What is the Commutative Property of Vector Multiplication?

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Homework Statement



I am trying to prove an identity and in order to finish the proof i need to show that:
for z a p-dimensional vector and A a non-singular square (pXp) matrix - also note that T is transpose - :
zTA^(-1)z = zzTA^(-1), where -1 denotes the inverse

I have looked everywhere to find vector multiplication that commutes but so far can't find a rule...


Thanks for your help!
 
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That doesn't look right to me...z^TA^{-1}z is a scalar (assuming z is a column vector), while zz^TA^{-1} is a p\times p matrix...what was the original problem?
 
Hey Gabby - thank you for pointing that out (i can't believe i didnt see that!).

I will try it again now - maybe I have done some stupid mistake.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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