Normalizing dual eigenvectors? Why only for trivial defects? The 2019 Stack Overflow Developer...
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Normalizing dual eigenvectors? Why only for trivial defects?
The 2019 Stack Overflow Developer Survey Results Are InShowing $ langle A^*Av|v rangle = langle Av | Avrangle$How to prove that $langle v, v rangle$ is the only scalar invariant for a vector under the isometry?An intuitive approach to the Jordan Normal formEigenvectors for normal operators and their adjointsQuestion about proof of Cauchy-Schwarz inequality.Does $langle f+h,grangle=langle f,grangle+langle h,grangle$ hold for all elements $f, g, h$ of an inner product space?Determining the exact one from all possible Jordan Canonical Forms of a matrixNon-similarity of possible Jordan formsJordan form of simple 2x2 matrixhomogeneous only trivial or infinite
$begingroup$
Two little questions to this passage:
(1) How can we normalize to get $langle u,u^*rangle =1$?
(2) Why is this possible if $lambda$ has trivial defect only (i.e. for trivial Jordan blocks)? I did not get it from the text.
linear-algebra inner-product-space
asked Mar 21 at 21:17
SalamoSalamo
359412
$endgroup$
add a comment |
$begingroup$
Two little questions to this passage:
(1) How can we normalize to get $langle u,u^*rangle =1$?
(2) Why is this possible if $lambda$ has trivial defect only (i.e. for trivial Jordan blocks)? I did not get it from the text.
linear-algebra inner-product-space
asked Mar 21 at 21:17
SalamoSalamo
359412
$endgroup$
add a comment |
$begingroup$
Two little questions to this passage:
(1) How can we normalize to get $langle u,u^*rangle =1$?
(2) Why is this possible if $lambda$ has trivial defect only (i.e. for trivial Jordan blocks)? I did not get it from the text.
linear-algebra inner-product-space
asked Mar 21 at 21:17
SalamoSalamo
359412
$endgroup$
Two little questions to this passage:
(1) How can we normalize to get $langle u,u^*rangle =1$?
(2) Why is this possible if $lambda$ has trivial defect only (i.e. for trivial Jordan blocks)? I did not get it from the text.
linear-algebra inner-product-space
linear-algebra inner-product-space
asked Mar 21 at 21:17
SalamoSalamo
359412
asked Mar 21 at 21:17
SalamoSalamo
359412
edited Mar 21 at 21:39
Salamo
asked Mar 21 at 21:17
SalamoSalamo
359412
asked Mar 21 at 21:17
SalamoSalamo
359412
asked Mar 21 at 21:17
SalamoSalamo
359412
359412
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The normalization is not deep: just that if $langle u, u^astrangle neq 0$, then $u^ast$ may be rescaled so that that inner product is 1. Asking whether we can find $u^ast$ such that $langle u, u^ast rangle neq 0$ is deeper.
I'm not sure whether you're working with a complex vector space with a Hermitian inner product, or more generally with a linear transformation and its dual. Below, I'll give an exposition of the latter; the former can be obtained by replacing duality with orthogonality, along with an appropriate sprinkling of complex conjugates.
Let $T: V to V$ be an endomorphism of a finite-dimensional vector space $V$ over an algebraically closed field $k$, and let $T^*: V^* to V^*$ denote the dual transformation. Let the Jordan blocks of $T$ acting on $V$ be $V = V_1 oplus ldots oplus V_m$.
Then the action of $T^*$ has Jordan decomposition $V^* = V_1^* oplus cdots oplus V_m^ast$, with the same eigenvalues. Thus, to understand the relationship between the eigenvectors of $T$ and the eigenvectors of $T^ast$, it suffices to analyze just one Jordan block. Further, the eigenvectors of $T$ and $T - lambda , mathrm{id}$ are the same, so it suffices to consider when $T$ has eigenvalue $0$, i.e. is nilpotent.
In the case when $T$ is nilpotent and has one Jordan block, $V$ has a basis ${e_1,ldots, e_n}$ with $T e_i = e_{i-1}$ for $i > 1$ and $Te_1 = 0$. That is, $T$ has matrix
$$ begin{bmatrix} 0 & 1 &&& \
& 0 & 1 & & \
& & ddots & ddots & \
& & & 0 & 1 \
& & & & 0
end{bmatrix}$$ Then $T^ast$ has only a one-dimensional eigenspace, spanned by $f$ defined by
$$ f(e_i) = begin{cases} 0 & i < n \ 1 & i = n.end{cases}$$
In terms of matrices, we can check this by taking the transpose of the matrix above. Then $f$ is our dual eigenvector and $e_1$ is our eigenvector, which has $f(e_1) neq 0$ if and only if $n = 1$, i.e. $T$ has trivial defect.
answered Mar 23 at 3:23
Joshua MundingerJoshua Mundinger
2,9321028
$endgroup$
$begingroup$
I see that $T^*$ has only one-dimensional eigenspace. But I don’t understand why it is spanned by the eigenfunction f which you have defined.
$endgroup$
– Salamo
Mar 23 at 7:47
$begingroup$
The transpose has kernel spanned by $(0, ldots, 0 ,1)$, which corresponds to the element in the dual basis of ${e_1,ldots, e_n}$ corresponding to $e_n$, which is $f$. Alternatively, you can check that $f circ T = 0$.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:42
$begingroup$
There was also a typo in the definition of $T$, which is now fixed.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:52
$begingroup$
Maybe one last question. Suppose $langle u,u^*rangleneq 0$. You say then it’s no deep thing to scale $u^*$ such that $langle u,u*rangle=1$. Why and how?
$endgroup$
– Salamo
Mar 23 at 19:26
$begingroup$
We have $langle u, cvrangle = c langle u, vrangle$ for all $c in k$, and $v$ is an eigenvector if and only if $cv$ is (for $c neq 0$), so if $v$ is an eigenvector we can set $u^ast = cv$ such that $langle u, u^astrangle = 1$ (what value of $c$ would this have to be?)
$endgroup$
– Joshua Mundinger
Mar 24 at 1:25
add a comment |
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$begingroup$
The normalization is not deep: just that if $langle u, u^astrangle neq 0$, then $u^ast$ may be rescaled so that that inner product is 1. Asking whether we can find $u^ast$ such that $langle u, u^ast rangle neq 0$ is deeper.
I'm not sure whether you're working with a complex vector space with a Hermitian inner product, or more generally with a linear transformation and its dual. Below, I'll give an exposition of the latter; the former can be obtained by replacing duality with orthogonality, along with an appropriate sprinkling of complex conjugates.
Let $T: V to V$ be an endomorphism of a finite-dimensional vector space $V$ over an algebraically closed field $k$, and let $T^*: V^* to V^*$ denote the dual transformation. Let the Jordan blocks of $T$ acting on $V$ be $V = V_1 oplus ldots oplus V_m$.
Then the action of $T^*$ has Jordan decomposition $V^* = V_1^* oplus cdots oplus V_m^ast$, with the same eigenvalues. Thus, to understand the relationship between the eigenvectors of $T$ and the eigenvectors of $T^ast$, it suffices to analyze just one Jordan block. Further, the eigenvectors of $T$ and $T - lambda , mathrm{id}$ are the same, so it suffices to consider when $T$ has eigenvalue $0$, i.e. is nilpotent.
In the case when $T$ is nilpotent and has one Jordan block, $V$ has a basis ${e_1,ldots, e_n}$ with $T e_i = e_{i-1}$ for $i > 1$ and $Te_1 = 0$. That is, $T$ has matrix
$$ begin{bmatrix} 0 & 1 &&& \
& 0 & 1 & & \
& & ddots & ddots & \
& & & 0 & 1 \
& & & & 0
end{bmatrix}$$ Then $T^ast$ has only a one-dimensional eigenspace, spanned by $f$ defined by
$$ f(e_i) = begin{cases} 0 & i < n \ 1 & i = n.end{cases}$$
In terms of matrices, we can check this by taking the transpose of the matrix above. Then $f$ is our dual eigenvector and $e_1$ is our eigenvector, which has $f(e_1) neq 0$ if and only if $n = 1$, i.e. $T$ has trivial defect.
answered Mar 23 at 3:23
Joshua MundingerJoshua Mundinger
2,9321028
$endgroup$
$begingroup$
I see that $T^*$ has only one-dimensional eigenspace. But I don’t understand why it is spanned by the eigenfunction f which you have defined.
$endgroup$
– Salamo
Mar 23 at 7:47
$begingroup$
The transpose has kernel spanned by $(0, ldots, 0 ,1)$, which corresponds to the element in the dual basis of ${e_1,ldots, e_n}$ corresponding to $e_n$, which is $f$. Alternatively, you can check that $f circ T = 0$.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:42
$begingroup$
There was also a typo in the definition of $T$, which is now fixed.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:52
$begingroup$
Maybe one last question. Suppose $langle u,u^*rangleneq 0$. You say then it’s no deep thing to scale $u^*$ such that $langle u,u*rangle=1$. Why and how?
$endgroup$
– Salamo
Mar 23 at 19:26
$begingroup$
We have $langle u, cvrangle = c langle u, vrangle$ for all $c in k$, and $v$ is an eigenvector if and only if $cv$ is (for $c neq 0$), so if $v$ is an eigenvector we can set $u^ast = cv$ such that $langle u, u^astrangle = 1$ (what value of $c$ would this have to be?)
$endgroup$
– Joshua Mundinger
Mar 24 at 1:25
add a comment |
$begingroup$
The normalization is not deep: just that if $langle u, u^astrangle neq 0$, then $u^ast$ may be rescaled so that that inner product is 1. Asking whether we can find $u^ast$ such that $langle u, u^ast rangle neq 0$ is deeper.
I'm not sure whether you're working with a complex vector space with a Hermitian inner product, or more generally with a linear transformation and its dual. Below, I'll give an exposition of the latter; the former can be obtained by replacing duality with orthogonality, along with an appropriate sprinkling of complex conjugates.
Let $T: V to V$ be an endomorphism of a finite-dimensional vector space $V$ over an algebraically closed field $k$, and let $T^*: V^* to V^*$ denote the dual transformation. Let the Jordan blocks of $T$ acting on $V$ be $V = V_1 oplus ldots oplus V_m$.
Then the action of $T^*$ has Jordan decomposition $V^* = V_1^* oplus cdots oplus V_m^ast$, with the same eigenvalues. Thus, to understand the relationship between the eigenvectors of $T$ and the eigenvectors of $T^ast$, it suffices to analyze just one Jordan block. Further, the eigenvectors of $T$ and $T - lambda , mathrm{id}$ are the same, so it suffices to consider when $T$ has eigenvalue $0$, i.e. is nilpotent.
In the case when $T$ is nilpotent and has one Jordan block, $V$ has a basis ${e_1,ldots, e_n}$ with $T e_i = e_{i-1}$ for $i > 1$ and $Te_1 = 0$. That is, $T$ has matrix
$$ begin{bmatrix} 0 & 1 &&& \
& 0 & 1 & & \
& & ddots & ddots & \
& & & 0 & 1 \
& & & & 0
end{bmatrix}$$ Then $T^ast$ has only a one-dimensional eigenspace, spanned by $f$ defined by
$$ f(e_i) = begin{cases} 0 & i < n \ 1 & i = n.end{cases}$$
In terms of matrices, we can check this by taking the transpose of the matrix above. Then $f$ is our dual eigenvector and $e_1$ is our eigenvector, which has $f(e_1) neq 0$ if and only if $n = 1$, i.e. $T$ has trivial defect.
answered Mar 23 at 3:23
Joshua MundingerJoshua Mundinger
2,9321028
$endgroup$
$begingroup$
I see that $T^*$ has only one-dimensional eigenspace. But I don’t understand why it is spanned by the eigenfunction f which you have defined.
$endgroup$
– Salamo
Mar 23 at 7:47
$begingroup$
The transpose has kernel spanned by $(0, ldots, 0 ,1)$, which corresponds to the element in the dual basis of ${e_1,ldots, e_n}$ corresponding to $e_n$, which is $f$. Alternatively, you can check that $f circ T = 0$.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:42
$begingroup$
There was also a typo in the definition of $T$, which is now fixed.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:52
$begingroup$
Maybe one last question. Suppose $langle u,u^*rangleneq 0$. You say then it’s no deep thing to scale $u^*$ such that $langle u,u*rangle=1$. Why and how?
$endgroup$
– Salamo
Mar 23 at 19:26
$begingroup$
We have $langle u, cvrangle = c langle u, vrangle$ for all $c in k$, and $v$ is an eigenvector if and only if $cv$ is (for $c neq 0$), so if $v$ is an eigenvector we can set $u^ast = cv$ such that $langle u, u^astrangle = 1$ (what value of $c$ would this have to be?)
$endgroup$
– Joshua Mundinger
Mar 24 at 1:25
add a comment |
$begingroup$
The normalization is not deep: just that if $langle u, u^astrangle neq 0$, then $u^ast$ may be rescaled so that that inner product is 1. Asking whether we can find $u^ast$ such that $langle u, u^ast rangle neq 0$ is deeper.
I'm not sure whether you're working with a complex vector space with a Hermitian inner product, or more generally with a linear transformation and its dual. Below, I'll give an exposition of the latter; the former can be obtained by replacing duality with orthogonality, along with an appropriate sprinkling of complex conjugates.
Let $T: V to V$ be an endomorphism of a finite-dimensional vector space $V$ over an algebraically closed field $k$, and let $T^*: V^* to V^*$ denote the dual transformation. Let the Jordan blocks of $T$ acting on $V$ be $V = V_1 oplus ldots oplus V_m$.
Then the action of $T^*$ has Jordan decomposition $V^* = V_1^* oplus cdots oplus V_m^ast$, with the same eigenvalues. Thus, to understand the relationship between the eigenvectors of $T$ and the eigenvectors of $T^ast$, it suffices to analyze just one Jordan block. Further, the eigenvectors of $T$ and $T - lambda , mathrm{id}$ are the same, so it suffices to consider when $T$ has eigenvalue $0$, i.e. is nilpotent.
In the case when $T$ is nilpotent and has one Jordan block, $V$ has a basis ${e_1,ldots, e_n}$ with $T e_i = e_{i-1}$ for $i > 1$ and $Te_1 = 0$. That is, $T$ has matrix
$$ begin{bmatrix} 0 & 1 &&& \
& 0 & 1 & & \
& & ddots & ddots & \
& & & 0 & 1 \
& & & & 0
end{bmatrix}$$ Then $T^ast$ has only a one-dimensional eigenspace, spanned by $f$ defined by
$$ f(e_i) = begin{cases} 0 & i < n \ 1 & i = n.end{cases}$$
In terms of matrices, we can check this by taking the transpose of the matrix above. Then $f$ is our dual eigenvector and $e_1$ is our eigenvector, which has $f(e_1) neq 0$ if and only if $n = 1$, i.e. $T$ has trivial defect.
answered Mar 23 at 3:23
Joshua MundingerJoshua Mundinger
2,9321028
$endgroup$
The normalization is not deep: just that if $langle u, u^astrangle neq 0$, then $u^ast$ may be rescaled so that that inner product is 1. Asking whether we can find $u^ast$ such that $langle u, u^ast rangle neq 0$ is deeper.
I'm not sure whether you're working with a complex vector space with a Hermitian inner product, or more generally with a linear transformation and its dual. Below, I'll give an exposition of the latter; the former can be obtained by replacing duality with orthogonality, along with an appropriate sprinkling of complex conjugates.
Let $T: V to V$ be an endomorphism of a finite-dimensional vector space $V$ over an algebraically closed field $k$, and let $T^*: V^* to V^*$ denote the dual transformation. Let the Jordan blocks of $T$ acting on $V$ be $V = V_1 oplus ldots oplus V_m$.
Then the action of $T^*$ has Jordan decomposition $V^* = V_1^* oplus cdots oplus V_m^ast$, with the same eigenvalues. Thus, to understand the relationship between the eigenvectors of $T$ and the eigenvectors of $T^ast$, it suffices to analyze just one Jordan block. Further, the eigenvectors of $T$ and $T - lambda , mathrm{id}$ are the same, so it suffices to consider when $T$ has eigenvalue $0$, i.e. is nilpotent.
In the case when $T$ is nilpotent and has one Jordan block, $V$ has a basis ${e_1,ldots, e_n}$ with $T e_i = e_{i-1}$ for $i > 1$ and $Te_1 = 0$. That is, $T$ has matrix
$$ begin{bmatrix} 0 & 1 &&& \
& 0 & 1 & & \
& & ddots & ddots & \
& & & 0 & 1 \
& & & & 0
end{bmatrix}$$ Then $T^ast$ has only a one-dimensional eigenspace, spanned by $f$ defined by
$$ f(e_i) = begin{cases} 0 & i < n \ 1 & i = n.end{cases}$$
In terms of matrices, we can check this by taking the transpose of the matrix above. Then $f$ is our dual eigenvector and $e_1$ is our eigenvector, which has $f(e_1) neq 0$ if and only if $n = 1$, i.e. $T$ has trivial defect.
answered Mar 23 at 3:23
Joshua MundingerJoshua Mundinger
2,9321028
edited Mar 23 at 14:52
answered Mar 23 at 3:23
Joshua MundingerJoshua Mundinger
2,9321028
answered Mar 23 at 3:23
Joshua MundingerJoshua Mundinger
2,9321028
answered Mar 23 at 3:23
Joshua MundingerJoshua Mundinger
2,9321028
2,9321028
$begingroup$
I see that $T^*$ has only one-dimensional eigenspace. But I don’t understand why it is spanned by the eigenfunction f which you have defined.
$endgroup$
– Salamo
Mar 23 at 7:47
$begingroup$
The transpose has kernel spanned by $(0, ldots, 0 ,1)$, which corresponds to the element in the dual basis of ${e_1,ldots, e_n}$ corresponding to $e_n$, which is $f$. Alternatively, you can check that $f circ T = 0$.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:42
$begingroup$
There was also a typo in the definition of $T$, which is now fixed.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:52
$begingroup$
Maybe one last question. Suppose $langle u,u^*rangleneq 0$. You say then it’s no deep thing to scale $u^*$ such that $langle u,u*rangle=1$. Why and how?
$endgroup$
– Salamo
Mar 23 at 19:26
$begingroup$
We have $langle u, cvrangle = c langle u, vrangle$ for all $c in k$, and $v$ is an eigenvector if and only if $cv$ is (for $c neq 0$), so if $v$ is an eigenvector we can set $u^ast = cv$ such that $langle u, u^astrangle = 1$ (what value of $c$ would this have to be?)
$endgroup$
– Joshua Mundinger
Mar 24 at 1:25
add a comment |
$begingroup$
I see that $T^*$ has only one-dimensional eigenspace. But I don’t understand why it is spanned by the eigenfunction f which you have defined.
$endgroup$
– Salamo
Mar 23 at 7:47
$begingroup$
The transpose has kernel spanned by $(0, ldots, 0 ,1)$, which corresponds to the element in the dual basis of ${e_1,ldots, e_n}$ corresponding to $e_n$, which is $f$. Alternatively, you can check that $f circ T = 0$.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:42
$begingroup$
There was also a typo in the definition of $T$, which is now fixed.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:52
$begingroup$
Maybe one last question. Suppose $langle u,u^*rangleneq 0$. You say then it’s no deep thing to scale $u^*$ such that $langle u,u*rangle=1$. Why and how?
$endgroup$
– Salamo
Mar 23 at 19:26
$begingroup$
We have $langle u, cvrangle = c langle u, vrangle$ for all $c in k$, and $v$ is an eigenvector if and only if $cv$ is (for $c neq 0$), so if $v$ is an eigenvector we can set $u^ast = cv$ such that $langle u, u^astrangle = 1$ (what value of $c$ would this have to be?)
$endgroup$
– Joshua Mundinger
Mar 24 at 1:25
$begingroup$
I see that $T^*$ has only one-dimensional eigenspace. But I don’t understand why it is spanned by the eigenfunction f which you have defined.
$endgroup$
– Salamo
Mar 23 at 7:47
$begingroup$
I see that $T^*$ has only one-dimensional eigenspace. But I don’t understand why it is spanned by the eigenfunction f which you have defined.
$endgroup$
– Salamo
Mar 23 at 7:47
$begingroup$
The transpose has kernel spanned by $(0, ldots, 0 ,1)$, which corresponds to the element in the dual basis of ${e_1,ldots, e_n}$ corresponding to $e_n$, which is $f$. Alternatively, you can check that $f circ T = 0$.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:42
$begingroup$
The transpose has kernel spanned by $(0, ldots, 0 ,1)$, which corresponds to the element in the dual basis of ${e_1,ldots, e_n}$ corresponding to $e_n$, which is $f$. Alternatively, you can check that $f circ T = 0$.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:42
$begingroup$
There was also a typo in the definition of $T$, which is now fixed.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:52
$begingroup$
There was also a typo in the definition of $T$, which is now fixed.
$endgroup$
– Joshua Mundinger
Mar 23 at 14:52
$begingroup$
Maybe one last question. Suppose $langle u,u^*rangleneq 0$. You say then it’s no deep thing to scale $u^*$ such that $langle u,u*rangle=1$. Why and how?
$endgroup$
– Salamo
Mar 23 at 19:26
$begingroup$
Maybe one last question. Suppose $langle u,u^*rangleneq 0$. You say then it’s no deep thing to scale $u^*$ such that $langle u,u*rangle=1$. Why and how?
$endgroup$
– Salamo
Mar 23 at 19:26
$begingroup$
We have $langle u, cvrangle = c langle u, vrangle$ for all $c in k$, and $v$ is an eigenvector if and only if $cv$ is (for $c neq 0$), so if $v$ is an eigenvector we can set $u^ast = cv$ such that $langle u, u^astrangle = 1$ (what value of $c$ would this have to be?)
$endgroup$
– Joshua Mundinger
Mar 24 at 1:25
$begingroup$
We have $langle u, cvrangle = c langle u, vrangle$ for all $c in k$, and $v$ is an eigenvector if and only if $cv$ is (for $c neq 0$), so if $v$ is an eigenvector we can set $u^ast = cv$ such that $langle u, u^astrangle = 1$ (what value of $c$ would this have to be?)
$endgroup$
– Joshua Mundinger
Mar 24 at 1:25
add a comment |
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