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Can someone please explain to me how for $Z$, $V \epsilon S^n$, $t \epsilon R$ $$(Z+tV) = Z^{1/2}(I + tZ^{-1/2}VZ^{-1/2})Z^{1/2}$$

I think I'm missing some fundamental aspect of matrix algebra. I've been using the matrix cookbook for help

edit: I removed the first part because in typing it out for the question, I realized my mistake!

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Just distribute the first $Z^{1/2}$, then the second $Z^{1/2}$. There's nothing complicated going on here. – Qiaochu Yuan May 30 '12 at 20:24
up vote 1 down vote accepted

First multiply the $Z^{\frac{1}{2}}$ on the left through the parentheses. $$ Z^\frac{1}{2}(I+tZ^{-\frac{1}{2}}VZ^\frac{1}{2})Z^\frac{1}{2}=(Z^{\frac{1}{2}}+tZ^{-\frac{1}{2}+\frac{1}{2}}VZ^{-\frac{1}{2}})Z^{\frac{1}{2}}=(Z^{\frac{1}{2}}+tVZ^{-\frac{1}{2}})Z^{\frac{1}{2}} $$ Remember that $Z^0=I$. Now multiply the $Z^{\frac{1}{2}}$ on the right through the parentheses. $$ Z^{\frac{1}{2}+\frac{1}{2}}+tVZ^{-\frac{1}{2}+\frac{1}{2}}=Z+tV $$

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