Differentiating an inner product w.r.t matrices
$begingroup$
Let $M_n(mathbb{R})$ denotes the space of all $n times n$ matrices identified with the Euclidean space $mathbb{R^{n^2}}$. Fix a column vector $x neq 0$. Define $f:M_n(mathbb{R}) rightarrowmathbb{R}$ by
begin{equation}
f(A)= langle A^2x,xrangle
end{equation}
I am trying to prove $f$ is differentiable.
It looks like you have to find the derivative w.r.t the matrix $A$. How do I do that? Any help is appreciated.
matrices functional-analysis multivariable-calculus
asked Dec 7 '18 at 1:27
TopologyTopology
770314
$endgroup$
add a comment |
$begingroup$
Let $M_n(mathbb{R})$ denotes the space of all $n times n$ matrices identified with the Euclidean space $mathbb{R^{n^2}}$. Fix a column vector $x neq 0$. Define $f:M_n(mathbb{R}) rightarrowmathbb{R}$ by
begin{equation}
f(A)= langle A^2x,xrangle
end{equation}
I am trying to prove $f$ is differentiable.
It looks like you have to find the derivative w.r.t the matrix $A$. How do I do that? Any help is appreciated.
matrices functional-analysis multivariable-calculus
asked Dec 7 '18 at 1:27
TopologyTopology
770314
$endgroup$
$begingroup$
Just to add another way to think about it. You can show that $f(A)$ is a polynomial in $a_{ij}$. So all $frac{partial f}{partial a_{ij}}$ exist and are continuous, hence, the function has a total derivative by this theorem ( calculus.subwiki.org/wiki/… )
$endgroup$
– user25959
Dec 7 '18 at 2:12
add a comment |
$begingroup$
Let $M_n(mathbb{R})$ denotes the space of all $n times n$ matrices identified with the Euclidean space $mathbb{R^{n^2}}$. Fix a column vector $x neq 0$. Define $f:M_n(mathbb{R}) rightarrowmathbb{R}$ by
begin{equation}
f(A)= langle A^2x,xrangle
end{equation}
I am trying to prove $f$ is differentiable.
It looks like you have to find the derivative w.r.t the matrix $A$. How do I do that? Any help is appreciated.
matrices functional-analysis multivariable-calculus
asked Dec 7 '18 at 1:27
TopologyTopology
770314
$endgroup$
Let $M_n(mathbb{R})$ denotes the space of all $n times n$ matrices identified with the Euclidean space $mathbb{R^{n^2}}$. Fix a column vector $x neq 0$. Define $f:M_n(mathbb{R}) rightarrowmathbb{R}$ by
begin{equation}
f(A)= langle A^2x,xrangle
end{equation}
I am trying to prove $f$ is differentiable.
It looks like you have to find the derivative w.r.t the matrix $A$. How do I do that? Any help is appreciated.
matrices functional-analysis multivariable-calculus
matrices functional-analysis multivariable-calculus
asked Dec 7 '18 at 1:27
TopologyTopology
770314
asked Dec 7 '18 at 1:27
TopologyTopology
770314
asked Dec 7 '18 at 1:27
TopologyTopology
770314
asked Dec 7 '18 at 1:27
TopologyTopology
770314
asked Dec 7 '18 at 1:27
TopologyTopology
770314
770314
$begingroup$
Just to add another way to think about it. You can show that $f(A)$ is a polynomial in $a_{ij}$. So all $frac{partial f}{partial a_{ij}}$ exist and are continuous, hence, the function has a total derivative by this theorem ( calculus.subwiki.org/wiki/… )
$endgroup$
– user25959
Dec 7 '18 at 2:12
add a comment |
$begingroup$
Just to add another way to think about it. You can show that $f(A)$ is a polynomial in $a_{ij}$. So all $frac{partial f}{partial a_{ij}}$ exist and are continuous, hence, the function has a total derivative by this theorem ( calculus.subwiki.org/wiki/… )
$endgroup$
– user25959
Dec 7 '18 at 2:12
$begingroup$
Just to add another way to think about it. You can show that $f(A)$ is a polynomial in $a_{ij}$. So all $frac{partial f}{partial a_{ij}}$ exist and are continuous, hence, the function has a total derivative by this theorem ( calculus.subwiki.org/wiki/… )
$endgroup$
– user25959
Dec 7 '18 at 2:12
$begingroup$
Just to add another way to think about it. You can show that $f(A)$ is a polynomial in $a_{ij}$. So all $frac{partial f}{partial a_{ij}}$ exist and are continuous, hence, the function has a total derivative by this theorem ( calculus.subwiki.org/wiki/… )
$endgroup$
– user25959
Dec 7 '18 at 2:12
add a comment |
1 Answer
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$begingroup$
For bilinear continuous functions $B,$ we have the product rule
$$B'(x,y) cdot (h, k) = B(x,k) + B(h,y).$$
Consider now the bilinear continuous function $theta(u, v) = (u mid v)$ (standard inner product in $mathbf{R}^2$), the bilinear function $varphi(A_1, A_2) = A_1A_2$ (matrix product), the function $psi(N) = (Nx, x),$ whose derivative is $psi'(N) cdot H = (Hx, 0)$ and the function $tau(A) = (A, A),$ which is linear, hence $tau'(A) = tau.$ The chain rule gives the derivative of $f = theta circ psi circ varphi circ tau$ (that is, $f$ is differentiable).
Do you want me to edit and spell-out the derivative?
answered Dec 7 '18 at 2:07
Will M.Will M.
2,505315
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
For bilinear continuous functions $B,$ we have the product rule
$$B'(x,y) cdot (h, k) = B(x,k) + B(h,y).$$
Consider now the bilinear continuous function $theta(u, v) = (u mid v)$ (standard inner product in $mathbf{R}^2$), the bilinear function $varphi(A_1, A_2) = A_1A_2$ (matrix product), the function $psi(N) = (Nx, x),$ whose derivative is $psi'(N) cdot H = (Hx, 0)$ and the function $tau(A) = (A, A),$ which is linear, hence $tau'(A) = tau.$ The chain rule gives the derivative of $f = theta circ psi circ varphi circ tau$ (that is, $f$ is differentiable).
Do you want me to edit and spell-out the derivative?
answered Dec 7 '18 at 2:07
Will M.Will M.
2,505315
$endgroup$
add a comment |
$begingroup$
For bilinear continuous functions $B,$ we have the product rule
$$B'(x,y) cdot (h, k) = B(x,k) + B(h,y).$$
Consider now the bilinear continuous function $theta(u, v) = (u mid v)$ (standard inner product in $mathbf{R}^2$), the bilinear function $varphi(A_1, A_2) = A_1A_2$ (matrix product), the function $psi(N) = (Nx, x),$ whose derivative is $psi'(N) cdot H = (Hx, 0)$ and the function $tau(A) = (A, A),$ which is linear, hence $tau'(A) = tau.$ The chain rule gives the derivative of $f = theta circ psi circ varphi circ tau$ (that is, $f$ is differentiable).
Do you want me to edit and spell-out the derivative?
answered Dec 7 '18 at 2:07
Will M.Will M.
2,505315
$endgroup$
add a comment |
$begingroup$
For bilinear continuous functions $B,$ we have the product rule
$$B'(x,y) cdot (h, k) = B(x,k) + B(h,y).$$
Consider now the bilinear continuous function $theta(u, v) = (u mid v)$ (standard inner product in $mathbf{R}^2$), the bilinear function $varphi(A_1, A_2) = A_1A_2$ (matrix product), the function $psi(N) = (Nx, x),$ whose derivative is $psi'(N) cdot H = (Hx, 0)$ and the function $tau(A) = (A, A),$ which is linear, hence $tau'(A) = tau.$ The chain rule gives the derivative of $f = theta circ psi circ varphi circ tau$ (that is, $f$ is differentiable).
Do you want me to edit and spell-out the derivative?
answered Dec 7 '18 at 2:07
Will M.Will M.
2,505315
$endgroup$
For bilinear continuous functions $B,$ we have the product rule
$$B'(x,y) cdot (h, k) = B(x,k) + B(h,y).$$
Consider now the bilinear continuous function $theta(u, v) = (u mid v)$ (standard inner product in $mathbf{R}^2$), the bilinear function $varphi(A_1, A_2) = A_1A_2$ (matrix product), the function $psi(N) = (Nx, x),$ whose derivative is $psi'(N) cdot H = (Hx, 0)$ and the function $tau(A) = (A, A),$ which is linear, hence $tau'(A) = tau.$ The chain rule gives the derivative of $f = theta circ psi circ varphi circ tau$ (that is, $f$ is differentiable).
Do you want me to edit and spell-out the derivative?
answered Dec 7 '18 at 2:07
Will M.Will M.
2,505315
edited Dec 7 '18 at 2:13
answered Dec 7 '18 at 2:07
Will M.Will M.
2,505315
answered Dec 7 '18 at 2:07
Will M.Will M.
2,505315
answered Dec 7 '18 at 2:07
Will M.Will M.
2,505315
2,505315
add a comment |
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$begingroup$
Just to add another way to think about it. You can show that $f(A)$ is a polynomial in $a_{ij}$. So all $frac{partial f}{partial a_{ij}}$ exist and are continuous, hence, the function has a total derivative by this theorem ( calculus.subwiki.org/wiki/… )
$endgroup$
– user25959
Dec 7 '18 at 2:12