Let B be a bilinear form on $M_n(\Bbb{C})$, $B(X,Y)=n\text{tr}(XY)-\text{tr}(X)\text{tr}(Y)$. Find $\text{rad }V$.

Question

There is another part of the problem which asks-
Let $V_1=\{X\in V|\text{tr}(X)=0\}$ (here $V=M_n(\Bbb{C})$). Prove that $V=V_1\oplus \text{rad }V$.
I can prove that $V_1$ is non-degenerate. And B is symmetric, hence reflexive. Thus $V=V_1\oplus V_1^{\bot}$.
$\text{rad }V=\text{rad }V_1^{\bot}$.
Now, my claim is $\text{rad }V=V_1^{\bot}$. To prove this I have $\text{rad }V_1^{\bot}=V_1^{\bot}$ or equivalently $B(X,Y)=0\ \forall X,Y\in V_1^{\bot}$
So, my target is show that for any $X,Y\in V_1^{\bot}$, we have $B(X,Y)=0$.
But I don't know how to prove this statement. Can anybody prove this clam? Thanks for assistance in advance.

Answer 1

$X$ is in rad $V$ if for all $Y$, we have $B(X,Y) = 0$. That is: for all $Y$, we have $$ n\operatorname{tr}(XY) - \operatorname{tr}(X)\operatorname{tr}(Y) = 0 \iff\\ \operatorname{tr}(n\,XY - \operatorname{tr}(X)Y) = 0 \iff\\ \operatorname{tr}((n\,X - \operatorname{tr}(X)I)Y) = 0. $$ In other words: if $C(X,Y)$ denotes the bilinear form $C(X,Y) = \operatorname{tr}(XY)$, then $X$ is in rad $V$ if and only if $C(n\,X - \operatorname{tr}(X)I,Y) = 0$ for all $Y$. Because $C$ is a non-degenerate bilinear form over $V$, this is equivalent to saying that $n\,X - \operatorname{tr}(X)I = 0$. This occurs if and only if $X$ is a multiple of $I$ (the identity matrix).

So, we find that the rad of $V$ is the one-dimensional space $\{a\,I: a \in \Bbb C\}$.