Skew Symmetric Determinant Proof

In summary, the proof shows that for a skew symmetric matrix with an odd number of rows, the determinant is always equal to zero. This is because when n is odd, the determinant can be rewritten as (-1)^n*det(A), which simplifies to -det(A). Since det(A) is equal to its negative, it must be zero.
  • #1
jolt527
8
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Hi all! I was working on some homework for the linear algebra section of my "Math Methods for Physicists" class and was studying skew symmetric matrices. There was a proof I saw on Wikipedia that proves that the determinant of a skew symmetric matrix is zero if the number of rows is an odd number.

[tex]det(A) = det(A^T) = det(-A) = (-1)^n*det(A)[/tex]

This is followed up by, "Hence, det(A) = 0 when n is odd." The problem is that I don't understand the proof too well. I understand that the determinant of a matrix is equal to the determinant of its transpose. That means that the determinant of the negation of a matrix is equal to those as well (-A = A^T). Looks like the (-1)^n*det(A) means that multiplying each row by (-1) will produce the same result as the other derivations so far.

If my logic is sound up to this point, then I get it all, until the big leap to, "Hence, det(A) = 0 when n is odd." Could someone point out either a flaw in my previous logic, or help me to understand how they get to the idea that det(A) must be zero when n is odd? Thank you! :)
 
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  • #2
Okay, my math skills must be low tonight, because I think I got it. :P If someone would confirm what I'm thinking, that'd be nice!

In the end we get that det(A) = (-1)^n*det(A). If n is odd, we get det(A) = -det(A), which is only possible when det(A) is zero. Does that sound right?
 
  • #3
jolt527 said:
In the end we get that det(A) = (-1)^n*det(A). If n is odd, we get det(A) = -det(A), which is only possible when det(A) is zero. Does that sound right?
Pretty much sums it up.
 

1. What is a skew symmetric matrix?

A skew symmetric matrix is a square matrix whose transpose is equal to its negative. This means that the elements below the main diagonal are the negatives of the elements above the main diagonal.

2. What is the determinant of a skew symmetric matrix?

The determinant of a skew symmetric matrix is always equal to 0. This can be proven using the properties of determinants and the fact that the transpose of a skew symmetric matrix is equal to its negative.

3. How do you prove that the determinant of a skew symmetric matrix is 0?

The proof involves using the properties of determinants and the fact that the transpose of a skew symmetric matrix is equal to its negative. By expanding the determinant, it can be shown that all the terms cancel out and the result is 0.

4. Why is the determinant of a skew symmetric matrix important?

The determinant of a skew symmetric matrix is important because it is a useful property in linear algebra. It can be used in various applications such as solving systems of linear equations, finding eigenvalues and eigenvectors, and determining if a matrix is invertible.

5. Can a skew symmetric matrix have a non-zero determinant?

No, a skew symmetric matrix cannot have a non-zero determinant. This is because the transpose of a skew symmetric matrix is equal to its negative, and the determinant of the transpose is equal to the determinant of the original matrix. Therefore, the determinant of a skew symmetric matrix is always 0.

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