I'm trying to show that for any square matrix (whose entries are all real) the eigenvalues are real or are complex conjugate pairs.

I've tried so far by stating that for 2*2 matrices, finding the determinant of (A-$\lambda$I) always leads to a quadratic which inherently yields real or complex roots of the form of $z=a\pm\ b$ although I doubt that this counts as sufficient proof for those cases anyway.

Am I on the right track at least? Am I also assuming incorrectly that any square matrix beyond 2*2 dimensions has eigenvalues that are either real or complex conjugate pairs?

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    $\begingroup$ If $A$ is a real $n \times n$ matrix, then the characteristic polynomial $P(\lambda) = \text{det}(A-\lambda I)$ is always a real polynomial of degree $n$ (You should be able to show this). Then, you just have to prove that the roots of any real polynomial are all real or come in complex conjugate pairs. $\endgroup$ – JimmyK4542 Aug 10 '14 at 6:20

The eigenvalues are the roots of the characteristic polynomial, and the coefficients of the characteristic polynomial are real since they depends on the element of the matrix. This gives that complex eigenvalues come in conjugate pairs.


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