Proof that linear combination of self adjoint maps is also self adjoint.

Question

I want to show that if $V$ is an inner product space and $S,T\in \mathcal{L}(V)$ are self-adjoint linear maps, then $aS+bT$ is a self-adjoint linear map for all $a,b\in \mathbb{R}$. From what I tried, I feel like I need to play with the properties of the inner product in order to shaw that $aS$ and $bT$ are self adjoint, but I am not sure how to proceed, if anyone could help.

Answer 1

An operator $T:V\rightarrow V$ is self adjoint if $$\left<Tu,v\right> = \left<u,Tv\right>$$ for all $u,v\in V$.

Fix $u,v\in V$ we want to prove that if $T$ and $S$ are self adjoint, then

$$\left<(aT+bS)u,v\right> = \left<u,(aT+bS)v\right>$$

Indeed, by the bi-linearity of the inner product

$\left<(aT+bS)u,v\right> = \left<aTu + bSu,v\right> = \left<aTu,v\right>+\left<bSu,v\right>=...=\left<u,(aT+bS)v\right>$

can you complete the "..." on your own?