Mixing
Find the maximum value of the function $G(u)=cfrac{u^TA u}{u^Tu}$ over $R^3$ {0}
Find the maximum value of the function of
$G(u)$=$cfrac{u^TAu}{u^Tu}$ over $Bbb R^3setminus 0_3$ ($0_3$ in this case is the zero vector in $Bbb R^3$
we know that $A$=$begin{bmatrix}3 & 2 &3 \2&3&2\2 & 2&3 end{bmatrix}$
To solve this, first I let $G(u)$=$begin{bmatrix}x\y\zend{bmatrix}$，
so $u^TAu$ = $begin{bmatrix}x&y&zend{bmatrix}$$begin{bmatrix}3 & 2 &3 \2&3&2\2 & 2&3 end{bmatrix}$$begin{bmatrix}x\y\zend{bmatrix}$
but next I don't know how to do. What is $u^T$ and $u$ in this question?
linear-algebra matrices eigenvalues-eigenvectors vectors
edited Nov 22 '18 at 7:44
Jean-Claude Arbaut
14.7k63464
asked Nov 20 '18 at 5:55
Anqi Luo
495
add a comment |
Find the maximum value of the function of
$G(u)$=$cfrac{u^TAu}{u^Tu}$ over $Bbb R^3setminus 0_3$ ($0_3$ in this case is the zero vector in $Bbb R^3$
we know that $A$=$begin{bmatrix}3 & 2 &3 \2&3&2\2 & 2&3 end{bmatrix}$
To solve this, first I let $G(u)$=$begin{bmatrix}x\y\zend{bmatrix}$，
so $u^TAu$ = $begin{bmatrix}x&y&zend{bmatrix}$$begin{bmatrix}3 & 2 &3 \2&3&2\2 & 2&3 end{bmatrix}$$begin{bmatrix}x\y\zend{bmatrix}$
but next I don't know how to do. What is $u^T$ and $u$ in this question?
linear-algebra matrices eigenvalues-eigenvectors vectors
edited Nov 22 '18 at 7:44
Jean-Claude Arbaut
14.7k63464
asked Nov 20 '18 at 5:55
Anqi Luo
495
The numerator of $G(u)$ should be $u^TAu$, no? $G(u)$ is a number, not a vector. You can let $u$ be the vector $[x y z]^T$. It is some vector in $Bbb R^3$. Determine the eigenvalues and eigenvectors of $A$. Why does this help?
– Ross Millikan
Nov 20 '18 at 6:05
What do you know about the eigenvalues of a symmetric matrix and their relationship to the quadratic form defined by that matrix?
– amd
Nov 21 '18 at 1:11
@amd: As currently written in the question, $A$ is not symmetric since $A_{3,1} = 2 neq 3 = A_{1,3}$. However, I do suspect that the OP intended for it to be symmetric.
– JimmyK4542
Nov 22 '18 at 8:00
@JimmyK4542 Only the symmetric part of $A$ contributes to the quadratic form.
– amd
Nov 22 '18 at 9:28
add a comment |
Find the maximum value of the function of
$G(u)$=$cfrac{u^TAu}{u^Tu}$ over $Bbb R^3setminus 0_3$ ($0_3$ in this case is the zero vector in $Bbb R^3$
we know that $A$=$begin{bmatrix}3 & 2 &3 \2&3&2\2 & 2&3 end{bmatrix}$
To solve this, first I let $G(u)$=$begin{bmatrix}x\y\zend{bmatrix}$，
so $u^TAu$ = $begin{bmatrix}x&y&zend{bmatrix}$$begin{bmatrix}3 & 2 &3 \2&3&2\2 & 2&3 end{bmatrix}$$begin{bmatrix}x\y\zend{bmatrix}$
but next I don't know how to do. What is $u^T$ and $u$ in this question?
linear-algebra matrices eigenvalues-eigenvectors vectors
edited Nov 22 '18 at 7:44
Jean-Claude Arbaut
14.7k63464
asked Nov 20 '18 at 5:55
Anqi Luo
495
Find the maximum value of the function of
$G(u)$=$cfrac{u^TAu}{u^Tu}$ over $Bbb R^3setminus 0_3$ ($0_3$ in this case is the zero vector in $Bbb R^3$
we know that $A$=$begin{bmatrix}3 & 2 &3 \2&3&2\2 & 2&3 end{bmatrix}$
To solve this, first I let $G(u)$=$begin{bmatrix}x\y\zend{bmatrix}$，
so $u^TAu$ = $begin{bmatrix}x&y&zend{bmatrix}$$begin{bmatrix}3 & 2 &3 \2&3&2\2 & 2&3 end{bmatrix}$$begin{bmatrix}x\y\zend{bmatrix}$
but next I don't know how to do. What is $u^T$ and $u$ in this question?
linear-algebra matrices eigenvalues-eigenvectors vectors
linear-algebra matrices eigenvalues-eigenvectors vectors
edited Nov 22 '18 at 7:44
Jean-Claude Arbaut
14.7k63464
asked Nov 20 '18 at 5:55
Anqi Luo
495
edited Nov 22 '18 at 7:44
Jean-Claude Arbaut
14.7k63464
asked Nov 20 '18 at 5:55
Anqi Luo
495
edited Nov 22 '18 at 7:44
Jean-Claude Arbaut
14.7k63464
edited Nov 22 '18 at 7:44
Jean-Claude Arbaut
14.7k63464
edited Nov 22 '18 at 7:44
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked Nov 20 '18 at 5:55
Anqi Luo
495
asked Nov 20 '18 at 5:55
Anqi Luo
495
asked Nov 20 '18 at 5:55
Anqi Luo
495
495
The numerator of $G(u)$ should be $u^TAu$, no? $G(u)$ is a number, not a vector. You can let $u$ be the vector $[x y z]^T$. It is some vector in $Bbb R^3$. Determine the eigenvalues and eigenvectors of $A$. Why does this help?
– Ross Millikan
Nov 20 '18 at 6:05
What do you know about the eigenvalues of a symmetric matrix and their relationship to the quadratic form defined by that matrix?
– amd
Nov 21 '18 at 1:11
@amd: As currently written in the question, $A$ is not symmetric since $A_{3,1} = 2 neq 3 = A_{1,3}$. However, I do suspect that the OP intended for it to be symmetric.
– JimmyK4542
Nov 22 '18 at 8:00
@JimmyK4542 Only the symmetric part of $A$ contributes to the quadratic form.
– amd
Nov 22 '18 at 9:28
add a comment |
The numerator of $G(u)$ should be $u^TAu$, no? $G(u)$ is a number, not a vector. You can let $u$ be the vector $[x y z]^T$. It is some vector in $Bbb R^3$. Determine the eigenvalues and eigenvectors of $A$. Why does this help?
– Ross Millikan
Nov 20 '18 at 6:05
What do you know about the eigenvalues of a symmetric matrix and their relationship to the quadratic form defined by that matrix?
– amd
Nov 21 '18 at 1:11
@amd: As currently written in the question, $A$ is not symmetric since $A_{3,1} = 2 neq 3 = A_{1,3}$. However, I do suspect that the OP intended for it to be symmetric.
– JimmyK4542
Nov 22 '18 at 8:00
@JimmyK4542 Only the symmetric part of $A$ contributes to the quadratic form.
– amd
Nov 22 '18 at 9:28
The numerator of $G(u)$ should be $u^TAu$, no? $G(u)$ is a number, not a vector. You can let $u$ be the vector $[x y z]^T$. It is some vector in $Bbb R^3$. Determine the eigenvalues and eigenvectors of $A$. Why does this help?
– Ross Millikan
Nov 20 '18 at 6:05
The numerator of $G(u)$ should be $u^TAu$, no? $G(u)$ is a number, not a vector. You can let $u$ be the vector $[x y z]^T$. It is some vector in $Bbb R^3$. Determine the eigenvalues and eigenvectors of $A$. Why does this help?
– Ross Millikan
Nov 20 '18 at 6:05
What do you know about the eigenvalues of a symmetric matrix and their relationship to the quadratic form defined by that matrix?
– amd
Nov 21 '18 at 1:11
What do you know about the eigenvalues of a symmetric matrix and their relationship to the quadratic form defined by that matrix?
– amd
Nov 21 '18 at 1:11
@amd: As currently written in the question, $A$ is not symmetric since $A_{3,1} = 2 neq 3 = A_{1,3}$. However, I do suspect that the OP intended for it to be symmetric.
– JimmyK4542
Nov 22 '18 at 8:00
@amd: As currently written in the question, $A$ is not symmetric since $A_{3,1} = 2 neq 3 = A_{1,3}$. However, I do suspect that the OP intended for it to be symmetric.
– JimmyK4542
Nov 22 '18 at 8:00
@JimmyK4542 Only the symmetric part of $A$ contributes to the quadratic form.
– amd
Nov 22 '18 at 9:28
@JimmyK4542 Only the symmetric part of $A$ contributes to the quadratic form.
– amd
Nov 22 '18 at 9:28
add a comment |
4 Answers
4
active
oldest
votes
Let's find a solution to the optimisation problem:
maximise $Phi(u) = u^TAu $ subject to $u^Tu=1$ for $u in mathbb{R}^3$ and some matrix $A$.
For this constrained optimisation problem, one can write the Lagrangian as:
$L(u,lambda) = u^TAu - lambda(u^Tu-1)$
For optimality,
$frac{partial}{partial u}L(u,lambda) = 2Au - 2lambda u = 0$, implying $Au = lambda u$.
$frac{partial}{partial lambda} L(u,lambda) = 1-u^Tu = 0$, implying $u^Tu=1$.
In other words, one of the solution(s) to $Au = lambda u$ subject to $u^Tu = 1$, i.e. one of the normalised eigenvectors of $A$ will maximise $Phi(u)$, and ultimately, $g(u)$.
answered Nov 20 '18 at 6:33
Aditya Dua
80418
add a comment |
Consider the slightly more general problem involving two symmetric matrices.
$$max bigg(frac{u^TAu}{u^TBu}bigg)$$
Define scalar variables for the numerator and denominator.
$$eqalign{
alpha &= u^TAu implies dalpha = 2(Au)^Tdu cr
beta &= u^TBu implies dbeta = 2(Bu)^Tdu cr
}$$
Write the quadratic form in terms of these variables, then find its differential and gradient.
$$eqalign{
lambda &= beta^{-1}alpha cr
dlambda
&= beta^{-2}(beta,dalpha-alpha,dbeta)
&= 2beta^{-2}(beta Au-alpha Bu)^Tdu cr
frac{partiallambda}{partial u} &= 2beta^{-2}(beta Au-alpha Bu) cr
}$$
Setting this gradient to zero leads to an eigenvalue equation.
$$eqalign{
beta Au &= alpha Bu cr
(B^{-1}A)u &= (beta^{-1}alpha)u cr
Lu &= lambda u cr
}$$
So the min/max value of the function $lambda$ is the min/max eigenvalue of $(B^{-1}A)$.
For your particular problem, $,,B=I,$ and $,lambda=G(u)$.
In your problem statement, I see that $A$ is not symmetric. If that's not a typo, then in the above derivation, replace the matrix with its symmetric part: $,Arightarrowbig(frac{A+A^T}{2}big)$
For the symmetrized matrix, $,lambda_{max}=7.3423292$
answered Nov 22 '18 at 5:36
greg
7,5251821
add a comment |
u and $u^T$ are the vectors $begin{bmatrix}x\y\z end{bmatrix}$ and $begin{bmatrix}x&y&z end{bmatrix}$. The function G(u) is a quadratic form, but are you sure about the restriction?
answered Nov 20 '18 at 6:28
Bayesian guy
47110
add a comment |
If $u_0$ is an answer to this problem, then so is $ku_0$ for $kne 0$. Therefore this problem can be converted to $$max x^TAx\s.t. \x^Tx=1$$using Lagrange's multiplier's method we obtain:$$2Ax=2lambda x$$which by substitution leads to $$x^TAx=x^Tlambda x=lambda x^Tx=lambda$$therefore $$max_{x^Tx=1} x^T Ax=max_{lambda text{ is an eigenvalue of }A}lambda=lambda_{max{ }}$$
answered Nov 22 '18 at 7:41
Mostafa Ayaz
13.6k3836
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let's find a solution to the optimisation problem:
maximise $Phi(u) = u^TAu $ subject to $u^Tu=1$ for $u in mathbb{R}^3$ and some matrix $A$.
For this constrained optimisation problem, one can write the Lagrangian as:
$L(u,lambda) = u^TAu - lambda(u^Tu-1)$
For optimality,
$frac{partial}{partial u}L(u,lambda) = 2Au - 2lambda u = 0$, implying $Au = lambda u$.
$frac{partial}{partial lambda} L(u,lambda) = 1-u^Tu = 0$, implying $u^Tu=1$.
In other words, one of the solution(s) to $Au = lambda u$ subject to $u^Tu = 1$, i.e. one of the normalised eigenvectors of $A$ will maximise $Phi(u)$, and ultimately, $g(u)$.
answered Nov 20 '18 at 6:33
Aditya Dua
80418
add a comment |
Let's find a solution to the optimisation problem:
maximise $Phi(u) = u^TAu $ subject to $u^Tu=1$ for $u in mathbb{R}^3$ and some matrix $A$.
For this constrained optimisation problem, one can write the Lagrangian as:
$L(u,lambda) = u^TAu - lambda(u^Tu-1)$
For optimality,
$frac{partial}{partial u}L(u,lambda) = 2Au - 2lambda u = 0$, implying $Au = lambda u$.
$frac{partial}{partial lambda} L(u,lambda) = 1-u^Tu = 0$, implying $u^Tu=1$.
In other words, one of the solution(s) to $Au = lambda u$ subject to $u^Tu = 1$, i.e. one of the normalised eigenvectors of $A$ will maximise $Phi(u)$, and ultimately, $g(u)$.
answered Nov 20 '18 at 6:33
Aditya Dua
80418
add a comment |
Let's find a solution to the optimisation problem:
maximise $Phi(u) = u^TAu $ subject to $u^Tu=1$ for $u in mathbb{R}^3$ and some matrix $A$.
For this constrained optimisation problem, one can write the Lagrangian as:
$L(u,lambda) = u^TAu - lambda(u^Tu-1)$
For optimality,
$frac{partial}{partial u}L(u,lambda) = 2Au - 2lambda u = 0$, implying $Au = lambda u$.
$frac{partial}{partial lambda} L(u,lambda) = 1-u^Tu = 0$, implying $u^Tu=1$.
In other words, one of the solution(s) to $Au = lambda u$ subject to $u^Tu = 1$, i.e. one of the normalised eigenvectors of $A$ will maximise $Phi(u)$, and ultimately, $g(u)$.
answered Nov 20 '18 at 6:33
Aditya Dua
80418
Let's find a solution to the optimisation problem:
maximise $Phi(u) = u^TAu $ subject to $u^Tu=1$ for $u in mathbb{R}^3$ and some matrix $A$.
For this constrained optimisation problem, one can write the Lagrangian as:
$L(u,lambda) = u^TAu - lambda(u^Tu-1)$
For optimality,
$frac{partial}{partial u}L(u,lambda) = 2Au - 2lambda u = 0$, implying $Au = lambda u$.
$frac{partial}{partial lambda} L(u,lambda) = 1-u^Tu = 0$, implying $u^Tu=1$.
In other words, one of the solution(s) to $Au = lambda u$ subject to $u^Tu = 1$, i.e. one of the normalised eigenvectors of $A$ will maximise $Phi(u)$, and ultimately, $g(u)$.
answered Nov 20 '18 at 6:33
Aditya Dua
80418
answered Nov 20 '18 at 6:33
Aditya Dua
80418
answered Nov 20 '18 at 6:33
Aditya Dua
80418
answered Nov 20 '18 at 6:33
Aditya Dua
80418
80418
add a comment |
add a comment |
Consider the slightly more general problem involving two symmetric matrices.
$$max bigg(frac{u^TAu}{u^TBu}bigg)$$
Define scalar variables for the numerator and denominator.
$$eqalign{
alpha &= u^TAu implies dalpha = 2(Au)^Tdu cr
beta &= u^TBu implies dbeta = 2(Bu)^Tdu cr
}$$
Write the quadratic form in terms of these variables, then find its differential and gradient.
$$eqalign{
lambda &= beta^{-1}alpha cr
dlambda
&= beta^{-2}(beta,dalpha-alpha,dbeta)
&= 2beta^{-2}(beta Au-alpha Bu)^Tdu cr
frac{partiallambda}{partial u} &= 2beta^{-2}(beta Au-alpha Bu) cr
}$$
Setting this gradient to zero leads to an eigenvalue equation.
$$eqalign{
beta Au &= alpha Bu cr
(B^{-1}A)u &= (beta^{-1}alpha)u cr
Lu &= lambda u cr
}$$
So the min/max value of the function $lambda$ is the min/max eigenvalue of $(B^{-1}A)$.
For your particular problem, $,,B=I,$ and $,lambda=G(u)$.
In your problem statement, I see that $A$ is not symmetric. If that's not a typo, then in the above derivation, replace the matrix with its symmetric part: $,Arightarrowbig(frac{A+A^T}{2}big)$
For the symmetrized matrix, $,lambda_{max}=7.3423292$
answered Nov 22 '18 at 5:36
greg
7,5251821
add a comment |
Consider the slightly more general problem involving two symmetric matrices.
$$max bigg(frac{u^TAu}{u^TBu}bigg)$$
Define scalar variables for the numerator and denominator.
$$eqalign{
alpha &= u^TAu implies dalpha = 2(Au)^Tdu cr
beta &= u^TBu implies dbeta = 2(Bu)^Tdu cr
}$$
Write the quadratic form in terms of these variables, then find its differential and gradient.
$$eqalign{
lambda &= beta^{-1}alpha cr
dlambda
&= beta^{-2}(beta,dalpha-alpha,dbeta)
&= 2beta^{-2}(beta Au-alpha Bu)^Tdu cr
frac{partiallambda}{partial u} &= 2beta^{-2}(beta Au-alpha Bu) cr
}$$
Setting this gradient to zero leads to an eigenvalue equation.
$$eqalign{
beta Au &= alpha Bu cr
(B^{-1}A)u &= (beta^{-1}alpha)u cr
Lu &= lambda u cr
}$$
So the min/max value of the function $lambda$ is the min/max eigenvalue of $(B^{-1}A)$.
For your particular problem, $,,B=I,$ and $,lambda=G(u)$.
In your problem statement, I see that $A$ is not symmetric. If that's not a typo, then in the above derivation, replace the matrix with its symmetric part: $,Arightarrowbig(frac{A+A^T}{2}big)$
For the symmetrized matrix, $,lambda_{max}=7.3423292$
answered Nov 22 '18 at 5:36
greg
7,5251821
add a comment |
Consider the slightly more general problem involving two symmetric matrices.
$$max bigg(frac{u^TAu}{u^TBu}bigg)$$
Define scalar variables for the numerator and denominator.
$$eqalign{
alpha &= u^TAu implies dalpha = 2(Au)^Tdu cr
beta &= u^TBu implies dbeta = 2(Bu)^Tdu cr
}$$
Write the quadratic form in terms of these variables, then find its differential and gradient.
$$eqalign{
lambda &= beta^{-1}alpha cr
dlambda
&= beta^{-2}(beta,dalpha-alpha,dbeta)
&= 2beta^{-2}(beta Au-alpha Bu)^Tdu cr
frac{partiallambda}{partial u} &= 2beta^{-2}(beta Au-alpha Bu) cr
}$$
Setting this gradient to zero leads to an eigenvalue equation.
$$eqalign{
beta Au &= alpha Bu cr
(B^{-1}A)u &= (beta^{-1}alpha)u cr
Lu &= lambda u cr
}$$
So the min/max value of the function $lambda$ is the min/max eigenvalue of $(B^{-1}A)$.
For your particular problem, $,,B=I,$ and $,lambda=G(u)$.
In your problem statement, I see that $A$ is not symmetric. If that's not a typo, then in the above derivation, replace the matrix with its symmetric part: $,Arightarrowbig(frac{A+A^T}{2}big)$
For the symmetrized matrix, $,lambda_{max}=7.3423292$
answered Nov 22 '18 at 5:36
greg
7,5251821
Consider the slightly more general problem involving two symmetric matrices.
$$max bigg(frac{u^TAu}{u^TBu}bigg)$$
Define scalar variables for the numerator and denominator.
$$eqalign{
alpha &= u^TAu implies dalpha = 2(Au)^Tdu cr
beta &= u^TBu implies dbeta = 2(Bu)^Tdu cr
}$$
Write the quadratic form in terms of these variables, then find its differential and gradient.
$$eqalign{
lambda &= beta^{-1}alpha cr
dlambda
&= beta^{-2}(beta,dalpha-alpha,dbeta)
&= 2beta^{-2}(beta Au-alpha Bu)^Tdu cr
frac{partiallambda}{partial u} &= 2beta^{-2}(beta Au-alpha Bu) cr
}$$
Setting this gradient to zero leads to an eigenvalue equation.
$$eqalign{
beta Au &= alpha Bu cr
(B^{-1}A)u &= (beta^{-1}alpha)u cr
Lu &= lambda u cr
}$$
So the min/max value of the function $lambda$ is the min/max eigenvalue of $(B^{-1}A)$.
For your particular problem, $,,B=I,$ and $,lambda=G(u)$.
In your problem statement, I see that $A$ is not symmetric. If that's not a typo, then in the above derivation, replace the matrix with its symmetric part: $,Arightarrowbig(frac{A+A^T}{2}big)$
For the symmetrized matrix, $,lambda_{max}=7.3423292$
answered Nov 22 '18 at 5:36
greg
7,5251821
edited Nov 22 '18 at 6:00
answered Nov 22 '18 at 5:36
greg
7,5251821
answered Nov 22 '18 at 5:36
greg
7,5251821
answered Nov 22 '18 at 5:36
greg
7,5251821
7,5251821
add a comment |
add a comment |
u and $u^T$ are the vectors $begin{bmatrix}x\y\z end{bmatrix}$ and $begin{bmatrix}x&y&z end{bmatrix}$. The function G(u) is a quadratic form, but are you sure about the restriction?
answered Nov 20 '18 at 6:28
Bayesian guy
47110
add a comment |
u and $u^T$ are the vectors $begin{bmatrix}x\y\z end{bmatrix}$ and $begin{bmatrix}x&y&z end{bmatrix}$. The function G(u) is a quadratic form, but are you sure about the restriction?
answered Nov 20 '18 at 6:28
Bayesian guy
47110
add a comment |
u and $u^T$ are the vectors $begin{bmatrix}x\y\z end{bmatrix}$ and $begin{bmatrix}x&y&z end{bmatrix}$. The function G(u) is a quadratic form, but are you sure about the restriction?
answered Nov 20 '18 at 6:28
Bayesian guy
47110
u and $u^T$ are the vectors $begin{bmatrix}x\y\z end{bmatrix}$ and $begin{bmatrix}x&y&z end{bmatrix}$. The function G(u) is a quadratic form, but are you sure about the restriction?
answered Nov 20 '18 at 6:28
Bayesian guy
47110
answered Nov 20 '18 at 6:28
Bayesian guy
47110
answered Nov 20 '18 at 6:28
Bayesian guy
47110
answered Nov 20 '18 at 6:28
Bayesian guy
47110
47110
add a comment |
add a comment |
If $u_0$ is an answer to this problem, then so is $ku_0$ for $kne 0$. Therefore this problem can be converted to $$max x^TAx\s.t. \x^Tx=1$$using Lagrange's multiplier's method we obtain:$$2Ax=2lambda x$$which by substitution leads to $$x^TAx=x^Tlambda x=lambda x^Tx=lambda$$therefore $$max_{x^Tx=1} x^T Ax=max_{lambda text{ is an eigenvalue of }A}lambda=lambda_{max{ }}$$
answered Nov 22 '18 at 7:41
Mostafa Ayaz
13.6k3836
add a comment |
If $u_0$ is an answer to this problem, then so is $ku_0$ for $kne 0$. Therefore this problem can be converted to $$max x^TAx\s.t. \x^Tx=1$$using Lagrange's multiplier's method we obtain:$$2Ax=2lambda x$$which by substitution leads to $$x^TAx=x^Tlambda x=lambda x^Tx=lambda$$therefore $$max_{x^Tx=1} x^T Ax=max_{lambda text{ is an eigenvalue of }A}lambda=lambda_{max{ }}$$
answered Nov 22 '18 at 7:41
Mostafa Ayaz
13.6k3836
add a comment |
If $u_0$ is an answer to this problem, then so is $ku_0$ for $kne 0$. Therefore this problem can be converted to $$max x^TAx\s.t. \x^Tx=1$$using Lagrange's multiplier's method we obtain:$$2Ax=2lambda x$$which by substitution leads to $$x^TAx=x^Tlambda x=lambda x^Tx=lambda$$therefore $$max_{x^Tx=1} x^T Ax=max_{lambda text{ is an eigenvalue of }A}lambda=lambda_{max{ }}$$
answered Nov 22 '18 at 7:41
Mostafa Ayaz
13.6k3836
If $u_0$ is an answer to this problem, then so is $ku_0$ for $kne 0$. Therefore this problem can be converted to $$max x^TAx\s.t. \x^Tx=1$$using Lagrange's multiplier's method we obtain:$$2Ax=2lambda x$$which by substitution leads to $$x^TAx=x^Tlambda x=lambda x^Tx=lambda$$therefore $$max_{x^Tx=1} x^T Ax=max_{lambda text{ is an eigenvalue of }A}lambda=lambda_{max{ }}$$
answered Nov 22 '18 at 7:41
Mostafa Ayaz
13.6k3836
answered Nov 22 '18 at 7:41
Mostafa Ayaz
13.6k3836
answered Nov 22 '18 at 7:41
Mostafa Ayaz
13.6k3836
answered Nov 22 '18 at 7:41
Mostafa Ayaz
13.6k3836
13.6k3836
add a comment |
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The numerator of $G(u)$ should be $u^TAu$, no? $G(u)$ is a number, not a vector. You can let $u$ be the vector $[x y z]^T$. It is some vector in $Bbb R^3$. Determine the eigenvalues and eigenvectors of $A$. Why does this help?
– Ross Millikan
Nov 20 '18 at 6:05
What do you know about the eigenvalues of a symmetric matrix and their relationship to the quadratic form defined by that matrix?
– amd
Nov 21 '18 at 1:11
@amd: As currently written in the question, $A$ is not symmetric since $A_{3,1} = 2 neq 3 = A_{1,3}$. However, I do suspect that the OP intended for it to be symmetric.
– JimmyK4542
Nov 22 '18 at 8:00
@JimmyK4542 Only the symmetric part of $A$ contributes to the quadratic form.
– amd
Nov 22 '18 at 9:28