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


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

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

Will someone give some way to solve it?

Thank you

  • $\begingroup$ 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 $S+T$. $\endgroup$ – Helmut Apr 19 at 13:06
  • $\begingroup$ 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! $\endgroup$ – EpsilonDelta Jul 20 at 10:42
  • $\begingroup$ @EpsilonDelta,I am extremely sorry but I wrote the wrong question.Kindly excuse for this $\endgroup$ – Math_Freak Jul 20 at 10:44

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