linear algebra problem?
A square matrix is called skew- symmetric. if A transpose= -A Prove that if A is an n x n skew- symmetric matrix, then I A I= (-1)^n I A I. Use the result to prove that I A I= 0
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A square matrix is called skew- symmetric. if A transpose= -A Prove that if A is an n x n skew- symmetric matrix, then I A I= (-1)^n I A I. Use the result to prove that I A I= 0
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Suppose that A is skew-symmetric. Then, A^t = -A.
Taking determinants, |A^t| = |-A|.
However, |A^t| = |A|, and |-A| = (-1)^n * |A|, since we are factoring a (-1) from each of the n rows of A.
Thus, |A| = (-1)^n |A|.
==> If n is odd, then |A| = -|A| ==> |A| = 0.
Note: This is not true if n is even. For example,
[0 1]
[-1 0] is skew symmetric, but its determinant is 1.
I hope this helps!