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.
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piero
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2What is your definition of $|A|_2$ for a matrix A? – krm2233 Apr 15 '23 at 20:38
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@krm2233 the maximum value of $| Ax |_2$ in the unit ball – piero Apr 15 '23 at 20:40
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A proof from first principles:
$$ \|A\|_2 = \sup_{x: \|x\| = 1} \|Ax\| = \sup_{x, y: \|x\|, \|y\| = 1} \langle y, Ax \rangle = \sup_{x, y: \|x\|, \|y\| = 1} \langle A^{\mathsf{T}} y, x \rangle = \sup_{y: \|y\| = 1} \|A^{\mathsf{T}} y\| = \|A^{\mathsf{T}}\|_2. $$
Above, I'm denoting $\langle x, y \rangle = x^{\mathsf{T}} y $.
VHarisop
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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 $\|A\|_2 = \|A^T\|_2$.
whpowell96
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