# 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) Commented Jul 18, 2018 at 0:05
• @ClementC. Sorry, my bad. Commented Jul 18, 2018 at 0:07
• Nice question ..............+1 Commented Feb 24, 2022 at 21:37

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! Commented Jul 18, 2018 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. Commented Jul 18, 2018 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. Commented Jul 18, 2018 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. Commented Jul 18, 2018 at 0:36

Although this question was answered over 5 years ago, I will add this answer, which gives an alternative explanation from a linear transformation perspective.

Consider a vector-induced matrix norm $$\Vert A \Vert = \mathrm{sup}_{\bar{x}\neq 0}\frac{\Vert A\bar{x}\Vert}{\Vert \bar{x} \Vert}$$ of an $$n\times n$$ real matrix $$A$$ (this explanation also works if $$A$$ is complex, but I will leave that out for brevity).

Let $$\bar{v}_1,\bar{v}_2,...\bar{v}_n$$ be the normalized eigenvectors of $$A$$, and $$\lambda_1,\lambda_2,...\lambda_n \in \mathbb R$$ be the corresponding eigenvalues (this explanation also works if $$\lambda_i$$ are complex conjugate pairs, but I will leave that out for brevity).

Any unit vector $$\bar{u}$$ in the space spanned by $$A$$ can be given by $$\bar{u}=(w_1 \bar{v}_1 +w_2\bar{v}_2+...+w_n\bar{v}_n)$$ where, $$w_i\in \mathbb R$$ are scaling factors.

Any non-zero vector $$\bar{x}$$ can be given as $$\bar{x}=k\bar{u}$$, where $$k>0\in \mathbb R$$

$$\therefore$$ $$A\bar{x}=k(\lambda_1w_1\bar{v}_1 +\lambda_2w_2\bar{v}_2+...+\lambda_nw_n\bar{v}_n)$$

Notice $$\Vert \bar{x} \Vert=k$$ and $$\Vert A\bar{x}\Vert=k\Vert \lambda_1w_1\bar{v}_1 +\lambda_2w_2\bar{v}_2+...+\lambda_nw_n\bar{v}_n \Vert$$

$$\Vert A \Vert = \mathrm{sup}_{\bar{x}\neq 0}\frac{\Vert A\bar{x}\Vert}{\Vert \bar{x} \Vert}=\mathrm{sup}_{w_i} \Vert \lambda_1w_1\bar{v}_1 +\lambda_2w_2\bar{v}_2+...+\lambda_nw_n\bar{v}_n \Vert$$

Now, notice that we can always select a unit vector $$\bar{u}$$ (with suitable $$w_1,w_2,...,w_n$$) such that $$\Vert \lambda_1w_1\bar{v}_1 +\lambda_2w_2\bar{v}_2+...+\lambda_nw_n\bar{v}_n \Vert \ge \vert \lambda_{max} \vert$$

where $$\lambda_{max}=max_i\{\lambda_i\}$$

$$\therefore \Vert A \Vert = \mathrm{sup}_{\bar{x}\neq 0}\frac{\Vert A\bar{x}\Vert}{\Vert \bar{x} \Vert}=\mathrm{sup}_{w_i} \Vert \lambda_1w_1\bar{v}_1 +\lambda_2w_2\bar{v}_2+...+\lambda_nw_n\bar{v}_n \Vert \ge\vert \lambda_{max} \vert \ge \vert \lambda_i \vert$$