Let A be real antisymmetric matrix with size of nxn, when n is odd. Prove that 0 is eigenvalue of A.
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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 $\displaystyle \det(A) = -\det(A)$, so that $\displaystyle \det(A)$ has to be 0.
Thanks, but how I continue from your hint please...? p(x)=det(Ix-A) =?
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.
0 is an eigenvalue iff there exist vector v!=0 so A(v)=0v=0... Thank you all...
The determinant of a matrix is the product of its eigenvalues. If the determinant is 0, then one of the eigenvalues must be 0.
http://www.mathhelpforum.com/math-he...rexamples.html Post 3 shows that the det is the product of the eigen values.
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