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I dont really understand what is going on in this question but I gave it a go anyway. I calculated $A^tB$ and got it's trace by adding the numbers on the diagonal. The trace was 0.

I then used the defintion of the dot product as $\langle a,b\rangle= ||a||||b||cos\theta$ I figured that as $\langle A,B\rangle = trace(A^tB)=0$, then $\theta$ must equal $\pi/2$.

This shows up as correct. However I dont know how to get the norm of A & B. I didnt even know a matrix had such a thing as a norm/length.

I also dont understand the wording of the question where they are saying $\langle A,B\rangle = trace(A^tB)$, how can you dot one matrix with another, I never heard of such a thing, we never covered it in class anyway.

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up vote 2 down vote accepted

I think you've misunderstood the inner product. Remember, not every inner product is the dot product; the dot product is a type of inner product.

You're told what $\langle A, B \rangle$ is here $-$ it's defined by $\langle A, B \rangle = \text{trace}(A^TB)$. You can't then just redefine it to be something else!

In general if $\langle\,,\rangle$ is an inner product then $\langle A, A \rangle = \lVert A \rVert^2$ and so on. This is how you calculate $\lVert A \rVert$ here.

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I was thinking it was something like that, defining the inner product to be something. So using that defintion of the inner product is $\langle A, A \rangle = trace(A^tA)?$ – Jim_CS Apr 11 '12 at 16:22
Yes.${}{}{}{}{}$ – Gerry Myerson Apr 12 '12 at 2:19

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