Mixing
Derivative of the matrix of eigenvalues of a real symmetric matrix
Given a real symmetric matrix $A$ with entries depending on $t$, the derivative $p$-th eigenvalue with respect to $t$ is given by
$$
lambda_p' = v_p^T A'v_p
$$
where $A'$ denotes the derivative of matrix $A$. This can be derived by premultiplying
$$
A' v_p + A v_p' = lambda_p' v_p + lambda_p v_p'
$$
with $v_p^T$, imposing that the eigenvectors have unit length (and thus $v_p cdot v_p' = 0$), and making use of the fact that $A$ is symmetric.
Say I want to use the same approach on the eigendecomposition of $A$
$$
A V = V Lambda \
A'V + A V' = V' Lambda + V Lambda' \
V^T A'V + V^T A V' = V^T V' Lambda + V^T V Lambda' \
V^T A'V + V^T A V' = V^T V' Lambda + Lambda' \
Lambda' = V^T A'V + V^T A V' - V^T V' Lambda \
Lambda' = V^T A'V + Lambda V^T V' - V^T V' Lambda \
$$
also
$$
I' = (V^T V)' = {V^T}'V + V^T V' = 0
$$
However, I don't see how I could use this statement to simplify the earlier expression. I feel like I'm missing something really obvious.
How can I express the derivative of the matrix of eigenvalues $Lambda$ in terms of the derivative of $A$?
linear-algebra matrices derivatives symmetric-matrices
asked Nov 22 '18 at 11:15
user495268user495268
183
|
show 2 more comments
Given a real symmetric matrix $A$ with entries depending on $t$, the derivative $p$-th eigenvalue with respect to $t$ is given by
$$
lambda_p' = v_p^T A'v_p
$$
where $A'$ denotes the derivative of matrix $A$. This can be derived by premultiplying
$$
A' v_p + A v_p' = lambda_p' v_p + lambda_p v_p'
$$
with $v_p^T$, imposing that the eigenvectors have unit length (and thus $v_p cdot v_p' = 0$), and making use of the fact that $A$ is symmetric.
Say I want to use the same approach on the eigendecomposition of $A$
$$
A V = V Lambda \
A'V + A V' = V' Lambda + V Lambda' \
V^T A'V + V^T A V' = V^T V' Lambda + V^T V Lambda' \
V^T A'V + V^T A V' = V^T V' Lambda + Lambda' \
Lambda' = V^T A'V + V^T A V' - V^T V' Lambda \
Lambda' = V^T A'V + Lambda V^T V' - V^T V' Lambda \
$$
also
$$
I' = (V^T V)' = {V^T}'V + V^T V' = 0
$$
However, I don't see how I could use this statement to simplify the earlier expression. I feel like I'm missing something really obvious.
How can I express the derivative of the matrix of eigenvalues $Lambda$ in terms of the derivative of $A$?
linear-algebra matrices derivatives symmetric-matrices
asked Nov 22 '18 at 11:15
user495268user495268
183
Is $A$ a function of $t$, or something like that? I ask because you differentiate, but you do not specify with respect to what. Also "eigenvalues have unit length" makes no sense.
– Giuseppe Negro
Nov 22 '18 at 11:20
Sorry, I edited the typo. Yes, for example the entries of $A$ could be a function of $t$ and we are taking the derivative with respect to $t$.
– user495268
Nov 22 '18 at 11:24
It is much better to specify the dependence on $t$ in the main text. That's an important point.
– Giuseppe Negro
Nov 22 '18 at 11:26
Now I finally understood your question. It is a nice one, +1. I have a semi-serious question; I suppose that $$Lambda'=mathrm{diag}(V_1^TA'V_1, V_2^TA'V_2,ldots, V_n^TA'V_n), $$ where $V_j$ denotes the $j$-th column of $V$, is not the answer you are looking for?
– Giuseppe Negro
Nov 22 '18 at 12:13
Oh, and wait a minute. If you impose that the eigenvectors are orthonormal, as you may since $A$ is symmetric, then $V$ is an orthogonal matrix, that is, $V^TV=I$. Differentiating this you should get something, just like in the beginning of the post you differentiated $v_icdot v_j=delta_{ij}$. (HTH)
– Giuseppe Negro
Nov 22 '18 at 12:15
|
show 2 more comments
Given a real symmetric matrix $A$ with entries depending on $t$, the derivative $p$-th eigenvalue with respect to $t$ is given by
$$
lambda_p' = v_p^T A'v_p
$$
where $A'$ denotes the derivative of matrix $A$. This can be derived by premultiplying
$$
A' v_p + A v_p' = lambda_p' v_p + lambda_p v_p'
$$
with $v_p^T$, imposing that the eigenvectors have unit length (and thus $v_p cdot v_p' = 0$), and making use of the fact that $A$ is symmetric.
Say I want to use the same approach on the eigendecomposition of $A$
$$
A V = V Lambda \
A'V + A V' = V' Lambda + V Lambda' \
V^T A'V + V^T A V' = V^T V' Lambda + V^T V Lambda' \
V^T A'V + V^T A V' = V^T V' Lambda + Lambda' \
Lambda' = V^T A'V + V^T A V' - V^T V' Lambda \
Lambda' = V^T A'V + Lambda V^T V' - V^T V' Lambda \
$$
also
$$
I' = (V^T V)' = {V^T}'V + V^T V' = 0
$$
However, I don't see how I could use this statement to simplify the earlier expression. I feel like I'm missing something really obvious.
How can I express the derivative of the matrix of eigenvalues $Lambda$ in terms of the derivative of $A$?
linear-algebra matrices derivatives symmetric-matrices
asked Nov 22 '18 at 11:15
user495268user495268
183
Given a real symmetric matrix $A$ with entries depending on $t$, the derivative $p$-th eigenvalue with respect to $t$ is given by
$$
lambda_p' = v_p^T A'v_p
$$
where $A'$ denotes the derivative of matrix $A$. This can be derived by premultiplying
$$
A' v_p + A v_p' = lambda_p' v_p + lambda_p v_p'
$$
with $v_p^T$, imposing that the eigenvectors have unit length (and thus $v_p cdot v_p' = 0$), and making use of the fact that $A$ is symmetric.
Say I want to use the same approach on the eigendecomposition of $A$
$$
A V = V Lambda \
A'V + A V' = V' Lambda + V Lambda' \
V^T A'V + V^T A V' = V^T V' Lambda + V^T V Lambda' \
V^T A'V + V^T A V' = V^T V' Lambda + Lambda' \
Lambda' = V^T A'V + V^T A V' - V^T V' Lambda \
Lambda' = V^T A'V + Lambda V^T V' - V^T V' Lambda \
$$
also
$$
I' = (V^T V)' = {V^T}'V + V^T V' = 0
$$
However, I don't see how I could use this statement to simplify the earlier expression. I feel like I'm missing something really obvious.
How can I express the derivative of the matrix of eigenvalues $Lambda$ in terms of the derivative of $A$?
linear-algebra matrices derivatives symmetric-matrices
linear-algebra matrices derivatives symmetric-matrices
asked Nov 22 '18 at 11:15
user495268user495268
183
asked Nov 22 '18 at 11:15
user495268user495268
183
edited Nov 22 '18 at 11:27
user495268
asked Nov 22 '18 at 11:15
user495268user495268
183
asked Nov 22 '18 at 11:15
user495268user495268
183
asked Nov 22 '18 at 11:15
user495268user495268
183
183
Is $A$ a function of $t$, or something like that? I ask because you differentiate, but you do not specify with respect to what. Also "eigenvalues have unit length" makes no sense.
– Giuseppe Negro
Nov 22 '18 at 11:20
Sorry, I edited the typo. Yes, for example the entries of $A$ could be a function of $t$ and we are taking the derivative with respect to $t$.
– user495268
Nov 22 '18 at 11:24
It is much better to specify the dependence on $t$ in the main text. That's an important point.
– Giuseppe Negro
Nov 22 '18 at 11:26
Now I finally understood your question. It is a nice one, +1. I have a semi-serious question; I suppose that $$Lambda'=mathrm{diag}(V_1^TA'V_1, V_2^TA'V_2,ldots, V_n^TA'V_n), $$ where $V_j$ denotes the $j$-th column of $V$, is not the answer you are looking for?
– Giuseppe Negro
Nov 22 '18 at 12:13
Oh, and wait a minute. If you impose that the eigenvectors are orthonormal, as you may since $A$ is symmetric, then $V$ is an orthogonal matrix, that is, $V^TV=I$. Differentiating this you should get something, just like in the beginning of the post you differentiated $v_icdot v_j=delta_{ij}$. (HTH)
– Giuseppe Negro
Nov 22 '18 at 12:15
|
show 2 more comments
Is $A$ a function of $t$, or something like that? I ask because you differentiate, but you do not specify with respect to what. Also "eigenvalues have unit length" makes no sense.
– Giuseppe Negro
Nov 22 '18 at 11:20
Sorry, I edited the typo. Yes, for example the entries of $A$ could be a function of $t$ and we are taking the derivative with respect to $t$.
– user495268
Nov 22 '18 at 11:24
It is much better to specify the dependence on $t$ in the main text. That's an important point.
– Giuseppe Negro
Nov 22 '18 at 11:26
Now I finally understood your question. It is a nice one, +1. I have a semi-serious question; I suppose that $$Lambda'=mathrm{diag}(V_1^TA'V_1, V_2^TA'V_2,ldots, V_n^TA'V_n), $$ where $V_j$ denotes the $j$-th column of $V$, is not the answer you are looking for?
– Giuseppe Negro
Nov 22 '18 at 12:13
Oh, and wait a minute. If you impose that the eigenvectors are orthonormal, as you may since $A$ is symmetric, then $V$ is an orthogonal matrix, that is, $V^TV=I$. Differentiating this you should get something, just like in the beginning of the post you differentiated $v_icdot v_j=delta_{ij}$. (HTH)
– Giuseppe Negro
Nov 22 '18 at 12:15
Is $A$ a function of $t$, or something like that? I ask because you differentiate, but you do not specify with respect to what. Also "eigenvalues have unit length" makes no sense.
– Giuseppe Negro
Nov 22 '18 at 11:20
Is $A$ a function of $t$, or something like that? I ask because you differentiate, but you do not specify with respect to what. Also "eigenvalues have unit length" makes no sense.
– Giuseppe Negro
Nov 22 '18 at 11:20
Sorry, I edited the typo. Yes, for example the entries of $A$ could be a function of $t$ and we are taking the derivative with respect to $t$.
– user495268
Nov 22 '18 at 11:24
Sorry, I edited the typo. Yes, for example the entries of $A$ could be a function of $t$ and we are taking the derivative with respect to $t$.
– user495268
Nov 22 '18 at 11:24
It is much better to specify the dependence on $t$ in the main text. That's an important point.
– Giuseppe Negro
Nov 22 '18 at 11:26
It is much better to specify the dependence on $t$ in the main text. That's an important point.
– Giuseppe Negro
Nov 22 '18 at 11:26
Now I finally understood your question. It is a nice one, +1. I have a semi-serious question; I suppose that $$Lambda'=mathrm{diag}(V_1^TA'V_1, V_2^TA'V_2,ldots, V_n^TA'V_n), $$ where $V_j$ denotes the $j$-th column of $V$, is not the answer you are looking for?
– Giuseppe Negro
Nov 22 '18 at 12:13
Now I finally understood your question. It is a nice one, +1. I have a semi-serious question; I suppose that $$Lambda'=mathrm{diag}(V_1^TA'V_1, V_2^TA'V_2,ldots, V_n^TA'V_n), $$ where $V_j$ denotes the $j$-th column of $V$, is not the answer you are looking for?
– Giuseppe Negro
Nov 22 '18 at 12:13
Oh, and wait a minute. If you impose that the eigenvectors are orthonormal, as you may since $A$ is symmetric, then $V$ is an orthogonal matrix, that is, $V^TV=I$. Differentiating this you should get something, just like in the beginning of the post you differentiated $v_icdot v_j=delta_{ij}$. (HTH)
– Giuseppe Negro
Nov 22 '18 at 12:15
Oh, and wait a minute. If you impose that the eigenvectors are orthonormal, as you may since $A$ is symmetric, then $V$ is an orthogonal matrix, that is, $V^TV=I$. Differentiating this you should get something, just like in the beginning of the post you differentiated $v_icdot v_j=delta_{ij}$. (HTH)
– Giuseppe Negro
Nov 22 '18 at 12:15
|
show 2 more comments
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Is $A$ a function of $t$, or something like that? I ask because you differentiate, but you do not specify with respect to what. Also "eigenvalues have unit length" makes no sense.
– Giuseppe Negro
Nov 22 '18 at 11:20
Sorry, I edited the typo. Yes, for example the entries of $A$ could be a function of $t$ and we are taking the derivative with respect to $t$.
– user495268
Nov 22 '18 at 11:24
It is much better to specify the dependence on $t$ in the main text. That's an important point.
– Giuseppe Negro
Nov 22 '18 at 11:26
Now I finally understood your question. It is a nice one, +1. I have a semi-serious question; I suppose that $$Lambda'=mathrm{diag}(V_1^TA'V_1, V_2^TA'V_2,ldots, V_n^TA'V_n), $$ where $V_j$ denotes the $j$-th column of $V$, is not the answer you are looking for?
– Giuseppe Negro
Nov 22 '18 at 12:13
Oh, and wait a minute. If you impose that the eigenvectors are orthonormal, as you may since $A$ is symmetric, then $V$ is an orthogonal matrix, that is, $V^TV=I$. Differentiating this you should get something, just like in the beginning of the post you differentiated $v_icdot v_j=delta_{ij}$. (HTH)
– Giuseppe Negro
Nov 22 '18 at 12:15