Proving the Spectral Theorem in Matrix Algebra

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Homework Statement



Prove that for any square matrix A, ker A= (im A*)\bot

Homework Equations



A* = A conjugate transpose

The Attempt at a Solution



so we want to show that all A that satisfies Ax=0 for some x, that set is equal to (im A*)\bot

But i really don't know what image of this could possibly be... we arent given a function.
 
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SNOOTCHIEBOOCHEE said:
But i really don't know what image of this could possibly be... we arent given a function.

it holds for any function.

prove it for any function (you should be able to)
 
So the image of the null space of A* is equal to the null space of A.

Im thinking this proof leads me in the direction that I am A* = A. then we are done. but i don't know how to show that.
 
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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