From Jeffrey Holts Linear Algebra with Applications 65 48 Pr

From Jeffrey Holt\'s Linear Algebra with Applications:

6.5 #48

Prove that if n is odd and A is a real n x n matrix, then there exists a nonzero vector u such that Au = cu, where c is a real number.

Solution

If n is odd then characteristic equation of A has an odd degree.

We know the complex roots occur in pairs ie if x , a complex number, is a root of characteristic polynomial then conjugate of x is also root of characteristic roots

So there are even number of complex roots of characteristic polynomial

But, polynomial has odd degree and has n roots

So, it must have at least one real root.

Hence it has at least one real eigenvalue, c

And u is it\'s corresponding eigenvector so that

Au=cu