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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?

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Yes. I also thought about how to prove this and it's not very hard. You choose an orthonormal basis of eigenvectors and you write it out. –  Bana Oct 20 '11 at 21:56
    
@Bana: Yes, exactly. –  Rasmus Oct 21 '11 at 7:42
    
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
    
@JonasMeyer: You are absolutely right. Thank you for the correction. I will edit to emphasize this. –  Rasmus Dec 6 '11 at 9:16
    
@Rasmus: Thanks (+1). –  Jonas Meyer Dec 6 '11 at 15:58
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