Positive definite operator/matrix question

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Homework Statement



Prove that T is positive definite if and only if

\sum_{i,j} A_{ij}a_{j}\bar{a_{i}} > 0
for any non-zero tuple (a1, ...... , an )

Let A be [ T ]_{\beta}

where \beta is an orthogonal basis for T

The Attempt at a Solution



the sum looked like the matrix multiplication of a n-tuple and a matrix A, so I looked into that and couldn't get anything.. any hints please? I'm also struggle to realize what significant that sum could have, right now it doesn't even mean anything to me.

Thanks!
 
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If the capital A's are supposed to be terms in matrix, you might want to think about

\overline{a}^tAa
 
Office_Shredder said:
If the capital A's are supposed to be terms in matrix, you might want to think about

\overline{a}^tAa

thanks, someone else told me to think about a^{*} A a , but I'm not sure why I'd be considering this. I basically have these things to work with: that if T is positive definite ( one way of the implication ), then all its eigenvalues are positive,T is self-adjoint and < Tx , x > > 0.

Thanks
 
I'm also confused a*A might not make sense.. if a is a column vector in F^n and A is a matrix?
 
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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