Mixing
When is the Frobenius norm equal to the spectral radius?
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I know that the spectral radius is $rho(A) = max |lambda_l| = S_{max}^2$ and that the Frobenius norm is $||A|| = sqrt{tr(A^*A)} = (sum_{k}S_k^2)^{1/2}$, which means I want to find the matrix A for which the following is true
$$ ||A||_F = sqrt{tr(A^*A)} = (sum_{k}S_k^2)^{1/2} = S_{max}^2 $$
So is the spectral radius equal to the frobenius norm if A is a square matrix with its largest eigenvalue equal to |1|?
(S are the singular values)
matrices norm
asked Jan 21 '16 at 22:34
SandraSandra
162
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add a comment |
$begingroup$
I know that the spectral radius is $rho(A) = max |lambda_l| = S_{max}^2$ and that the Frobenius norm is $||A|| = sqrt{tr(A^*A)} = (sum_{k}S_k^2)^{1/2}$, which means I want to find the matrix A for which the following is true
$$ ||A||_F = sqrt{tr(A^*A)} = (sum_{k}S_k^2)^{1/2} = S_{max}^2 $$
So is the spectral radius equal to the frobenius norm if A is a square matrix with its largest eigenvalue equal to |1|?
(S are the singular values)
matrices norm
asked Jan 21 '16 at 22:34
SandraSandra
162
$endgroup$
add a comment |
$begingroup$
I know that the spectral radius is $rho(A) = max |lambda_l| = S_{max}^2$ and that the Frobenius norm is $||A|| = sqrt{tr(A^*A)} = (sum_{k}S_k^2)^{1/2}$, which means I want to find the matrix A for which the following is true
$$ ||A||_F = sqrt{tr(A^*A)} = (sum_{k}S_k^2)^{1/2} = S_{max}^2 $$
So is the spectral radius equal to the frobenius norm if A is a square matrix with its largest eigenvalue equal to |1|?
(S are the singular values)
matrices norm
asked Jan 21 '16 at 22:34
SandraSandra
162
$endgroup$
I know that the spectral radius is $rho(A) = max |lambda_l| = S_{max}^2$ and that the Frobenius norm is $||A|| = sqrt{tr(A^*A)} = (sum_{k}S_k^2)^{1/2}$, which means I want to find the matrix A for which the following is true
$$ ||A||_F = sqrt{tr(A^*A)} = (sum_{k}S_k^2)^{1/2} = S_{max}^2 $$
So is the spectral radius equal to the frobenius norm if A is a square matrix with its largest eigenvalue equal to |1|?
(S are the singular values)
matrices norm
matrices norm
asked Jan 21 '16 at 22:34
SandraSandra
162
asked Jan 21 '16 at 22:34
SandraSandra
162
edited Jan 23 '16 at 10:31
Sandra
asked Jan 21 '16 at 22:34
SandraSandra
162
asked Jan 21 '16 at 22:34
SandraSandra
162
asked Jan 21 '16 at 22:34
SandraSandra
162
162
add a comment |
add a comment |
1 Answer
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$begingroup$
We assume that $Anot= 0$.
Let $spectrum(A)=(lambda_i)$ where $|lambda_1|geq |lambda_2|geqcdots$ and $Sigma(A)=(sigma_i)$ where $sigma_1geq sigma_2geqcdots$. Then $rho(A)=[lambda_1|leq sigma_1=||A||_F=sqrt{sum_i sigma_i^2}$. Thus $|lambda_1|=sigma_1$ and $sigma_2=cdots=sigma_n=0$.
Finally $rank(A)=1$ and $A=uv^*$ where $u,v$ are non-zero vectors. One has $AA^*=||v||^2uu^*$, $lambda_1=trace(uv^*)=v^*u$ and $sigma_1^2=||v||^2trace(uu^*)=||v||^2||u||^2$. Moreover $|lambda_1|=|v^*u|leq ||u||||v||=sigma_1$, one deduces that the Schwartz inequality is an equality, that implies that $u,v$ are parallel.
Conclusion: $A=0$ or $A=alpha uu^*$ where $alphain mathbb{C}^*$ and $u$ is a non-zero vector. (The converse is easy)
EDIT. @ Sandra , I see your edit. I think that you did not understand one word of my answer.
answered Jan 23 '16 at 14:18
loup blancloup blanc
24.1k21851
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1 Answer
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1 Answer
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$begingroup$
We assume that $Anot= 0$.
Let $spectrum(A)=(lambda_i)$ where $|lambda_1|geq |lambda_2|geqcdots$ and $Sigma(A)=(sigma_i)$ where $sigma_1geq sigma_2geqcdots$. Then $rho(A)=[lambda_1|leq sigma_1=||A||_F=sqrt{sum_i sigma_i^2}$. Thus $|lambda_1|=sigma_1$ and $sigma_2=cdots=sigma_n=0$.
Finally $rank(A)=1$ and $A=uv^*$ where $u,v$ are non-zero vectors. One has $AA^*=||v||^2uu^*$, $lambda_1=trace(uv^*)=v^*u$ and $sigma_1^2=||v||^2trace(uu^*)=||v||^2||u||^2$. Moreover $|lambda_1|=|v^*u|leq ||u||||v||=sigma_1$, one deduces that the Schwartz inequality is an equality, that implies that $u,v$ are parallel.
Conclusion: $A=0$ or $A=alpha uu^*$ where $alphain mathbb{C}^*$ and $u$ is a non-zero vector. (The converse is easy)
EDIT. @ Sandra , I see your edit. I think that you did not understand one word of my answer.
answered Jan 23 '16 at 14:18
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
$begingroup$
We assume that $Anot= 0$.
Let $spectrum(A)=(lambda_i)$ where $|lambda_1|geq |lambda_2|geqcdots$ and $Sigma(A)=(sigma_i)$ where $sigma_1geq sigma_2geqcdots$. Then $rho(A)=[lambda_1|leq sigma_1=||A||_F=sqrt{sum_i sigma_i^2}$. Thus $|lambda_1|=sigma_1$ and $sigma_2=cdots=sigma_n=0$.
Finally $rank(A)=1$ and $A=uv^*$ where $u,v$ are non-zero vectors. One has $AA^*=||v||^2uu^*$, $lambda_1=trace(uv^*)=v^*u$ and $sigma_1^2=||v||^2trace(uu^*)=||v||^2||u||^2$. Moreover $|lambda_1|=|v^*u|leq ||u||||v||=sigma_1$, one deduces that the Schwartz inequality is an equality, that implies that $u,v$ are parallel.
Conclusion: $A=0$ or $A=alpha uu^*$ where $alphain mathbb{C}^*$ and $u$ is a non-zero vector. (The converse is easy)
EDIT. @ Sandra , I see your edit. I think that you did not understand one word of my answer.
answered Jan 23 '16 at 14:18
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
$begingroup$
We assume that $Anot= 0$.
Let $spectrum(A)=(lambda_i)$ where $|lambda_1|geq |lambda_2|geqcdots$ and $Sigma(A)=(sigma_i)$ where $sigma_1geq sigma_2geqcdots$. Then $rho(A)=[lambda_1|leq sigma_1=||A||_F=sqrt{sum_i sigma_i^2}$. Thus $|lambda_1|=sigma_1$ and $sigma_2=cdots=sigma_n=0$.
Finally $rank(A)=1$ and $A=uv^*$ where $u,v$ are non-zero vectors. One has $AA^*=||v||^2uu^*$, $lambda_1=trace(uv^*)=v^*u$ and $sigma_1^2=||v||^2trace(uu^*)=||v||^2||u||^2$. Moreover $|lambda_1|=|v^*u|leq ||u||||v||=sigma_1$, one deduces that the Schwartz inequality is an equality, that implies that $u,v$ are parallel.
Conclusion: $A=0$ or $A=alpha uu^*$ where $alphain mathbb{C}^*$ and $u$ is a non-zero vector. (The converse is easy)
EDIT. @ Sandra , I see your edit. I think that you did not understand one word of my answer.
answered Jan 23 '16 at 14:18
loup blancloup blanc
24.1k21851
$endgroup$
We assume that $Anot= 0$.
Let $spectrum(A)=(lambda_i)$ where $|lambda_1|geq |lambda_2|geqcdots$ and $Sigma(A)=(sigma_i)$ where $sigma_1geq sigma_2geqcdots$. Then $rho(A)=[lambda_1|leq sigma_1=||A||_F=sqrt{sum_i sigma_i^2}$. Thus $|lambda_1|=sigma_1$ and $sigma_2=cdots=sigma_n=0$.
Finally $rank(A)=1$ and $A=uv^*$ where $u,v$ are non-zero vectors. One has $AA^*=||v||^2uu^*$, $lambda_1=trace(uv^*)=v^*u$ and $sigma_1^2=||v||^2trace(uu^*)=||v||^2||u||^2$. Moreover $|lambda_1|=|v^*u|leq ||u||||v||=sigma_1$, one deduces that the Schwartz inequality is an equality, that implies that $u,v$ are parallel.
Conclusion: $A=0$ or $A=alpha uu^*$ where $alphain mathbb{C}^*$ and $u$ is a non-zero vector. (The converse is easy)
EDIT. @ Sandra , I see your edit. I think that you did not understand one word of my answer.
answered Jan 23 '16 at 14:18
loup blancloup blanc
24.1k21851
edited Jan 26 '16 at 4:50
answered Jan 23 '16 at 14:18
loup blancloup blanc
24.1k21851
answered Jan 23 '16 at 14:18
loup blancloup blanc
24.1k21851
answered Jan 23 '16 at 14:18
loup blancloup blanc
24.1k21851
24.1k21851
add a comment |
add a comment |
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