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We know that the spectral norm of a matrix $A \in \Bbb K(n,n)$

$$\left \| A \right \| _2=\sqrt{\lambda_{\text{max}}(A^{^*}A)}$$

I need to prove that multiplying with an unitary matrix $U \in U(n)$ from the left or right does not change the value of the norm i.e. $$\left \| A \right \| _2 = \left \| UA \right \| _2 = \left \| AU \right \| _2$$ I was able to prove that $$\left \| UA \right \| _2=\sqrt{\lambda_{\text{max}}((UA)^{^*}(UA))}$$ $$=\sqrt{\lambda_{\text{max}}(A^{^*}U^{^*}UA)}$$ $$=\sqrt{\lambda_{\text{max}}(A^{^*}A)}=\left \| A \right \| _2$$ since for unitary matrices $$U^{^*}U = I$$

I have no idea how to prove the the argument when multiplying from the right. This has been my approach. $$\left \| AU \right \| _2=\sqrt{\lambda_{\text{max}}((AU)^{^*}(AU))}$$ $$=\sqrt{\lambda_{\text{max}}(U^{^*}A^{^*}AU)}$$ I've really got no idea on what to do next. Any help is much appreciated.

thErgonomic
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1 Answers1

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The key fact is that the nonzero eigenvalues of $AB$ and $BA$ are always the same. So in particular $$\lambda_{\text{max}}(U^* A^* A U) = \lambda_{\text{max}}(U U^* A^* A) = \lambda_{\text{max}}(A^* A)$$

Robert Israel
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