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I don't understand why one should take transpose of $\operatorname{tr}(AB^T)$ and why we use the fact that $\operatorname{tr}(M)=\operatorname{tr}(M^T)$ for any $M$ that is a square matrix to solve the problem.

Community
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Kesha
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    Your question lacks context. – mvw May 01 '16 at 04:55
  • Go ahead and remove the transpose. Try to prove that the resulting expression is an inner product. Then you'll understand. (For non-square matrices it won't even be defined, and for square matrices positivity fails.) –  May 01 '16 at 04:56
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    If you don't have the transpose there, $AB$ is not defined unless $n=m$. – Batman May 01 '16 at 04:59

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If you grind through the details, you will see that $\operatorname{tr} (A B^T) = \sum_{i,j} [A]_{ij} [B_{ij}]$, hence this is the 'standard' inner product if you view the matrices $A,B$ as giant columns.

copper.hat
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