 Solution

Choose a basis $u_1,\dots ,u_r$ of $U$, extend it to a basis $u_1,\dots ,u_r,w_1,\dots ,w_s$ of $W$, and extend again to a basis of $V$ by adding $v_1,\dots ,v_t$. DefineEvery member of the full basis lies in $W$ or $W^{\prime }$, so the sum of vector subspaces is $W+W^{\prime }=V$. If $x\in W\cap W^{\prime }$, 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 $W\cap W^{\prime }=U$. This is a subspace supplement with a prescribed intersection.

For the matrix example, letThe skew-symmetric matrices are exactly $U=\mathrm{span}\{J\}$, and the traceless matrices form $W$. A suitable answer isThe matrix trace of $aI+bJ$ is $2a$, so $W\cap W^{\prime }=\mathrm{span}\{J\}=U$. Every $2\times 2$ real matrix has the decompositionwhose first term lies in $W$ and second in $W^{\prime }$. Therefore $W+W^{\prime }=M_2(\mathbb{R})$, as required.