Let $V = M_{n x n} (\Bbb C)$ and B be a bilinear form on V defined by ,
$B(X,Y) = n\,\text{tr}(XY) - \text{tr}(x) \text{tr}(Y)$ where tr$(X)$ is the trace of X. Prove that B is non-degenerate.
Can someone please give an example of $X \in M_{n x n} (\Bbb C)$ such that $B(X,Y)=0,\,\forall Y \in M_{n x n} (\Bbb C)$. Thanks in advance for help..
You have $B(I_n, Y) = 0$ for all $Y$. So it seems to be a degenerate form.
However, it will be non-degenerate if restricted to the subspace of matrices of trace $0$.
$\bf{Added:}$ Let's show that the form is non-degenerate when restricted to the subspace of matrices of trace $0$. The restriction of the form is just $n \operatorname{tr}(XY)$.
Let $X$ with $\operatorname{tr} (X)=0$ so that $n \operatorname{tr}(XY ) = 0$, for all $Y$ so that $\operatorname{tr}(Y)=0$. Let $Y\in M_{n\times n}(\mathbb{C})$. Then $Y_1=Y - \frac{\operatorname{tr}(Y)}{n} \cdot I_n$ has trace $0$. We have $Y = Y_1 + c I_n$. So $\operatorname{tr}(XY) = \operatorname{tr}(XY_1) + \operatorname{tr}(X(c I_n))$. The first term is $0$. So is the second term, since $\operatorname{tr}(X)=0$. Therefore, $\operatorname{tr}(X Y) = 0$ for all $Y\in M_{n\times n}(\mathbb{C})$. But that implies easily $X = 0$. (Indeed, take for $Y$ matrices of form $E_{ij}$).
Obs: the proof works for $n\times n$ matrices over any field $F$, provided $n$ is invertible in $F$. However, in the case of $\mathbb{C}$, we could use another, quicker proof. Indeed, if $\operatorname{tr}(X)= 0$, so is $\operatorname{tr}(\bar{X}^{t})=0$. Moreover, if $\operatorname{tr}(X \bar{X}^{t}) = \sum_{i,j=1}^n |a_{ij}|^2$, so if that is $0$, $X$ must also be $0$.
This is the Killing form for the reductive Lie algebra $L=\mathfrak{gl}(n)$, up to a factor $1/2$. Since the Killing form is non-degenerate if and only if the Lie algebra is semisimple, it cannot be non-degenerate for $L$, because $L$ is not semisimple. For references see this MSE-question, and the related links. However, for $L=\mathfrak{sl}(n)$, it is non-degenerate.
