# Subordinated matrix norm of diagonalizable matrix is its spectral radius

I want to solve the following problem

Let $$A$$ be a matrix that admits a basis of eigenvectors (i.e. a diagonaliazable matrix). Find a norm $$|||\cdot |||$$ subordinated to a vector norm $$|||\cdot |||$$ such that $$\rho (A)=|||A|||$$, where $$\rho (A)$$ is the spectral radius of $$A$$.

I know that for a symmetric matrix we can use the euclidean norm. If $$A=S^{-1}DS$$, then it is not always true that $$\|A\|_2=\|D\|_2$$ (it is of course true if $$S$$ is orthogonal), I think that maybe the supremum norm $$\|\cdot \|_\infty$$ works, but I'm not sure how could I prove it.

Let $$(X_1, \ldots, X_n) \in \mathcal{M}_{n,1}(\mathbb{K})^n$$ be a base of eigenvectors of $$A$$ associated respectively to the eigenvalues $$(\lambda_1, \ldots, \lambda_n) \in \mathbb{K}^n$$.
Consider the norm-1 associated to this base i.e. for any $$X \in \mathcal{M}_{n,1}(\mathbb{K})$$, there exists a unique $$(\alpha_1, \ldots, \alpha_n) \in \mathbb{K}^n$$ such that $$X = \sum_{i = 1}^n \alpha_i X_i$$, we then define $$\| X \| = \sum_{i = 1}^n |\alpha_i|$$ (observation: for all $$1 \leqslant i \leqslant n$$, $$\| X_i \| = 1$$) : the defined application $$\| \cdot \|$$ is a norm (it is homogeneous, well-defined and satisfy the triangle inequality).
Now, consider the norm $$||| \cdot |||$$ defined over $$\mathcal{M}_n(\mathbb{K})$$ subordinated to $$\| \cdot \|$$ then, for all integer $$1 \leqslant i \leqslant n$$, $$\| A X_i \| = |\lambda_i|$$ so $$||| A ||| \geqslant \rho(A) = \max\limits_{1 \leqslant i \leqslant n}(|\lambda_i|)$$. Furthermore, for any matrix column $$X = \sum\limits_{i = 1}^n \alpha_i X_i$$ with $$\| X \| = 1$$ we have:
$$\| AX \| = \left\| \sum_{i = 1}^n \alpha_iAX_i \right\| = \left\| \sum_{i = 1}^n \alpha_i\lambda_iX_i \right\| = \sum_{i = 1}^n |\alpha_i||\lambda_i| \leqslant \rho(A)\sum_{i = 1}^n |\alpha_i| = \rho(A)||X|| = \rho(A)$$