Exercise 7.A.15

Proof. Let $w_1,w_2\in V$. We have

Hence $T^{\text{∗}}v=⟨v,x⟩u$.

(a) Suppose $T$ is selft-adjoint. Then

for all $v\in V$. We can assume $u$ and $x$ are non-zero (otherwise there is nothing to prove). Taking $v=u$ forces $⟨v,u⟩\ne 0$, showing that $x$ and $u$ are linearly dependent.

Conversely, suppose $x$ and $u$ are linearly dependent. We can assume $x$ and $u$ are non-zero, otherwise $T$ would equal $0$, which already is self-adjoint. Then $u=\mathit{cx}$, for some non-zero $c\in ℝ$. Thus

Therefore $T=T^{\text{∗}}$.

(b) Again, we can assume $u$ and $x$ are both non-zero in both directions of the proof.

We have

Taking $v=u$ ensures $⟨⟨v,u⟩x,x⟩\ne 0$, showing that $u$ and $x$ are linearly dependent.

Conversely, suppose $x$ and $u$ are linearly dependent. Then $u=\mathit{cx}$ for some non-zero $c\in 𝔽$. Then

Hence $T{T}^{\text{∗}}=T^{\text{∗}}T$. □