# Prove $\| A\|_2=\|A^T\|_2$

Let $$A \in \mathbb{R}^{n \times n},$$ Prove $$\| A\|_2=\|A^T\|_2$$

Similarly as the 1-norm I could show that $$\| A\|_2=\max_{i} \sqrt{\sum |a_{ij}|^2}$$. But this result does not help me to prove this identity.

• What is your definition of $\|A\|_2$ for a matrix A? Apr 15, 2023 at 20:38
• @krm2233 the maximum value of $\| Ax \|_2$ in the unit ball Apr 15, 2023 at 20:40

$$\|A\|_2 = \sup_{x: \|x\| = 1} \|Ax\| = \sup_{x, y: \|x\|, \|y\| = 1} \langle y, Ax \rangle = \sup_{x, y: \|x\|, \|y\| = 1} \langle A^{\mathsf{T}} y, x \rangle = \sup_{y: \|y\| = 1} \|A^{\mathsf{T}} y\| = \|A^{\mathsf{T}}\|_2.$$
Above, I'm denoting $$\langle x, y \rangle = x^{\mathsf{T}} y$$.
Another way to see this is that $$\|A\|_2^2 = \max_{\|x\|=1} \|Ax\|^2 = \max_{\|x\|=1}(Ax)^T(Ax) = \max_{\|x\|=1}x^T(A^TA)x = \max\{\lambda\in\sigma(A^TA)\}$$ and $$\|A^T\|_2^2 = \max_{\|x\|=1} \|A^Tx\|^2 = \max_{\|x\|=1}(A^Tx)^T(A^Tx) = \max_{\|x\|=1}x^T(AA^T)x = \max\{\lambda\in\sigma(AA^T)\}.$$
Now let $$v, \lambda\neq 0$$ be an eigenpair of $$A^TA$$, then $$u = Av\neq 0$$ and satisfies $$AA^Tu = AA^T(Av) = A(A^TA)v = \lambda Av = \lambda u,$$ so $$u, \lambda$$ is an eigenpair of $$AA^T$$. Therefore, $$A^TA$$ have the same nonzero eigenvalues and the maximum in both norm equations will be the same, so $$\|A\|_2 = \|A^T\|_2$$.