= Solution
{c}
Choose a <basis> $u_1,\ldots,u_r$ of $U$, extend it to a <basis> $u_1,\ldots,u_r,w_1,\ldots,w_s$ of $W$, and extend again to a <basis> of $V$ by adding $v_1,\ldots,v_t$. Define
$$
\boxed{W'=\operatorname{span}\{u_1,\ldots,u_r,v_1,\ldots,v_t\}.}
$$
Every member of the full <basis> lies in $W$ or $W'$, so the <sum of vector subspaces> is $W+W'=V$. If $x\in W\cap W'$, express it in both displayed <bases>. Uniqueness of the coordinates in the full <basis> forces its $w_j$ and $v_j$ coefficients to vanish. Thus $x\in U$. The reverse inclusion is built into the construction, proving $W\cap W'=U$. This is a <subspace supplement with a prescribed intersection>.
For the matrix example, let
$$
J=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\qquad
I=\begin{pmatrix}1&0\\0&1\end{pmatrix}.
$$
The <skew-symmetric matrices> are exactly $U=\operatorname{span}\{J\}$, and the traceless <matrices> form $W$. A suitable answer is
$$
\boxed{W'=\operatorname{span}\{I,J\}
=\left\{\begin{pmatrix}a&b\\-b&a\end{pmatrix}:a,b\in\mathbb R\right\}.}
$$
The <matrix trace> of $aI+bJ$ is $2a$, so $W\cap W'=\operatorname{span}\{J\}=U$. Every $2\times2$ real <matrix> has the decomposition
$$
A=\left(A-\frac{\operatorname{tr}A}{2}I\right)+\frac{\operatorname{tr}A}{2}I,
$$
whose first term lies in $W$ and second in $W'$. Therefore $W+W'=M_2(\mathbb R)$, as required.
Back to article page
