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Given a $N\times N$ matrix $A$ over $\mathbb R$. Let $ \rho\left( A \right) = \max \left\{ {\left| \lambda \right|;\lambda \mbox{ eigenvalue of }A} \right\}$. Someone told me that, the following holds:

$$ \rho\left( A \right) = \inf \left\{\lVert| A \rVert|,\lVert \cdot \rVert\mbox{ matrix norm}\right\} . $$

Here is the definition of a matrix norm Wikipedia.

I'm not sure if this is true, and in that case, how to prove it :S

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up vote 3 down vote accepted

It's true and the proof is classical, in fact in a standard course about numerical linear algebra . We can see $A$ as a matrix over the field of complex numbers. If $|\lambda|=\rho(A)$, $x$ is an eigenvector associated to $\lambda$ and $N$ a matrix norm then $|\lambda|\leq \sup_{x\neq 0}\frac{N'(Ax)}{N'(x)}=N(A)$ where $N'$ is some norm over $\mathbb C^n$, so $\rho(A)\leq N(A)$ for all matrix norm $N$.

Now we fix $\varepsilon>0$. Let $U$ such that $U^{-1}AU=\pmatrix{\lambda_1&t_{12}&\ldots&t_{1n}\\\ 0&\lambda_2&&t_{2,n}\\\ \vdots&&\ddots&\vdots\\\ 0&0&\ldots&\lambda_n}$ where $\lambda_j$ are the eigenvalues of $A$. For $\delta>0$, denote $D_{\delta}=\operatorname{diag}(1,\delta,\ldots,\delta^{n-1})$, and put $V_{\delta}:=UD_{\delta}$. We have $$V_{\delta}^{—1}AV_{\delta}=\pmatrix{\lambda_1&\delta \cdot t_{12}&\ldots&\delta^{n-1}\cdot t_{1n}\\\ 0&\lambda_2&&\delta^{n-2}\cdot t_{2,n}\\\ \vdots&&\ddots&\vdots\\\ 0&0&\ldots&\lambda_n}.$$ We fix $\delta$ such that for all $k\in\{1,\ldots,n-1\}$, $\sum_{j=k+1}^n|t_{k,j}\delta^{j-k}|<\varepsilon$ and we define $N(M)=\sup_{x\neq 0}\frac{\lVert V_{\delta}^{-1}MV_{\delta}x\rVert_{\infty}}{\lVert V_{\delta}^{-1}x\rVert_{\infty}}$.

Thanks to the characterisation of the norm $\lVert\cdot\rVert_{\infty}$ of a matrix as the maximum of the sum of the absolute values of the coefficients of a column we can check that $N(A)\leq \rho(A)+\varepsilon$.

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