MHB Can You Solve These Advanced Determinant Problems Involving Matrices?

[Yankel]

Let A be an nXn real matrix

(a) show that if the transpose of A equals -A, and n is odd, then the determinant of A is 0.

(b) show that if (A*A)+I=0, then n must be even.

(c) if all the values of A are either 1 or -1, show that the determinant of A is divisible by (2^n-1).

these are hard questions, I have no clue of to even start them...a solution to any of the questions will be appreciated !

thanks
(Yes)

 

[Fantini]

For the first, apply the determinant to both sides, using the facts that $\det (kA) = k^n \det A$ and $\det A^t = \det A$.

As for the second, what do you mean with $A^{\ast} A$? Would it be $A^t A$?

 

[Yankel]

For the first, apply the determinant to both sides, using the facts that $\det (kA) = k^n \det A$ and $\det A^t = \det A$.

As for the second, what do you mean with $A^{\ast} A$? Would it be $A^t A$?

nein, I meant A squared, I just don't know how to do it with Latex...sorry

 

[Fantini]

Oh, you meant $A \ast A$ as $A \cdot A = A^2$. Well, try writing $A^2 = - I$ and taking the determinant. It follows that $\det (A^2) = (\det A)^2 = \det(-I) = (-1)^n \det I = (-1)^n$. This equation will be true if and only if $n$ is even, because the determinant takes on real values and there are no real numbers whose square is negative.

Don't know about the third, sorry.

 

[awkward]

Let A be an nXn real matrix

(a) show that if the transpose of A equals -A, and n is odd, then the determinant of A is 0.

(b) show that if (A*A)+I=0, then n must be even.

(c) if all the values of A are either 1 or -1, show that the determinant of A is divisible by \( 2^{n-1} \).

these are hard questions, I have no clue of to even start them...a solution to any of the questions will be appreciated !

thanks
(Yes)

Here is a sketch of a proof of (c) by induction on n. The idea of filling in all the details (in LaTex) makes me feel faint, so I will leave that to you.

The case n=1 is easy.

Now suppose (c) holds for some value of n, and let A be an n+1 by n+1 matrix all of whose elements are either -1 or 1. Add the second row of A to the first row. This does not change the determinant. But in the new matrix, every element of the first row is the sum of two numbers, each of which is either -1 or 1, hence is either -2, 0, or 2. Therefore each element of the new first row is divisible by 2. Now expand the determinant by minors about the first row, and apply the inductive hypothesis.

 

[Opalg]

This is a modification of awkward's idea in the previous comment. Given an $n\times n$ matrix in which each entry is $\pm1$, subtract the top row from each of the other rows. You can then take a factor 2 out of each of those rows, giving a factor of $2^{n-1}$ in the determinant.