Exercise 16

When we decompose a vector space as a direct sum, the dimensions of the subspaces add to the dimension of the space. The situation with a space that is given as the sum of its subspaces is not as simple. This exercise considers the two-subspace special case.

• For these subspaces of $\mathcal{M}_{2\!\times\!2}$ find $W_{1}\cap W_{2}$, $\dim(W_{1}\cap W_{2})$, $W_{1}+W_{2}$, and $\dim(W_{1}+W_{2})$.

  $$W_{1}=\text{$\{\hskip 1.00006pt\begin{pmatrix}0&0\\ c&d\end{pmatrix}\mid c,d\in\mathbb{R}\hskip 1.00006pt\}$}\qquad W_{2}=\text{$\{\hskip 1.00006pt\begin{pmatrix}0&b\\ c&0\end{pmatrix}\mid b,c\in\mathbb{R}\hskip 1.00006pt\}$}$$

• Suppose that $U$ and $W$ are subspaces of a vector space. Suppose that the sequence $\langle\vec{\beta}_{1},\dots,\vec{\beta}_{k}\rangle$ is a basis for $U\cap W$. Finally, suppose that the prior sequence has been expanded to give a sequence $\langle\vec{\mu}_{1},\dots,\vec{\mu}_{j},\vec{\beta}_{1},\dots,\vec{\beta}_{k}\rangle$ that is a basis for $U$, and a sequence $\langle\vec{\beta}_{1},\dots,\vec{\beta}_{k},\vec{\omega}_{1},\dots,\vec{\omega}_{p}\rangle$ that is a basis for $W$. Prove that this sequence

  $$\langle\vec{\mu}_{1},\dots,\vec{\mu}_{j},\vec{\beta}_{1},\dots,\vec{\beta}_{k},\vec{\omega}_{1},\dots,\vec{\omega}_{p}\rangle$$

  is a basis for the sum $U+W$.

• Conclude that $\dim(U+W)=\dim(U)+\dim(W)-\dim(U\cap W)$.

• Let $W_{1}$ and $W_{2}$ be eight-dimensional subspaces of a ten-dimensional space. List all values possible for $\dim(W_{1}\cap W_{2})$.