Killing form and trace form on a simple Lie algebra Let $k$ be the Killing form on a finite dimensional simple Lie subalgebra  $L$ of $\mathfrak{gl}(m,\mathbb{C})$.
Let $(x,y)\mapsto Trace(xy)$ denote the trace form on $L$. 
Claim: $k$ and trace form are proportional.
Proof: (1) Take a basis $B=\{x_1,\cdots,x_n\}$ of $L$.
(2) Let $p(\lambda)= det[k(x_i,x_j)-\lambda Trace(x_i,x_j)]$ be the determinant of the form $k-\lambda. Trace$ on $L$ w.r.t. above basis.
(3) Since $k$ and $Trace$ are symmetric and associative, so is $k-\lambda. Trace$. 
(4) The polynomial $p(\lambda)$ has a zero in $\mathbb{C}$, say $\lambda_0$. 
(5) Then $k-\lambda_i Trace$ is associative degenerate form on $L$; its radical is non-zero ideal of $L$.
(6) Since $L$ is simple, it follows that radical should be $L$, i.e. $k=\lambda Trace$.
Question: Of course $p(\lambda)$ is non-zero polynomial of degree $n$ over $\mathbb{C}$, but why it shouldn't be $\lambda^n$ so that only $0$ will be its root?

If $\lambda=0$ was a zero, you'd deduce that the Killing form  of $L$ is degenerate, contrary the assumed simplicity of $L$.
