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Is it true that for any $n \times n$ matrix $A$, with real entries, and where all eigenvalues are real, and non zero, the operator norm ( where$||A|| = \max_{|x|=1} |Ax|$ with |.| is the standard length in $\mathbb{R}^{n}) is the same as the largest eigenvalue in absolute value? If it is true, can someone please give me a reference for this result?

If it is not true, what if we assume that eigenvalues are also distinct, would that make it true?

Thank you for any help.

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No, it's not true. Let $A = \begin{bmatrix} 2 & 1000 \\ 0 & 1 \end{bmatrix}$. The eigenvalues of $A$ are $1$ and $2$, but the norm of $A$ is not $2$.

(The norm of $A$ is larger than $2$, because $\begin{bmatrix} 2 & 1000 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1000 \\ 1 \end{bmatrix}$.)

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  • $\begingroup$ Sorry, I edited to assume eigenvalues are not zero. $\endgroup$ – Jessica Jun 2 '15 at 9:06

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