Assume T is an operator on a finite dimensional real vector space and all the eigen values are real. Prove that Tr($T^2$) $\ge 0$.

Question

Assume T is an operator on a finite dimensional real vector space and all the eigen values are real. Prove that Tr($T^2$) $\ge 0$.
I considered using the fact that the sum of all eigenvalues equal to trace($T$). But I can't connected it to trace($T^2$). May anyone help?

Answer 1

If all eigenvalues are real, then the matrix $M$ of $T$ with respect to some basis is an upper triangular matrix such that the entries of the main diagonal are the eigenvalues $\lambda_1,\ldots,\lambda_n$ of $T$. So\begin{align}\operatorname{tr}(T^2)&=\operatorname{tr}(M^2)\\&=\sum_{k=1}^n{\lambda_k}^2\\&\geqslant0.\end{align}

Answer 2

If $ \lambda $ is an eigenvalue of $T^2$, then there is an eigenvalue $ \mu$ of $T$ such that $\lambda = \mu^2$. Since $ \mu $ is real, we have that $ \lambda \ge 0$.