Show that $S+T$ has all eigen values non-negative.

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Let $V$ be a finite dimensional vector space and $S,T\in \mathcal L(V)$ where $\mathcal L(V)$ denotes the space of linear operators.

If $S,T$ are self-adjoint and have all eigen values positive show that $S+T$ has all eigen values non-negative.

I tried like this:

Let $\lambda $ be an eigen value of $S+T$ and let $(S+T)v=\lambda v$

Now since $S^{*}=S$ and $T^{*}=T\implies (S+T)^{*}=S+T$

Thus $\langle (S+T)v,(S+T)v\rangle =\langle \lambda v,\lambda v\rangle=|\lambda|^2\langle v,v\rangle$

Also

$\langle (S+T)v,(S+T)v\rangle=\langle v,(S+T)^2v\rangle$

How to show that $\lambda\ge 0 $ from above?

Will someone give some way to solve it?

Thank you