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

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

• You should show first that $S,T$ are positiv definite: $\langle x,Sx\rangle>0$ if $x\neq0$ (for example using diagonalisation). Then apply this to your eigenvalue equation for $S+T$. – Helmut Apr 19 at 13:06
• Hey. I write an answer to your question and then you delete this. Don't you think this is rude? I take the time to help you and then you do that! – EpsilonDelta Jul 20 at 10:42
• @EpsilonDelta,I am extremely sorry but I wrote the wrong question.Kindly excuse for this – Math_Freak Jul 20 at 10:44