# Singular and eigen values properties…

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?

• For part 1 to make sense, the definition of $\lambda_{\max}(A)$ should be the biggest eigenvalue of $A$ not in absolute value. – angryavian Dec 6 '18 at 17:55

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.

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 $$\Sigma^t \Sigma$$.
• 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 $B$? – C. Cristi Dec 6 '18 at 17:51
• How did you make the note that the norm of $U^tv=1$ is the same? – C. Cristi Dec 6 '18 at 18:08
• Hey, why $\lambda_{max}(\frac {C+C^t}{2})=max_{\|v\|=1}v^tCv$ and not $=max_{\|v\|=1}v^t\frac{C+C^t}{2}v$? – C. Cristi Dec 6 '18 at 18:16
• Orthogonal matrix are norm-preserving because $\|Uv\|=\sqrt{\langle Uv,Uv\rangle}=\sqrt{\langle U^tv,Uv\rangle}=\sqrt{UU^t\langle v, v\rangle}= \sqrt{\langle v, v\rangle}$? – C. Cristi Dec 6 '18 at 18:22