What are the possible values of determinant of a nxn matrix if n odd vs even

Question

Assume that A is an n x n matrix with complex entries such that $$A^TA = -I_n$$What are the possible values of det(A) if (i) n is even, (ii) n odd?

Justify your answers, clearly stating any properties of determinants that you use in your solution.

I have been struggling with this question for a long time and still can't make a start. I have been toying with $$det(A^t) = det(A)$$

$$det(A)(B) = det(A) det(B)$$

but I cannot make a plausible start. Please guide a bit. Thanks in advance.

Answer 1

I cannot make a plausible start

I'd say you're nearly finished. Of course, you haven't gotten anywhere on writing a solution yet, but that's just the final step. Finding the solution is the largest and most important step (sadly, it's mostly invisible if you read books or attend lectures), and here you have done plenty of progress.

You've found that $\det(A^t) = \det(A)$, and you've found that $\det(AB) = \det(A)\det(B)$. Combining these, we get that $\det(A^tA) = \det(A)^2$.

The determinant of $-I_n$ is $(-1)^n$, which differs depending on whether $n$ is even or odd.

If $n$ is even, the determinant of $-I_n$ is $1$, and we have $\det(A)^2 = 1$. What are the two possible values of $\det(A)$ in this case? If, on the other hand, $n$ is odd, the determinant of $-I_n$ is $-1$, and we get $\det(A)^2 = -1$. What are the two possible values in this case?

Answer 2

$$ \big(\det(A)\big)^2=\det(A^T)\det(A)=\det(A^TA)=\det (-I)=(-1)^n $$ hence $$ \det(A)=\left\{ \begin{array}{ccc} \pm i & \text{if $n$ odd}, \\ \pm 1 & \text{if $n$ even.} \end{array} \right. $$