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If I am given $Y\in \mathbb{C}^{n \times n}$ is nonsingular and a matrix norm defined as $\|A\|_Y = \|Y^{-1}AY\|_2$, how do I show that it has the property: $$\|AB\|_Y \leq \|A\|_Y\|B\|_Y$$

At first I was thinking that $ \|Y^{-1}AY\|_2$ is equal to the max singular value of $A$ because it looks like a similarity transform; however, I don't think that is valid.

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  • $\begingroup$ Do you already have submultiplicativity for the $2$-norm? $\endgroup$ – Chappers Apr 19 '18 at 16:59
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Hint:

Note that\begin{align} \|AB\|_Y &= \|Y^{-1}ABY\|_2\\ &=\|(Y^{-1}AY)(Y^{-1}BY)\|_2 \end{align}

Try to complete it using submultiplicative property.

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