Do real matrices always have real eigenvalues?

I was trying to show that orthogonal matrices have eigenvalues $1$ or $-1$.

Let $u$ be an eigenvector of $A$ (orthogonal) corresponding to eigenvalue $\lambda$. Since orthogonal matrices preserve length, $ \|Au\|=|\lambda|\cdot\|u\|=\|u\|$. Since $\|u\|\ne0$, $|\lambda|=1$.

Now I am stuck to show that lambda is only a real number. Can any one help with this?

Solution 1:

The eigenvalues of $$ \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} $$ are $\cos\theta \pm i\sin\theta= e^{\pm i\theta}$. This is an orthogonal matrix.

If a matrix with real entries is symmetric (equal to its own transpose) then its eigenvalues are real (and its eigenvectors are orthogonal). Every $n\times n$ matrix whose entries are real has at least one real eigenvalue if $n$ is odd. That is because the characteristic polynomial has real coefficients so the complex conjugate of a root is another root, and you can't have an odd number of roots if they come in pairs of distinct entries.

But generally, the eigenvalues of matrices with real entries need not be real.

Solution 2:

No, a real matrix does not necessarily have real eigenvalues; an example is $\pmatrix{0&1\\-1&0}$.

On the other hand, since this matrix happens to be orthogonal and has the eigenvalues $\pm i$ -- for eigenvectors $(1\mp i, 1\pm i)$ -- I think you're supposed to consider only real eigenvalues in the first place.

Solution 3:

I guess it depends whether you are working with vector spaces over the real numbers or vector spaces over the complex numbers.In the latter case the answer is no, however in the former the answer has to be yes.Is it not guys ?