Mixing
Optimizing linear combination of matrices and vectors
$begingroup$
Given the function $G(X,y)$ = $y^T$$X$$y+$$b^T$$b$, where $y$ is a vector in $R^n$, $b$ is a constant vector in $R^n$ and $X$ is a constant symmetric, square, and invertible matrix, what value of $y$ such that $||y||=1$ minimizes the function $G$?
To minimize the function, I first thought to take the partial derivative wrt $y$ and set it equal to 0, so $y^T$$(A^T+A)=0$. But with the restriction of the magnitude of $y$, I'm not sure how to approach the rest. Any help would be appreciated!
linear-algebra matrices optimization matrix-calculus
asked Jan 27 at 0:24
AnthonyAnthony
35519
$endgroup$
add a comment |
$begingroup$
Given the function $G(X,y)$ = $y^T$$X$$y+$$b^T$$b$, where $y$ is a vector in $R^n$, $b$ is a constant vector in $R^n$ and $X$ is a constant symmetric, square, and invertible matrix, what value of $y$ such that $||y||=1$ minimizes the function $G$?
To minimize the function, I first thought to take the partial derivative wrt $y$ and set it equal to 0, so $y^T$$(A^T+A)=0$. But with the restriction of the magnitude of $y$, I'm not sure how to approach the rest. Any help would be appreciated!
linear-algebra matrices optimization matrix-calculus
asked Jan 27 at 0:24
AnthonyAnthony
35519
$endgroup$
add a comment |
$begingroup$
Given the function $G(X,y)$ = $y^T$$X$$y+$$b^T$$b$, where $y$ is a vector in $R^n$, $b$ is a constant vector in $R^n$ and $X$ is a constant symmetric, square, and invertible matrix, what value of $y$ such that $||y||=1$ minimizes the function $G$?
To minimize the function, I first thought to take the partial derivative wrt $y$ and set it equal to 0, so $y^T$$(A^T+A)=0$. But with the restriction of the magnitude of $y$, I'm not sure how to approach the rest. Any help would be appreciated!
linear-algebra matrices optimization matrix-calculus
asked Jan 27 at 0:24
AnthonyAnthony
35519
$endgroup$
Given the function $G(X,y)$ = $y^T$$X$$y+$$b^T$$b$, where $y$ is a vector in $R^n$, $b$ is a constant vector in $R^n$ and $X$ is a constant symmetric, square, and invertible matrix, what value of $y$ such that $||y||=1$ minimizes the function $G$?
To minimize the function, I first thought to take the partial derivative wrt $y$ and set it equal to 0, so $y^T$$(A^T+A)=0$. But with the restriction of the magnitude of $y$, I'm not sure how to approach the rest. Any help would be appreciated!
linear-algebra matrices optimization matrix-calculus
linear-algebra matrices optimization matrix-calculus
asked Jan 27 at 0:24
AnthonyAnthony
35519
asked Jan 27 at 0:24
AnthonyAnthony
35519
asked Jan 27 at 0:24
AnthonyAnthony
35519
asked Jan 27 at 0:24
AnthonyAnthony
35519
asked Jan 27 at 0:24
AnthonyAnthony
35519
35519
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $w$ be an unconstrained vector and define $y$ in terms of it
$$y = frac{w}{sqrt{w^Tw}}$$
Consider the function
$$eqalign{
lambda &= y^TXy = frac{w^TXw}{w^Tw} cr
dlambda &= bigg(frac{2Xw-2lambda w}{w^Tw}bigg)^Tdw cr
frac{partiallambda}{partial w} &= frac{2}{w^Tw}Big(Xw-lambda wBig) cr
}$$
Setting this gradient to zero yields the eigenvalue equation
$$Xw = lambda w$$
So, in terms of the original variables:
$,y$ is simply the (normalize) eigenvector corresponding to the smallest eigenvalue of $X$, and $G=lambda+b^Tb$.
answered Jan 27 at 19:12
greggreg
8,9551824
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add a comment |
$begingroup$
I take it that the problem is
$$
min_{y} quad f(y) = y^T A y \ text{s.t.}quad |y|_2 = 1
tag{1}
$$
for $A$ symmetric, invertible.
If this is correct, then consider using the eigendecomposition of $A$ using the Spectral Theorem. Can you see what we might want $y$ to be? We can disregard $b$ since it is independent of $y$.
As a start, we have $y^{top} A y = sum_{i=1, k=1}^n y_i, left( sum_{j=1}^n lambda_j , [v_j]_i , [v_j]_k right) , y_k$ $= sum_{i=1, j=1, k=1}^n y_j, y_k, lambda_i , [v_i]_j , [v_i]_k$.
Then from the last expression in the equality, we see that if we let $y_j = [v_i]_j$, that is, we let the $j$-th component of $y$ be the $j$-th component of the (unit length) eigenvector $i$, then we recover $lambda_i$. So, to minimize $f$, we can set $y$ equal to the eigenvalue associated with the smallest eigenvector.
answered Jan 27 at 0:41
jjjjjjjjjjjj
1,204516
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $w$ be an unconstrained vector and define $y$ in terms of it
$$y = frac{w}{sqrt{w^Tw}}$$
Consider the function
$$eqalign{
lambda &= y^TXy = frac{w^TXw}{w^Tw} cr
dlambda &= bigg(frac{2Xw-2lambda w}{w^Tw}bigg)^Tdw cr
frac{partiallambda}{partial w} &= frac{2}{w^Tw}Big(Xw-lambda wBig) cr
}$$
Setting this gradient to zero yields the eigenvalue equation
$$Xw = lambda w$$
So, in terms of the original variables:
$,y$ is simply the (normalize) eigenvector corresponding to the smallest eigenvalue of $X$, and $G=lambda+b^Tb$.
answered Jan 27 at 19:12
greggreg
8,9551824
$endgroup$
add a comment |
$begingroup$
Let $w$ be an unconstrained vector and define $y$ in terms of it
$$y = frac{w}{sqrt{w^Tw}}$$
Consider the function
$$eqalign{
lambda &= y^TXy = frac{w^TXw}{w^Tw} cr
dlambda &= bigg(frac{2Xw-2lambda w}{w^Tw}bigg)^Tdw cr
frac{partiallambda}{partial w} &= frac{2}{w^Tw}Big(Xw-lambda wBig) cr
}$$
Setting this gradient to zero yields the eigenvalue equation
$$Xw = lambda w$$
So, in terms of the original variables:
$,y$ is simply the (normalize) eigenvector corresponding to the smallest eigenvalue of $X$, and $G=lambda+b^Tb$.
answered Jan 27 at 19:12
greggreg
8,9551824
$endgroup$
add a comment |
$begingroup$
Let $w$ be an unconstrained vector and define $y$ in terms of it
$$y = frac{w}{sqrt{w^Tw}}$$
Consider the function
$$eqalign{
lambda &= y^TXy = frac{w^TXw}{w^Tw} cr
dlambda &= bigg(frac{2Xw-2lambda w}{w^Tw}bigg)^Tdw cr
frac{partiallambda}{partial w} &= frac{2}{w^Tw}Big(Xw-lambda wBig) cr
}$$
Setting this gradient to zero yields the eigenvalue equation
$$Xw = lambda w$$
So, in terms of the original variables:
$,y$ is simply the (normalize) eigenvector corresponding to the smallest eigenvalue of $X$, and $G=lambda+b^Tb$.
answered Jan 27 at 19:12
greggreg
8,9551824
$endgroup$
Let $w$ be an unconstrained vector and define $y$ in terms of it
$$y = frac{w}{sqrt{w^Tw}}$$
Consider the function
$$eqalign{
lambda &= y^TXy = frac{w^TXw}{w^Tw} cr
dlambda &= bigg(frac{2Xw-2lambda w}{w^Tw}bigg)^Tdw cr
frac{partiallambda}{partial w} &= frac{2}{w^Tw}Big(Xw-lambda wBig) cr
}$$
Setting this gradient to zero yields the eigenvalue equation
$$Xw = lambda w$$
So, in terms of the original variables:
$,y$ is simply the (normalize) eigenvector corresponding to the smallest eigenvalue of $X$, and $G=lambda+b^Tb$.
answered Jan 27 at 19:12
greggreg
8,9551824
answered Jan 27 at 19:12
greggreg
8,9551824
answered Jan 27 at 19:12
greggreg
8,9551824
answered Jan 27 at 19:12
greggreg
8,9551824
8,9551824
add a comment |
add a comment |
$begingroup$
I take it that the problem is
$$
min_{y} quad f(y) = y^T A y \ text{s.t.}quad |y|_2 = 1
tag{1}
$$
for $A$ symmetric, invertible.
If this is correct, then consider using the eigendecomposition of $A$ using the Spectral Theorem. Can you see what we might want $y$ to be? We can disregard $b$ since it is independent of $y$.
As a start, we have $y^{top} A y = sum_{i=1, k=1}^n y_i, left( sum_{j=1}^n lambda_j , [v_j]_i , [v_j]_k right) , y_k$ $= sum_{i=1, j=1, k=1}^n y_j, y_k, lambda_i , [v_i]_j , [v_i]_k$.
Then from the last expression in the equality, we see that if we let $y_j = [v_i]_j$, that is, we let the $j$-th component of $y$ be the $j$-th component of the (unit length) eigenvector $i$, then we recover $lambda_i$. So, to minimize $f$, we can set $y$ equal to the eigenvalue associated with the smallest eigenvector.
answered Jan 27 at 0:41
jjjjjjjjjjjj
1,204516
$endgroup$
add a comment |
$begingroup$
I take it that the problem is
$$
min_{y} quad f(y) = y^T A y \ text{s.t.}quad |y|_2 = 1
tag{1}
$$
for $A$ symmetric, invertible.
If this is correct, then consider using the eigendecomposition of $A$ using the Spectral Theorem. Can you see what we might want $y$ to be? We can disregard $b$ since it is independent of $y$.
As a start, we have $y^{top} A y = sum_{i=1, k=1}^n y_i, left( sum_{j=1}^n lambda_j , [v_j]_i , [v_j]_k right) , y_k$ $= sum_{i=1, j=1, k=1}^n y_j, y_k, lambda_i , [v_i]_j , [v_i]_k$.
Then from the last expression in the equality, we see that if we let $y_j = [v_i]_j$, that is, we let the $j$-th component of $y$ be the $j$-th component of the (unit length) eigenvector $i$, then we recover $lambda_i$. So, to minimize $f$, we can set $y$ equal to the eigenvalue associated with the smallest eigenvector.
answered Jan 27 at 0:41
jjjjjjjjjjjj
1,204516
$endgroup$
add a comment |
$begingroup$
I take it that the problem is
$$
min_{y} quad f(y) = y^T A y \ text{s.t.}quad |y|_2 = 1
tag{1}
$$
for $A$ symmetric, invertible.
If this is correct, then consider using the eigendecomposition of $A$ using the Spectral Theorem. Can you see what we might want $y$ to be? We can disregard $b$ since it is independent of $y$.
As a start, we have $y^{top} A y = sum_{i=1, k=1}^n y_i, left( sum_{j=1}^n lambda_j , [v_j]_i , [v_j]_k right) , y_k$ $= sum_{i=1, j=1, k=1}^n y_j, y_k, lambda_i , [v_i]_j , [v_i]_k$.
Then from the last expression in the equality, we see that if we let $y_j = [v_i]_j$, that is, we let the $j$-th component of $y$ be the $j$-th component of the (unit length) eigenvector $i$, then we recover $lambda_i$. So, to minimize $f$, we can set $y$ equal to the eigenvalue associated with the smallest eigenvector.
answered Jan 27 at 0:41
jjjjjjjjjjjj
1,204516
$endgroup$
I take it that the problem is
$$
min_{y} quad f(y) = y^T A y \ text{s.t.}quad |y|_2 = 1
tag{1}
$$
for $A$ symmetric, invertible.
If this is correct, then consider using the eigendecomposition of $A$ using the Spectral Theorem. Can you see what we might want $y$ to be? We can disregard $b$ since it is independent of $y$.
As a start, we have $y^{top} A y = sum_{i=1, k=1}^n y_i, left( sum_{j=1}^n lambda_j , [v_j]_i , [v_j]_k right) , y_k$ $= sum_{i=1, j=1, k=1}^n y_j, y_k, lambda_i , [v_i]_j , [v_i]_k$.
Then from the last expression in the equality, we see that if we let $y_j = [v_i]_j$, that is, we let the $j$-th component of $y$ be the $j$-th component of the (unit length) eigenvector $i$, then we recover $lambda_i$. So, to minimize $f$, we can set $y$ equal to the eigenvalue associated with the smallest eigenvector.
answered Jan 27 at 0:41
jjjjjjjjjjjj
1,204516
edited Jan 28 at 20:29
answered Jan 27 at 0:41
jjjjjjjjjjjj
1,204516
answered Jan 27 at 0:41
jjjjjjjjjjjj
1,204516
answered Jan 27 at 0:41
jjjjjjjjjjjj
1,204516
1,204516
add a comment |
add a comment |
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