# How to prove that the spectral radius of a linear operator is the infimum over all subordinate norms of the corresponding norm of the operator.

I am trying to understand a proof I have seen of the following theorem:

$$\rho(A)=\inf_{\|\cdot\|}\|A\|.$$

I understand that to do this, the idea is to show that 1) $\rho(A)\leq\|A\|$ for any norm, and then to show that, a norm exists for which, 2) $\|A\|_{\epsilon}\lt\rho(A)+\epsilon,\,\,\, \forall\, \epsilon\gt 0$, which would then indicate that, indeed, the statement in the theorem is true.

I can do the first part:
Let $V$ be a vector space. Let $x$ be an eigenvector of $A\in\mathbb{R}^{n\times n}$, with corresponding eigenvalue $\lambda$, such that $|\lambda|=\rho(A)$, where $\rho$ denotes the spectral radius. Then, by the definition of a subordinate norm, $$\|A\|=\sup_{v\in V}\frac{\|Av\|}{\|v\|}\ge\frac{\|Ax\|}{\|x\|}=\frac{|\lambda|\|x\|}{\|x\|}=|\lambda|,$$ or $$\rho(A)\le\|A\|.$$

In the next part of the proof, I can make it most of the way:
Define $$D=\begin{pmatrix} \delta&0&0&0&\cdots&0 \\ 0&\delta^2&0&0&\cdots&0 \\ 0 & 0 & \delta^3 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \delta^4 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \delta^n\end{pmatrix},$$ where $\delta>0$ is small. $D^{-1}$ exists, and it is a diagonal matrix with entries consisting of the reciprocals of the entries on the diagonal of $D$. By the Schur Decomposition theorem, we know that $A$ can be decomposed as $Q^T AQ=\Lambda+U$, where $Q$ is an orthogonal matrix, $\Lambda$ is a diagonal matrix with the eigenvalues of $A$ on the diagonal, and $U$ is a strict upper-triangular matrix. Now, investigate $$D^{-1}Q^T A Q D=D^{-1}( \Lambda +U) D=D^{-1}\Lambda D + D^{-1} U D.$$ On the right hand side, $D^{-1}\Lambda D = \Lambda,$ since these three matrices are diagonal, so $$D^{-1}Q^T A Q D = \Lambda + D^{-1}UD.$$ Now, consider $$\|D^{-1}Q^TAQD\|_\infty = \|\Lambda + D^{-1}UD\|_\infty\le\|\Lambda\|_\infty+\|D^{-1}UD\|_\infty$$ by the triangle inequality. Since $\Lambda$ has the eigenvalues of $A$ on its diagonal, I know $\|\Lambda\|_\infty=\rho(A)$. If we look at the entries of $D^{-1}UD$, we find values of $(D^{-1}UD)_{ij}=\delta^{j-i}u_{ij}$ for $j>i$, where $u_{ij}$ is the $(i,j)$-entry of $U$. The entries for $j\le i$ are $0$, since $U$ is upper triangular. Thus, we know that $\|D^{-1}UD\|_\infty\le \delta\|U\|_\infty$. The "less than" part will almost always hold, but it seems that if $u_{n-1,n}$ is huge, maybe $\|U\|_\infty=u_{n-1,n}$ (recall the $\|\cdot\|_\infty$ is the max row sum). In that case, $\|D^{-1}UD\|_\infty=\delta\|U\|_\infty$. Ignoring this for now, we have

$$\|D^{-1}Q^TAQD\|_\infty \le\|\Lambda\|_\infty+\|D^{-1}UD\|_\infty \le \rho(A) + \delta\|U\|_\infty.$$ Now, define $\delta\equiv\frac{\epsilon}{\|U\|_\infty},$ and $\|A\|_\epsilon=\|D^{-1}Q^TAQD\|_\infty$. (It can be shown that $\|A\|_\epsilon$ is a subordinate norm, but as this post is already quite long, I will skip that proof.) Then, finally, we have $$\|A\|_\epsilon\le \rho(A)+\epsilon.$$

However, this is not quite what I wanted to show, because of the $\le$ sign in the last inequality. Does anyone know how I can change this to a strict inequality? Thanks a lot.

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You need, for each $\epsilon > 0$ a norm $\|\cdot\|_{\epsilon}$ for which $\|A\|_{\epsilon} < \rho(A) + \epsilon$.

But what you have shown is sufficient. Indeed let $\delta > 0$.

Choose, by the above, a norm $\|\cdot\|_{\frac{\delta}{2}}$ such that $\|A\|_{\frac{\delta}{2}}\leq \rho(A) + \frac{\delta}{2}$.

But then $\|A\|_{\frac{\delta}{2}}\leq \rho(A) + \frac{\delta}{2} < \rho(A) + \delta$. So take $\|\cdot\|_{\delta}$ to be $\|\cdot\|_{\frac{\delta}{2}}$ and the claim is satisfied.

$\bf{\text{Similar Example:}}$ A sequence $(x_{n})_{n=1}^{\infty}$ in a metric space $(X,d)$ converges to $x\in X$ if for all $\epsilon > 0$ there is $N\geq 1$ such that $d(x_{n}, x)<\epsilon$ whenever $n\geq N$.

This definition does not change if we replace $< \epsilon$ with $\leq \epsilon$.

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