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If $A^T . A = I$, prove that determinant A = +-1.

I don't even know where to start. Can somebody please give me a good start at least.

Hafiz Temuri
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1 Answers1

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Since $\det(A)=\det(A^T)$ we have $\det(AA^T)=\det(I)=1$. This factors to $\det(A)\det(A^T)=1$ which in the real case we know means $\det(A)=\det(A^T)= \pm 1$.

Eoin
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  • Sorry, I don't understand how you prove that $det(A)=det(A^T)=+-1$ – Hafiz Temuri Nov 03 '14 at 05:49
  • Let $a=\det(A)$. We know, from here that $\det(A)=\det(A^T)$. Then we may substitute to obtain $a^2=1$. Then we have $a=\sqrt{1}=\pm 1$. – Eoin Nov 03 '14 at 05:52
  • gotcher, sorry I am kinda slow in MATHS. And its asking me to wait for 6min. I'll mark your answer as correct once the wait time over. – Hafiz Temuri Nov 03 '14 at 05:55
  • @TalhaTemuri Don't apologize! If you're ever confused you should ask. I tend to forget to add details. – Eoin Nov 03 '14 at 05:56
  • @TalhaTemuri No problem. Thanks for asking something I've learned before ;) – Eoin Nov 03 '14 at 05:56