Question-73

subspace

Let \(\displaystyle U\) and \(\displaystyle V\) be two-dimensional subspaces of \(\displaystyle \mathbb{R}^{5}\) and let \(\displaystyle W=U\cap V\). Find the number of possible dimensions of \(\displaystyle W\).

Answer

\(3\)

Solution

We know that \(\displaystyle \text{dim}( W) \leqslant 2\) since \(\displaystyle W\subseteq U\). It follows that \(\displaystyle W\) can have dimensions \(\displaystyle 0,1,2\). A simple example to demonstrate this. Let \(\displaystyle B=\{e_{1} ,\cdots ,e_{5}\}\) be the standard basis for \(\displaystyle \mathbb{R}^{5}\).

\(\displaystyle U=\text{span}\{e_{1} ,e_{2}\} ,V=\text{span}\{e_{3} ,e_{4}\}\) results in \(\displaystyle W=\{0\}\) which has dimension \(\displaystyle 0\).
\(\displaystyle U=\text{span}\{e_{1} ,e_{2}\} ,V=\text{span}\{e_{2} ,e_{3}\}\) results in \(\displaystyle W=\text{span}\{w_{2}\}\) which has dimension \(\displaystyle 1\).
\(\displaystyle U=V=\text{span}\{e_{1} ,e_{2}\}\) results in \(\displaystyle W=U=V\), which has dimension \(\displaystyle 2\).