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

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.

• Hint: the adjoint operation $(-)^*:\mathcal{L}(V)\to\mathcal{L}(V)$ is linear. Nov 30, 2022 at 4:35

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

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

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

Indeed, by the bi-linearity of the inner product

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

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

• Thank you very much, I think I can complete it on my own now Nov 30, 2022 at 4:41