Problem for proving a subspace

Question

I have this problem:

Let (V,⟨−, −⟩) be a finite dimensional real inner product space. Prove that the set of self-adjoint linear maps T : V → V is a subspace of the set of all linear maps T : V → V .

I know I must prove 1. non-empty 2. closure under scalar mult. 3. closure under vec. add.

I proved non-empty using the definition of adjoint and the 0 map.

I am kind of stuck for the other two. I know, for example, for scalar multiplication, by definition $\lambda T(v) = (\lambda T)(v)$, and I am not sure what there is to show. My professor had stated that I could use the properties of the adjoint, but I am not sure where they come in use.

All I can really say is that I know that if T is self-adjoint then it corresponds to a symmetric matrix, and the scalar multiple of a symmetric matrix is still symmetric. Is this the right path?

Answer 1

Here are some hints.

Suppose $T,U$ are self-adjoint operators on the (finite dimensional) real inner product space $V$ and $\lambda$ is a real number.

What you have to show are thee two following facts：

• $(T+U)^*=T^*+U^*$
• $(\lambda T)^*=\lambda T^*$

where $T^*$ is the adjoint operator of $T$.

(actually these are the standard properties of adjoint operators)

And then together with the assumptions $T=T^*,U=U^*$, you may make your conclusion.