# The sup-norm of a diagonalizable operator

I want to get familiar with computing sup-norms of diagonalizable operators on $\mathbf{R}^n$. Suppose that I have a diagonalizable linear map $T:\mathbf{R}^n\to \mathbf{R}^n$ and I consider $\mathbf{R}^n$ with its standard norm. Then, what is the norm of $T$ in terms of its eigenvalues?

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If $\mathbb R^n$ is equipped with the euclidean norm, then the norm of a orthogonally diagonizable operator $T\colon\mathbb R^n\to\mathbb R^n$ is the supremum of the absolute values of the eigenvalues of $T$. Is this what you were asking for?
As the problem is currently stated, this is not correct in general. That is true if $T$ is orthogonally diagonalizable (equivalently, symmetric), which based on Bana's comment seems to be implicitly assumed. For example, $\begin{bmatrix}1&1\\0&0\end{bmatrix}$ is diagonalizable with eigenvalues $0$ and $1$, but the norm of the matrix is $\sqrt 2$. –  Jonas Meyer Dec 6 '11 at 6:41