A real nxn matrix always has a real eigenvalue when n is odd

Solution 1:

No, this is a consequence of the fact that an odd-degree polynomial always has a real root. The same statement is not true for even-degree polynomials.

Notice that the eigenvalues are the solutions to the equation $\det(A-xI)=0$, and that the LHS of this equation is just a polynomial in $x$ with degree $n$.

Solution 2:

Some polynomials don't have any real root. For example $X^2+1$ cannot be zero when $x$ is real.

It can be proven, using the intermediate value theorem or using fundamental theorem of algebra, that every polynomial of odd degree has at least 1 real root.

But the same is not always true for polynomials of even degree.

Solution 3:

No -- if $n$ is even, then an $n\times n$ matrix does not necessarily have a real eigenvalue -- for example, $$ \begin{pmatrix}0&1\\-1&0\end{pmatrix} $$ doesn't.

The reason why even/odd matters is that the characteristic polynomial of an $n\times n$ matrix has degree $n$, and real polynomials of odd degree always have at least one real root.