Spectral radius, second induced norm

In my textbook there are few facts left without any sign of a proof, which really bugs me, and I was thinking maybe someone can help me:

1. $A\in \mathbb{R}^{m,n} \Rightarrow \ \|A\|_2 = \sqrt{\rho(A^TA)}$ where $\rho(A^TA)$ denotes spectral radius of a matrix $A^TA$ i.e. maximum absolute value of an eigenvalue of $A^TA$ (in this case all eigenvalues are real and non-negative)
2. If $A\in\mathbb{R}^{n,n}$ and $A^T=A$ then $\|A\|_2=\rho(A)$
3. $\|A\|_2\le\sqrt{\|A^TA\|_{\infty}}$

I suppose these are well known facts, yet I couldn't find their proofs on the Internet. Eigenvalues are very new for me, maybe that is why I couldn't (and why I am not able to prove it alone, I hope they are not too complicated, I prefer simple proofs). In first also fact that eigenvalues of $A^TA$ are real and non-negative is not easy for me.

Additionally I have several questions:

1. I suppose that (yet I don't know a lot about it) $\|A\|_2=\max_{\vec{x}\neq\vec{0}} \frac{\|A\vec{x}\|_2}{\|\vec{x}\|_2}$ is not easy to compute. Is this second definition of $\|A\|_2$ (with spectral radius) helpful? I mean easier to compute? Can one easy and quite fast find eigenvalues, at least numerically?
2. Is this inequality in the third point very good, that is $\|A\|_2$ is close to given upper bound?

1. This was asked here already many times: $$\|A\|_2^2 = \max_{x\neq 0}\frac{\|Ax\|_2^2}{\|x\|_2^2}=\max_{x\neq 0}\frac{x^TA^TAx}{x^Tx}=\lambda_{\max}(A^TA)=\rho(A^TA).$$ The next to last equality is due to the Courant-Fischer theorem.

2. If $A$ is symmetric, then $\rho(A^TA)=\rho(A^2)=\rho(A)^2$. (Notice that symmetric matrices have real eigenvalues and if a $\lambda$ is an eigenvalue of a symmetric $A$, then $\lambda^2$ is a nonnegative eigenvalue of $A^2$.)

3. For any (square) $B$ and any matrix norm $\|\cdot\|$ subordinate with respect to some vector norm, we have $\rho(B)\leq\|B\|$. Plug that into the point 1) with the special choice $\|\cdot\|=\|\cdot\|_{\infty}$ to get $\|A\|_2=\sqrt{\rho(A^TA)}\leq\sqrt{\|A^TA\|_{\infty}}$.

IMHO it does not need to be the case generally (sometimes the bound can be tight, sometimes not). However, unlike $\rho(A^TA)$, the $\infty$-norm is almost trivial to compute.
• Thank you very very much! I have only one more question. Why we have for any square $B$ and any matrix norm $\|\cdot\|, \rho(B)\le\|B\|$? Does this also follow from some well known theory? Or is just a simple observation that I don't see? – xan Dec 4 '13 at 10:10
• @xan It's simple: for any eigenpair $(\lambda,x)$ of $A$, we have from $Ax=\lambda x$ that $\|\lambda x\|=|\lambda|\|x\|=\|Ax\|\leq\|A\|\|x\|$ and hence $|\lambda|\leq\|A\|$. It holds for any vector norm and any matrix norm subordinate w.r.t. that vector norm (that is, a pair of matrix/vector norms satisfying $\|Ax\|\leq\|A\|\|x\|$ for all $x$). In particular, the statement holds for any operator norm such as the $\infty$-norm. – Algebraic Pavel Dec 4 '13 at 10:54