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.
 A: 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:
$$\| AX \| = \left\| \sum_{i = 1}^n \alpha_iAX_i \right\| = \left\| \sum_{i = 1}^n \alpha_i\lambda_iX_i \right\| = \sum_{i = 1}^n |\alpha_i||\lambda_i| \leqslant \rho(A)\sum_{i = 1}^n |\alpha_i| = \rho(A)||X|| = \rho(A)$$
