^dagger - bar lambda I$ is not onto, we may conclude it is also not injective by the preceding basic demonstration; but not injective implies the null space is not $ 0 $, since if $x_1 ne x_2$ but $Sx_1 = Sx_2$, we have $x_1 - x_2 ne 0$ but
$S(x_1 - x_2) = Sx_1 - Sx_2 = 0, tag18$
whence $0 ne x_1 - x_2 in ker S ne 0 $.
answered Jul 24 at 0:28
Robert Lewis
36.9k22255
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Lauds to our colleague Vizag for his elegant demonstration that
$lambda ; textan eigenvalue of ; T Longrightarrow bar lambda ; textan eigenvalue of ; T^dagger; tag 1$
however, his work on this subject leaves unaddressed the title question, that is,
$text"Why is ; T^dagger - bar lambda I ; textnot one-to-one?" tag 2$
I wish to take up this specific topic here, and provide a sort of "classic" answer; specifically, I wish to demonstrate the essential and well-known result,
"A linear map $S:V to V$ from a finite dimensional vector space to itself is one-to-one if and only if it is onto."
Note: in what follows we allow $V$ to be a vector space over any base field $Bbb F$.
Proof: The argument is based upon elementary notions of basis and linear independence.
We first assume $S:V to V$ is onto. Then we let $w_1, w_2, dots w_n$ be a basis for $V$ over the field $Bbb F$, and we see, since $S$ is surjective, that there must be a set of vectors $v_i$, $1 le i le n$, with
$Sv_i = w_i, ; 1 le i le n; tag 3$
I claim the set $ v_i mid 1 le i le n $ is linearly independent over $Bbb F$; for if not, there would exist $alpha_i in Bbb F$, not all zero, with
$displaystyle sum_1^n alpha_i v_i = 0; tag 4$
then
$displaystyle sum_1^n alpha_i w_i = sum_1^n alpha_i Sv_i = S left (sum_1^n alpha_i v_i right ) = S(0) = 0; tag 5$
but this contradicts the linear independence of the $w_i$ unless
$alpha_i = 0, ; 1 le i le n; tag 6$
but condition (6) is precluded by our assumption that not all the $alpha_i = 0$; therefore the $v_i$ are linearly independent over $Bbb F$ and hence form a basis for $V$; then any $x in V$ may be written
$x = displaystyle sum_1^n x_i v_i, ; x_i in Bbb F; tag 7$
now suppose $S$ were not injective. Then we could find $x_1, x_2 in V$ with
$Sx_1 = Sx_2; tag 8$
if, in accord with (7) we set
$x_1 = displaystyle sum_1^n alpha_i v_i, tag 9$
$x_2 = displaystyle sum_1^n beta_i v_i, tag10$
then from (8)-(10),
$displaystyle sum_1^n alpha_i w_i = sum_1^n alpha_i Sv_i = S left (sum_1^n
alpha_i v_i right ) = S left (sum_1^n beta_i v_i right) = sum_1^n beta_i Sv_i = sum_1^n beta_i w_i, tag11$
whence
$displaystyle sum_1^n (alpha_i - beta_i) w_i = 0; tag12$
now the linear independence of the $w_i$ forces
$alpha_i = beta_i, ; 1 le i le n, tag13$
whence again via (9)-(10)
$x_1 = x_2, tag14$
and we see that $S$ is injective.
Going the other way, we now suppose $S$ is injective; and let the set
$v_i mid 1 le i le n $ form a basis for $V$. I claim that the vectors $Sv_1, Sv_2, ldots, Sv_n$ also form a basis; for if not, they must be linearly dependent and we may find $alpha_i in Bbb F$ such that
$displaystyle S left ( sum_1^n alpha_i v_i right ) = sum_1^n alpha_i Sv_i = 0; tag15$
now with $S$ injective this forces
$displaystyle sum_1^n alpha_i v_i = 0, tag16$
impossible by the assumed linear independence of the $v_i$; thus the $Sv_i$ do form a basis and hence any $y in V$ may be written
$y = displaystyle sum_1^n beta_i Sv_i = S left ( sum_1^n beta_i v_i right ); tag17$
thus every $y in V$ lies in the image of $S$ which at last seen to be onto.
End: Proof.
If we apply this result to $T^dagger - bar lambda I$ as in the body of the question, we see that, having shown that $T^dagger - bar lambda I$ is not onto, we may conclude it is also not injective by the preceding basic demonstration; but not injective implies the null space is not $ 0 $, since if $x_1 ne x_2$ but $Sx_1 = Sx_2$, we have $x_1 - x_2 ne 0$ but
$S(x_1 - x_2) = Sx_1 - Sx_2 = 0, tag18$
whence $0 ne x_1 - x_2 in ker S ne 0 $.
answered Jul 24 at 0:28
Robert Lewis
36.9k22255
Lauds to our colleague Vizag for his elegant demonstration that
$lambda ; textan eigenvalue of ; T Longrightarrow bar lambda ; textan eigenvalue of ; T^dagger; tag 1$
however, his work on this subject leaves unaddressed the title question, that is,
$text"Why is ; T^dagger - bar lambda I ; textnot one-to-one?" tag 2$
I wish to take up this specific topic here, and provide a sort of "classic" answer; specifically, I wish to demonstrate the essential and well-known result,
"A linear map $S:V to V$ from a finite dimensional vector space to itself is one-to-one if and only if it is onto."
Note: in what follows we allow $V$ to be a vector space over any base field $Bbb F$.
Proof: The argument is based upon elementary notions of basis and linear independence.
We first assume $S:V to V$ is onto. Then we let $w_1, w_2, dots w_n$ be a basis for $V$ over the field $Bbb F$, and we see, since $S$ is surjective, that there must be a set of vectors $v_i$, $1 le i le n$, with
$Sv_i = w_i, ; 1 le i le n; tag 3$
I claim the set $ v_i mid 1 le i le n $ is linearly independent over $Bbb F$; for if not, there would exist $alpha_i in Bbb F$, not all zero, with
$displaystyle sum_1^n alpha_i v_i = 0; tag 4$
then
$displaystyle sum_1^n alpha_i w_i = sum_1^n alpha_i Sv_i = S left (sum_1^n alpha_i v_i right ) = S(0) = 0; tag 5$
but this contradicts the linear independence of the $w_i$ unless
$alpha_i = 0, ; 1 le i le n; tag 6$
but condition (6) is precluded by our assumption that not all the $alpha_i = 0$; therefore the $v_i$ are linearly independent over $Bbb F$ and hence form a basis for $V$; then any $x in V$ may be written
$x = displaystyle sum_1^n x_i v_i, ; x_i in Bbb F; tag 7$
now suppose $S$ were not injective. Then we could find $x_1, x_2 in V$ with
$Sx_1 = Sx_2; tag 8$
if, in accord with (7) we set
$x_1 = displaystyle sum_1^n alpha_i v_i, tag 9$
$x_2 = displaystyle sum_1^n beta_i v_i, tag10$
then from (8)-(10),
$displaystyle sum_1^n alpha_i w_i = sum_1^n alpha_i Sv_i = S left (sum_1^n
alpha_i v_i right ) = S left (sum_1^n beta_i v_i right) = sum_1^n beta_i Sv_i = sum_1^n beta_i w_i, tag11$
whence
$displaystyle sum_1^n (alpha_i - beta_i) w_i = 0; tag12$
now the linear independence of the $w_i$ forces
$alpha_i = beta_i, ; 1 le i le n, tag13$
whence again via (9)-(10)
$x_1 = x_2, tag14$
and we see that $S$ is injective.
Going the other way, we now suppose $S$ is injective; and let the set
$v_i mid 1 le i le n $ form a basis for $V$. I claim that the vectors $Sv_1, Sv_2, ldots, Sv_n$ also form a basis; for if not, they must be linearly dependent and we may find $alpha_i in Bbb F$ such that
$displaystyle S left ( sum_1^n alpha_i v_i right ) = sum_1^n alpha_i Sv_i = 0; tag15$
now with $S$ injective this forces
$displaystyle sum_1^n alpha_i v_i = 0, tag16$
impossible by the assumed linear independence of the $v_i$; thus the $Sv_i$ do form a basis and hence any $y in V$ may be written
$y = displaystyle sum_1^n beta_i Sv_i = S left ( sum_1^n beta_i v_i right ); tag17$
thus every $y in V$ lies in the image of $S$ which at last seen to be onto.
End: Proof.
If we apply this result to $T^dagger - bar lambda I$ as in the body of the question, we see that, having shown that $T^dagger - bar lambda I$ is not onto, we may conclude it is also not injective by the preceding basic demonstration; but not injective implies the null space is not $ 0 $, since if $x_1 ne x_2$ but $Sx_1 = Sx_2$, we have $x_1 - x_2 ne 0$ but
$S(x_1 - x_2) = Sx_1 - Sx_2 = 0, tag18$
whence $0 ne x_1 - x_2 in ker S ne 0 $.
answered Jul 24 at 0:28
Robert Lewis
36.9k22255
edited Jul 24 at 0:46
answered Jul 24 at 0:28
Robert Lewis
36.9k22255
answered Jul 24 at 0:28
Robert Lewis
36.9k22255
answered Jul 24 at 0:28
Robert Lewis
36.9k22255
36.9k22255
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1
If the range of $T* -barlambdaI$ was all of $V$, then how can $v$ be orthogonal to all of it? Unless $v$ is zero, but it cannot be zero because it is an eigenvector of something, and zero vectors are not eigenvectors of anything.
– Marcus Aurelius
Jul 22 at 16:49
I thought that would mean that it's not onto, but I still can't see why it's not one-to-one.
– K.M
Jul 22 at 17:14
This is just the dimension formula. Dimension of kernel + dimension of image = dimension of V (or more generally, the domain).
– Marcus Aurelius
Jul 22 at 17:15