# Why is the norm of a matrix larger than its eigenvalue?

I know there are different definitions of Matrix Norm, but I want to use the definition on WolframMathWorld, and Wikipedia also gives a similar definition.

The definition states as below:

Given a square complex or real $n\times n$ matrix $A$, a matrix norm $\|A\|$ is a nonnegative number associated with $A$ having the properties

1.$\|A\|>0$ when $A\neq0$ and $\|A\|=0$ iff $A=0$,

2.$\|kA\|=|k|\|A\|$ for any scalar $k$,

3.$\|A+B\|\leq\|A\|+\|B\|$, for $n \times n$ matrix $B$

4.$\|AB\|\leq\|A\|\|B\|$.

Then, as the website states, we have $\|A\|\geq|\lambda|$, here $\lambda$ is an eigenvalue of $A$. I don't know how to prove it, by using just these four properties.

• @mechanodroid Is it? Can you verify point 4? (The submultiplicative property of matrix norms) – Clement C. Jul 18 '18 at 0:05
• @ClementC. Sorry, my bad. – mechanodroid Jul 18 '18 at 0:07

Suppose $v$ is an eigenvector for $A$ corresponding to $\lambda$. Form the "eigenmatrix" $B$ by putting $v$ in all the columns. Then $AB = \lambda B$. So, by properties $2$ and $4$ (and $1$, to make sure $\|B\| > 0$), $$|\lambda| \|B\| = \|\lambda B\| = \|AB\| \le \|A\| \|B\|.$$ Hence, $\|A\| \ge |\lambda|$ for all eigenvalues $\lambda$.

• Nice trick. Thank you! – Eric Yewen Sun Jul 18 '18 at 0:12

Let $\|\cdot\|$ be a matrix norm.

It is known that the spectral radius $r(A) = \lim_{n\to\infty} \|A^n\|^{\frac1n}$ has the property $|\lambda| \le r(A)$ for all $\lambda\in \sigma(A)$.

Indeed, let $\lambda \in \mathbb{C}$ such that $|\lambda| > r(A)$.

Then $I - \frac1{\lambda} A$ is invertible. Namely, check that the inverse is given by $\sum_{n=0}^\infty\frac1{\lambda^n}A^n$.

This series converges absolutely because $\frac1{|\lambda|}$ is less than the radius of convergence of the power series $\sum_{n=1}^\infty \|A\|^nx^n$, which is $\frac1{\limsup_{n\to\infty} \|A^n\|^{\frac1n}} = \frac1{r(A)}$.

Hence $$\lambda I - A = \lambda\left(I - \frac1{\lambda} A\right)$$

is also invertible so $\lambda \notin \sigma(A)$.

Now using submultiplicativity we get $\|A^n\| \le \|A\|^n$ so

$$|\lambda| \le r(A) = \lim_{n\to\infty} \|A^n\|^{\frac1n} \le \lim_{n\to\infty} \|A\|^{n\cdot\frac1n} = \|A\|$$

• I had a look at the Wikipedia page. It seems that the identity $r(A) = \lim_{n\to\infty} \|A^n\|^{\frac{1}{n}}$ holds for natural matrix norms, i.e. operator norms induced by norms on $\mathbb{R}^n$. I don't have a counterexample, but I suspect it doesn't hold in general. – Theo Bendit Jul 18 '18 at 0:27
• @TheoBendit I used the abstract spectral radius defined in Banach algebras simply as $\lim_{n\to\infty} \|A^n\|^{\frac{1}{n}}$. Perhaps the name is not appropriate for matrices. Secondly, every two matrix norms are equivalent so if you take a natural matrix norm $\|\cdot\|_1$ we have $m\|\cdot\|_1 \le \|\cdot\| \le M\|\cdot\|_1$ so $$m^{1/n}\|A^n\|_1^{1/n} \le \|A^n\|^{1/n} \le M^{1/n}\|A^n\|_1^{1/n}$$ Letting $n\to\infty$ gives $\lim_{n\to\infty} \|A^n\|_1^{\frac{1}{n}} = \lim_{n\to\infty} \|A^n\|^{\frac{1}{n}}$. So it should hold for every matrix norm. – mechanodroid Jul 18 '18 at 0:32
• @TheoBendit Actually it is there in the Wikipedia page under the name Gelfand's Formula. I guess the only nontrivial thing in my answer is to show that the sequence $(\|A^n\|^{1/n})_n$ indeed converges. – mechanodroid Jul 18 '18 at 0:36