= Solution
The <Pauli matrix commutator identity> gives
$$
[T^a,T^b]=\varepsilon^{abc}T^c
$$
for the anti-Hermitian basis of the <SU(2) Lie algebra>. In the ordered basis $(T^1,T^2,T^3)$ the <Adjoint representation> matrices, with columns recording the images of basis elements, are
$$
\operatorname{ad}_{T^1}=\begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix},\quad
\operatorname{ad}_{T^2}=\begin{pmatrix}0&0&1\\0&0&0\\-1&0&0\end{pmatrix},\quad
\operatorname{ad}_{T^3}=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}.
$$
Their squared <matrix traces> are $-2$, and the mixed products have zero trace. Therefore the <Killing form of the SU(2) Lie algebra> is
$$
\boxed{\kappa(T^a,T^b)=-2\delta^{ab},\qquad\kappa(X,Y)=-2\sum_{a=1}^3x_ay_a.}
$$
It is negative definite and <nondegenerate>, so this real <Lie algebra> is semisimple. It is also simple: the displayed bracket is the three-dimensional <cross product>; if an <ideal of a Lie algebra> contains nonzero $x$, the vectors $[y,x]$ span the plane perpendicular to $x$, and together with $x$ span the whole algebra. This confirms that the simplicity implication applies with a nonzero <Killing form>.
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