Let $\mathfrak{g} \subset gl(V)$ be a semisimple Lie algebra. I already know that symmetric bilinear form $f(x,y)=\mathbf{Trace}(XY)$ is nonsingular on $\mathfrak{g}$. And I've read that any ideal $\mathfrak{g_1}$ of $\mathfrak{g}$ is also semisimple, from which it also follows that $\mathfrak{g_1}$ is nonsingular space with respect to this bilinear form.
Could someone help me to show that $\mathfrak{g_1}$ is actually nonsingular?
I suppose that $f(x,y)=\mathbf{Trace}(XY)$ means in fact $$f(x,y)=\mathbf{Trace}(ad_xad_y)$$ since I don't get otherwise what $X,Y$ mean since $f$ is function of $x,y$... in other words I assume that we are speaking about the Killing form. Then the thing is quite easy: suppose $\mathfrak{g_1}$ is an ideal, then consider $\mathfrak{g}=\mathfrak{g_1}\oplus\mathfrak{g}_{2}$, find a base for $\mathfrak{g}$ made by vector of $\mathfrak{g}_1$ followed by vector of $\mathfrak{g}_2$ and then in this base consider the block matrix form of the adjoint matrix $ad_x$. Since $\mathfrak{g_1}$ is an ideal you have that $[x,\mathfrak{g}]\in\mathfrak{g_1}$ whenever $x\in\mathfrak{g_1}$ so you have $$ad_x=[x,.]=\begin{array}{c} \begin{array}{cc} \mathfrak{g}_{1} & \mathfrak{g}_{2}\end{array}\\ \left(\begin{array}{cc} * & *\\ 0 & 0 \end{array}\right) \end{array}$$ Which means that for every $y \in \mathfrak{g}$ you have $ad_xad_y$ of the form $$ad_xad_y=\begin{array}{c} \begin{array}{cc} \mathfrak{g}_{1} & \mathfrak{g}_{2}\end{array}\\ \left(\begin{array}{cc} * & *\\ 0 & 0 \end{array}\right) \end{array}$$ So for every vector $v \in \mathfrak{g}$ let us decompose it as $v=v_1+v_2$ where $v_1 \in \mathfrak{g}_{1}$ and $v_2 \in \mathfrak{g}_{2}$. Then, given the form of the matrix above, you have that for every vector $x \in \mathfrak{g}_{1}$ the ${Trace}(ad_xad_v)=0$ if and only if ${Trace}(ad_xad_{v_1})=0$. This means that $$ker\left(f_{|\mathfrak{g}_{1}}\right)\subseteq ker\left(f\right) $$ and since $ker\left(f\right)=0$ therefore $ker\left(f_{|\mathfrak{g}_{1}}\right)=0$