A real nxn matrix always has a real eigenvalue when n is odd what do it mean by "A real nxn matrix always has a real eigenvalue when n is odd?" Shouldnt it be that regardless of odd or even there will always have a real root?
hope to hear from you guys soon:)
 A: 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 $n$.
A: 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. 
A: 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.
A: The eigenvalues of a matrix are roots of the characteristic polynomial $p(x)$ of the matrix. If $n$ is even, then the degree of $p(x)$ is even, which means each complex root has its conjugate as a root as well (that's not to say there can't be real roots, there would just have to be an even number of real roots). If $n$ is odd, then the degree of $p(x)$ is odd, which means at least one root would have to be real. This left over root is a real eigenvalue of the matrix.
