Prove that for $A \in \mathbb R^{m \times n}$, $\|A\|_2 = σ_{\max}$ using the Schur-Jordan decomposition of the appropriate matrix.
I have to apply the Schur-Jordan decomposition on $AA^T$ since it is symmetric, and because from it, we get the singular values.
So $AA^T = QDQ^T$, where $Q$ is an orthogonal matrix, and $D$ is a diagonal matrix with diagonal entries the eigenvalues of $A$.
So $||AA^T||_2 = ||QDQ^T||_2 \leq ||QD||_2 × ||Q^T||_2 = ||D||_2 × ||Q^T||_2$ (since $Q$ is orthogonal, then $||QD||_2 = ||D||_2$, and $||Q||_2 = 1$).
Hence, $||AA^T||_2 \leq max|\lambda_i| × 1$, since $||D||_2$ is the max of the its eigenvalues. But how to continue from there to reach what's required? I know that the singualr values of $A$ are the square root of the eigenvalues of $AA^T$. Any help please?