$\left \langle A|B \right \rangle=\text{tr}(A+B)$ defines an inner product

Question

Prove if the following statement is true or false.

Let $V=M_{n}(\mathbb{R})$. Then $\left \langle A|B \right \rangle=\text{tr}(A+B)$ defines an inner product.

Attempt:

Consider a nilpotent matrix $A$. $\left \langle A,A \right \rangle=2\text{tr}(A)=0$, but $A$ is not necessarily $0$. Therefore, $\left \langle A|B \right \rangle=\text{tr}(A+B)$ does not define an inner product because it does not comply with the positive definiteness.

Answer 1

You can also consider the matrix $-I_{n}$. Then,

$$\left \langle -I_{n}|-I_{n} \right \rangle=\text{tr}(-2I_{n})=-2n<0,\forall n \in \mathbb{N}.$$

Answer 2

Alternatively, you could just note that $\varphi$, where $\varphi(A,B) = \operatorname{tr}(A+B)$, is not bilinear.

Rather, for any $c\in \mathbb{R}$, we have $$c\varphi(A,B) = c\operatorname{tr}(A+B) = \operatorname{tr}(cA+cB) = \varphi(cA,cB)$$ so if it were bilinear, this would $= c^2\varphi(A,B)$, and choosing any $c \neq 0,1$ would imply $\varphi(A,B) = 0$ for every $A,B$, which is apparently false.