I came across a claim that given $A\in\mathbb{R}^{n\times n}$ the following equality holds $$\left\Vert A\right\Vert _{2}=\sup\left\{ \text{trace}\left(ZA\right)\ :\ Z\succeq0\right\}$$ Where $\left\Vert A\right\Vert _{2}=\sup_{\left\Vert x\right\Vert _{2}=1}\left\Vert Ax\right\Vert _{2}$ w.r.t the Euclidian norm on $\mathbb{R}^{n}$. I can't figure out whether this is actually correct, if it is I would be happy to see a proof.
Characterization of spectral norm: $\left\Vert A\right\Vert _{2}=\sup\left\{ \text{trace}\left(ZA\right)\ :\ Z\succeq0\right\}$
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linear-algebra
matrices
norm
1 Answers
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Do you have some restrictions on $Z$? otherwise if we take $Z$ to be the identity matrix, then your proposition would say $||A||_2 \geq Trace(A)$ which is untrue.
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0Yeap, that easily shows the claim is untrue. I think the person that wrote the claim thought the induced 2-norm is self-dual like the vector 2-norm. – 2017-01-18