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I'm going through my class notes and trying to follow along with the math here:

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However, in the last step I'm unsure why you are able to 'divide' the $\mathbf{y}$ vector out of the norm $||A^{-1}\mathbf{y}||$, is that a valid operation since it's a scalar value?

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    Given a square matrix $M$, the definition of $\|M\|$ being used is $$\| M \| = \max_{\mathbf{y} \neq 0} \frac{\| M \mathbf{y} |}{\| \mathbf{y}\|} $$ Or if the norm has been defined othewise, this must be an ensuing property. So the last step is not a "division", but simply an application of the definition of norm. – 2017-01-16
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    okay, i didnt see that thank you! – 2017-01-16

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