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I am reading Franz Hohn's Elementary Matrix Algebra (1973) and having trouble solving the following exercise:

Prove that, if $A$ and $B$ are both of order $n$,

(a) $\det A^{T}B = \det A B^T = \det A^T B^T = \det AB$

(b) $\det A^*B^* = \overline{\det AB}$.

My trouble is that the author has not yet proven the multiplicative property $\det AB = \det A \det B.$ If I could use that property (together with $\det A^T = \det A$ and $(AB)^T = B^T A^T$) then the exercise would be trivial. If I could get the first equality in (a) then I could get the rest of the problem.

While attempting to solve this exercise I ended up just proving the multiplicative property, but I don't think that's what the author intends. Am I missing something simple? Any hint is greatly appreciated.

Community
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Leonard Blackburn
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1 Answers1

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We can use that $\det(C^T)=\det(C)$ and $\det(AB)=\det(BA)$ as follows: $$ \det(A^TB)=\det(A^TB)^T=\det(B^TA)=\det(AB^T) $$ At some point we need to prove the "classical" properties, like $\det(AB)=\det(BA)$ or even better, $\det(AB)=\det(A)\det(B)$. And then we do not care too much whether the text has it at page $xy$ or not.

Dietrich Burde
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  • Hmm, isn't $(A^T B)^T = B^T A$ and not $A B^T$? Or are you using another property, that $\det AB = \det BA$ (which also hasn't been stated or proved in the text)? Does that have a separate proof from the multiplicative property? – Leonard Blackburn Feb 26 '19 at 19:33
  • Ok, I was just wondering if there was a separate simpler proof of the equations in my question that didn't utilize det(AB) = det(A)det(B). But since I'm not getting an answer I'm guessing there probably isn't. It just seems like a misplaced exercise. – Leonard Blackburn Feb 27 '19 at 14:00
  • But I can't accept your answer since it doesn't really answer my question, which was to prove the equations without using the multiplicative property. But I think my question is a bad question that I never should have asked. It doesn't seem right to ask what the author's intent was. – Leonard Blackburn Feb 27 '19 at 14:03
  • Don't worry about acceptance. I think the best thing to do is to study the proof that $\det(AB)=\det(A)\det(B)$ holds somewhere else, e.g., at this duplicate. Then your question is, as you said, a trivial consequence. – Dietrich Burde Feb 27 '19 at 14:06
  • Thanks for your help! (I've already proved it myself and then three different proofs in the book by Hohn I mention in the question.) – Leonard Blackburn Mar 01 '19 at 17:39