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Let $A\in\mathcal {M}_n(\mathbb{R})$, we will denote $\lambda_{\max}(A)$ the biggest eigenvalue of $A$ in absolute value, as for $B\in\mathcal M_{m,n}(\mathbb{R})$ we will denote $\sigma_{\max}(B)$ the highest singular value of $B$.

I have $3$ things to show from which I have shown 1 and for the last one I only have an idea.

$1$. with $A$ a symetric matrix show that $\lambda_{\max}(A)= \max_{\|v\|=1}v^tAv$

$$A=A^t\implies \max_{\|v\|=1}(v^tAv) =\max_{\|v\|=1}(v^tA^tv) =\max_{\|v\|=1}((Av)^tv) =\max_{\|v\|=1}((\lambda v)^tv) =\max_{\|v\|=1}(\lambda v^tv) =\max_{\|v\|=1}(\lambda \|v\|^2) =\max_{\|v\|=1}(\lambda) =\max(\lambda).$$

$2$. Show that $\sigma_{\max}(B) = \sqrt{\lambda_{\max}(B^tB)}$.

$3$. with $C\in\mathcal M_n(\mathbb{R})\implies\lambda_{\max}(\frac{C+C^t}{2})\leq\sigma_{\max}(C)$.

Here I think is something relating with AM-GM Mean right?

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  • $\begingroup$ For part 1 to make sense, the definition of $\lambda_{\max}(A)$ should be the biggest eigenvalue of $A$ not in absolute value. $\endgroup$ – angryavian Dec 6 '18 at 17:55
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  1. There are a few errors in your attempt. When you use $Av = \lambda v$ you are assuming $v$ is an eigenvector of $A$, even though the maximum is over all unit norm $v$ which may include non-eigenvectors. Also your attempt does not really use symmetry; note that you could have proceeded as $v^t A v = v^t (\lambda v) = \lambda \|v\|^2$ directly (disregarding the eigenvalue issue mentioned above). Symmetry is important here because of the spectral theorem. This may help you fix your proof.

  2. I suppose you are defining singular values from the SVD? Then write $B=U\Sigma V^t$ and note $B^t B = V \Sigma^t \Sigma V^t$. Since $\Sigma^t \Sigma$ is a diagonal matrix, this is a diagonalization of $B^t B$, so you can write down the eigenvalues of $B^t B$ in terms of the singular values of $B$.

  3. Let $U\Sigma V^t$ be the SVD of $C$. Note $(C+C^t)/2$ is symmetric. Using the first part, $$\lambda_{\max}(\frac{C+C^t}{2}) = \max_{\|v\|=1} v^t C v = \max_{\|v\|=1} v^t U \Sigma V^t v \le \max_{\|x\|=\|y\|=1} x^t \Sigma y = \sigma_{\max}(C),$$ where the inequality comes from the change of variables $x=U^t v$ and $y = V^t v$ and noting that orthogonal matrices are norm-preserving (i.e. $\|x\|=\|v\|$).


Edit:

As I mentioned in my comment, I think $\lambda_{\max}$ should be the largest eigenvalue, not the largest in absolute value.

Response to your comment:

  1. Let $UDU^t$ be the eigendecomposition of $A$. Then $\max_{\|v\|=1} v^t A v = \max_{\|v\|=1} v^t UDU^t v = \max_{\|w\| = 1} w^t D w = \lambda_{\max}(A)$, where we have made the change of variables $w = U^t v$ and noted that orthogonal matrices are norm-preserving.

  2. I already told you that the eigenvalues of $B^t B$ are the diagonal entries of the diagonal matrix $\Sigma^t \Sigma$.

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  • $\begingroup$ 1. Yeah I saw, okay to use the spectral theorem we have that: if $A$ is symetric then there is $Q$ and $D$ such that $A=QDQ^t$ then $A$ is orto-diagonalizable... but how do I use that here, since if I plug in $A$ in there I don't solve anything, can you be more explicit, also, for 2, yes, indeed it's about the SVD, how do I write down the eigenvalues of $B^tB$ in terms of singular values of $B$? $\endgroup$ – C. Cristi Dec 6 '18 at 17:51
  • $\begingroup$ @C.Cristi See my edit. $\endgroup$ – angryavian Dec 6 '18 at 18:05
  • $\begingroup$ How did you make the note that the norm of $U^tv=1$ is the same? $\endgroup$ – C. Cristi Dec 6 '18 at 18:08
  • $\begingroup$ Hey, why $\lambda_{max}(\frac {C+C^t}{2})=max_{\|v\|=1}v^tCv$ and not $=max_{\|v\|=1}v^t\frac{C+C^t}{2}v$? $\endgroup$ – C. Cristi Dec 6 '18 at 18:16
  • $\begingroup$ Orthogonal matrix are norm-preserving because $\|Uv\|=\sqrt{\langle Uv,Uv\rangle}=\sqrt{\langle U^tv,Uv\rangle}=\sqrt{UU^t\langle v, v\rangle}= \sqrt{\langle v, v\rangle}$? $\endgroup$ – C. Cristi Dec 6 '18 at 18:22

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