# Thread: Antisymmetric matrix.

1. ## Antisymmetric matrix.

Let A be real antisymmetric matrix with size of nxn, when n is odd.

Prove that 0 is eigenvalue of A.

2. Originally Posted by Also sprach Zarathustra
Let A be real antisymmetric matrix with size of nxn, when n is odd.

Prove that 0 is eigenvalue of A.
Hint: Show that $\det(A) = -\det(A)$, so that $\det(A)$ has to be 0.

3. Thanks, but how I continue from your hint please...?

p(x)=det(Ix-A) =?

4. The idea is that if a matrix has determinant 0, then it has 0 as an eigenvalue. Try to prove it if you can't see it.

5. 0 is an eigenvalue iff there exist vector v!=0 so A(v)=0v=0...

Thank you all...

6. The determinant of a matrix is the product of its eigenvalues. If the determinant is 0, then one of the eigenvalues must be 0.

7. http://www.mathhelpforum.com/math-he...rexamples.html

Post 3 shows that the det is the product of the eigen values.