Maximum "stretch factor" of linear map Let $A \in M_n (\mathbb R)$ be diagonalizable matrix, let $\lambda_1, \dots, \lambda_n$ be its eigenvalues.
I want to know if the maximum "stretch factor" of $A$ is the maximum of its unsigned eigenvalues i.e.
$$\sup_{\Vert X \Vert = 1} \Vert AX\Vert = \max_{1\leq i\leq n} |\lambda_i|$$
Any help would be appreciated.
$\Vert \cdot \Vert$ will be the euclidean norm of $\mathbb R^n$.

In general, the maximum stretch factor is the largest singular value $\sigma_1$ of $A$, i.e. the square root of the largest eigenvalue $\lambda_1$ of $A^TA$. Note that if you assume $A$ is symmetric, then $\sigma_1$ is exactly $\sqrt{\lambda^2} = |\lambda|$ where $\lambda$ is the largest eigenvalue of $A$.
This comes from our knowledge of quadratic forms. Since maximizing $\|Ax\|$ is equivalently maximizing the square root of $\|Ax\|^2 = (Ax)\cdot(Ax) = x^TA^TAx$, we’re really maximizing the square root of the quadratic form given by $A^TA$. The restriction $\|x\|=1$ tells us that this is exactly the largest singular value of $A$, $\sigma_1$.
Note that if $x$ is the eigenvector for $A^TA$ corresponding to $\lambda_1$, then 
$$\sqrt{x^TA^TAx} = \sqrt{x^T(\lambda_1x)} = \sqrt{\lambda_1\|x\|^2} = \sigma_1\|x\|.$$
Thus, if you want to maximize the length of vectors on a sphere of any fixed radius, i.e. constraining yourself to $\|x\| = r$ for a fixed $r$, you just scale by $r$.
