## Real symmetric matrices have real eigenvalues

A real matrix is symmetric if $A^t=A$. I will show in this post that a real symmetric matrix have real eigenvalues.

I will need a dot product for the prof and I’ll use the basic dot product for two vectors $X$ and $Y$: $=X^t\overline{Y}$, where $\overline{Y}$ is the complex conjugate of the vector $Y$.

The useful property of this dot product is that $=$, for any matrix $A$

And considering that $A$ is real, a simple proof is:

$=(AX)^t\overline{Y}=X^tA^t\overline{Y}=X^t\overline{A^tY}=$

An eigenvalue have a correspondent eigenvector: $AX=\lambda X$.

We have $=<\lambda X, X>=\lambda $ and considering that A is symmetric $====\overline{\lambda}$.

From $\lambda=\overline{\lambda}$ and because $X$ is not a zero vector it results that the imaginary part of $\lambda$ is zero, so the eigenvalue is a real number.