Prove this matrix inequality
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I am reading a math paper where a proof uses the following result without proof. I have been stuck on this part for various days, and have talked to others about it but still can't figure out why it is true. There is no assumption that V is normal. I have tried change of variable, but this has not worked. ANy hints or a solution woul dbe appreciated!
Let $W$ be $n$ by $n$ real symmetric positive definite matrix, let $V$ be an invertible $n$ by $n$ square matrix. Then the matrix $V^TWV$ is also symetric and positive definite. Now let $|V|$ denote the largest singular value of $V$. That is, the square root of the largest eigenvalue of $V^TV$.
Then, show that:
begin{equation}
min_{|x| = 1} x^TVWV^Tx leq |V|^2 min_{|y| = 1} y^TWy
end{equation}
linear-algebra linear-transformations positive-definite symmetric-matrices
asked Nov 23 at 2:43
jmsac
9611
add a comment |
up vote
1
down vote
favorite
I am reading a math paper where a proof uses the following result without proof. I have been stuck on this part for various days, and have talked to others about it but still can't figure out why it is true. There is no assumption that V is normal. I have tried change of variable, but this has not worked. ANy hints or a solution woul dbe appreciated!
Let $W$ be $n$ by $n$ real symmetric positive definite matrix, let $V$ be an invertible $n$ by $n$ square matrix. Then the matrix $V^TWV$ is also symetric and positive definite. Now let $|V|$ denote the largest singular value of $V$. That is, the square root of the largest eigenvalue of $V^TV$.
Then, show that:
begin{equation}
min_{|x| = 1} x^TVWV^Tx leq |V|^2 min_{|y| = 1} y^TWy
end{equation}
linear-algebra linear-transformations positive-definite symmetric-matrices
asked Nov 23 at 2:43
jmsac
9611
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading a math paper where a proof uses the following result without proof. I have been stuck on this part for various days, and have talked to others about it but still can't figure out why it is true. There is no assumption that V is normal. I have tried change of variable, but this has not worked. ANy hints or a solution woul dbe appreciated!
Let $W$ be $n$ by $n$ real symmetric positive definite matrix, let $V$ be an invertible $n$ by $n$ square matrix. Then the matrix $V^TWV$ is also symetric and positive definite. Now let $|V|$ denote the largest singular value of $V$. That is, the square root of the largest eigenvalue of $V^TV$.
Then, show that:
begin{equation}
min_{|x| = 1} x^TVWV^Tx leq |V|^2 min_{|y| = 1} y^TWy
end{equation}
linear-algebra linear-transformations positive-definite symmetric-matrices
asked Nov 23 at 2:43
jmsac
9611
I am reading a math paper where a proof uses the following result without proof. I have been stuck on this part for various days, and have talked to others about it but still can't figure out why it is true. There is no assumption that V is normal. I have tried change of variable, but this has not worked. ANy hints or a solution woul dbe appreciated!
Let $W$ be $n$ by $n$ real symmetric positive definite matrix, let $V$ be an invertible $n$ by $n$ square matrix. Then the matrix $V^TWV$ is also symetric and positive definite. Now let $|V|$ denote the largest singular value of $V$. That is, the square root of the largest eigenvalue of $V^TV$.
Then, show that:
begin{equation}
min_{|x| = 1} x^TVWV^Tx leq |V|^2 min_{|y| = 1} y^TWy
end{equation}
linear-algebra linear-transformations positive-definite symmetric-matrices
linear-algebra linear-transformations positive-definite symmetric-matrices
asked Nov 23 at 2:43
jmsac
9611
asked Nov 23 at 2:43
jmsac
9611
asked Nov 23 at 2:43
jmsac
9611
asked Nov 23 at 2:43
jmsac
9611
asked Nov 23 at 2:43
jmsac
9611
9611
add a comment |
add a comment |
1 Answer
1
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up vote
2
down vote
accepted
Fix $x$ such that $|x| = 1$.
Let $y = V^top x / |V^top x|$. Note that $|y| = 1$ and that $|V^top x| le |V^top| = |V|$.
Then
$$x^top V W V^top x = |V^top x|^2 cdot y^top W y le |V|^2 cdot y^top W y.$$
answered Nov 23 at 2:57
angryavian
38k23180
Why does $|V^{T}x| leq |V^T|$ Hold? The second is the largest e value of $V^TV$ not of $V^T$
– jmsac
Nov 23 at 3:06
Sorry the second quantity is the square root of the largest e-value of $V^TV$
– jmsac
Nov 23 at 3:20
@jmsac $|V^top x| = sqrt{x^top VV^top x} le lambda_{max}(VV^top) = lambda_{max}(V^top V)$
– angryavian
Nov 23 at 3:21
Ahh ok thank you!
– jmsac
Nov 23 at 3:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Fix $x$ such that $|x| = 1$.
Let $y = V^top x / |V^top x|$. Note that $|y| = 1$ and that $|V^top x| le |V^top| = |V|$.
Then
$$x^top V W V^top x = |V^top x|^2 cdot y^top W y le |V|^2 cdot y^top W y.$$
answered Nov 23 at 2:57
angryavian
38k23180
Why does $|V^{T}x| leq |V^T|$ Hold? The second is the largest e value of $V^TV$ not of $V^T$
– jmsac
Nov 23 at 3:06
Sorry the second quantity is the square root of the largest e-value of $V^TV$
– jmsac
Nov 23 at 3:20
@jmsac $|V^top x| = sqrt{x^top VV^top x} le lambda_{max}(VV^top) = lambda_{max}(V^top V)$
– angryavian
Nov 23 at 3:21
Ahh ok thank you!
– jmsac
Nov 23 at 3:23
add a comment |
up vote
2
down vote
accepted
Fix $x$ such that $|x| = 1$.
Let $y = V^top x / |V^top x|$. Note that $|y| = 1$ and that $|V^top x| le |V^top| = |V|$.
Then
$$x^top V W V^top x = |V^top x|^2 cdot y^top W y le |V|^2 cdot y^top W y.$$
answered Nov 23 at 2:57
angryavian
38k23180
Why does $|V^{T}x| leq |V^T|$ Hold? The second is the largest e value of $V^TV$ not of $V^T$
– jmsac
Nov 23 at 3:06
Sorry the second quantity is the square root of the largest e-value of $V^TV$
– jmsac
Nov 23 at 3:20
@jmsac $|V^top x| = sqrt{x^top VV^top x} le lambda_{max}(VV^top) = lambda_{max}(V^top V)$
– angryavian
Nov 23 at 3:21
Ahh ok thank you!
– jmsac
Nov 23 at 3:23
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Fix $x$ such that $|x| = 1$.
Let $y = V^top x / |V^top x|$. Note that $|y| = 1$ and that $|V^top x| le |V^top| = |V|$.
Then
$$x^top V W V^top x = |V^top x|^2 cdot y^top W y le |V|^2 cdot y^top W y.$$
answered Nov 23 at 2:57
angryavian
38k23180
Fix $x$ such that $|x| = 1$.
Let $y = V^top x / |V^top x|$. Note that $|y| = 1$ and that $|V^top x| le |V^top| = |V|$.
Then
$$x^top V W V^top x = |V^top x|^2 cdot y^top W y le |V|^2 cdot y^top W y.$$
answered Nov 23 at 2:57
angryavian
38k23180
answered Nov 23 at 2:57
angryavian
38k23180
answered Nov 23 at 2:57
angryavian
38k23180
answered Nov 23 at 2:57
angryavian
38k23180
38k23180
Why does $|V^{T}x| leq |V^T|$ Hold? The second is the largest e value of $V^TV$ not of $V^T$
– jmsac
Nov 23 at 3:06
Sorry the second quantity is the square root of the largest e-value of $V^TV$
– jmsac
Nov 23 at 3:20
@jmsac $|V^top x| = sqrt{x^top VV^top x} le lambda_{max}(VV^top) = lambda_{max}(V^top V)$
– angryavian
Nov 23 at 3:21
Ahh ok thank you!
– jmsac
Nov 23 at 3:23
add a comment |
Why does $|V^{T}x| leq |V^T|$ Hold? The second is the largest e value of $V^TV$ not of $V^T$
– jmsac
Nov 23 at 3:06
Sorry the second quantity is the square root of the largest e-value of $V^TV$
– jmsac
Nov 23 at 3:20
@jmsac $|V^top x| = sqrt{x^top VV^top x} le lambda_{max}(VV^top) = lambda_{max}(V^top V)$
– angryavian
Nov 23 at 3:21
Ahh ok thank you!
– jmsac
Nov 23 at 3:23
Why does $|V^{T}x| leq |V^T|$ Hold? The second is the largest e value of $V^TV$ not of $V^T$
– jmsac
Nov 23 at 3:06
Why does $|V^{T}x| leq |V^T|$ Hold? The second is the largest e value of $V^TV$ not of $V^T$
– jmsac
Nov 23 at 3:06
Sorry the second quantity is the square root of the largest e-value of $V^TV$
– jmsac
Nov 23 at 3:20
Sorry the second quantity is the square root of the largest e-value of $V^TV$
– jmsac
Nov 23 at 3:20
@jmsac $|V^top x| = sqrt{x^top VV^top x} le lambda_{max}(VV^top) = lambda_{max}(V^top V)$
– angryavian
Nov 23 at 3:21
@jmsac $|V^top x| = sqrt{x^top VV^top x} le lambda_{max}(VV^top) = lambda_{max}(V^top V)$
– angryavian
Nov 23 at 3:21
Ahh ok thank you!
– jmsac
Nov 23 at 3:23
Ahh ok thank you!
– jmsac
Nov 23 at 3:23
add a comment |
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