Problem-1

Let \(W\) be a finite dimensional real vector space, and let \(U\) and \(V\) be two subspaces of \(W\). Let \(U+V\) be the space

\[ U+V=\{u+v\,:\,u\in U\text{ and }v\in V\} \]

Show the following:

We see that \(0\in U+V\). If \(u_{1}+v_{1}\in U+V\) and \(u_{2}+v_{2}\in U+V\), then \((u_{1}+u_{2})+(v_{1}+v_{2})\in U+V\). If \(u+v\in U+V\) and \(\lambda\in\mathbb{R}\), then \(\lambda(u+v)=\lambda u+\lambda v\in U+V\). Using these three observations, we conclude that \(U+V\) is a subspace of \(W\).

We know that \(U\cap V\) is a subspace of \(W\). Let \(\{w_{1},\cdots,w_{k}\}\) be a basis for \(U\cap V\). We can now extend this to a basis for \(U\) and a basis for \(V\). The two bases are \(\{w_{1},\cdots,w_{k},u_{1},\cdots,u_{m}\}\) for \(U\) and \(\{w_{1},\cdots,w_{k},v_{1},\cdots,v_{n}\}\) for \(V\). Consider the set \(\{w_{1},\cdots,w_{k},u_{1},\cdots,u_{m},v_{1},\cdots,v_{n}\}.\) It is quite easy to see that this set spans \(U+V\). If we can show that this set is also linearly independent, then we have a basis for \(U+V\).

Let us take a linear combination of this collection and set it to zero:

\[ (a_{1}w_{1}+\cdots+a_{k}w_{k})+(b_{1}u_{1}+\cdots+b_{m}u_{m})+(c_{1}v_{1}+\cdots+c_{n}v_{n})=0 \]

We can now group the terms in the following way:

\[ a_{1}w_{1}+\cdots+a_{k}w_{k}+b_{1}u_{1}+\cdots+b_{m}u_{m}=-(c_{1}v_{1}+\cdots+c_{n}v_{n}) \]

The LHS is a vector in \(U\) and the RHS is a vector in \(V\). Since the two are equal, the vector in question is in \(U\cap V\). We can now express this using the basis for \(U\cap V\):

\[ \begin{aligned} -(c_{1}v_{1}+\cdots+c_{n}v_{n}) & =d_{1}w_{1}+\cdots+d_{k}w_{k}\\ c_{1}v_{1}+\cdots+c_{n}v_{n}+d_{1}w_{1}+\cdots+d_{k}w_{k} & =0 \end{aligned} \]

This is actually a linear combination of the basis vectors of \(V\) set to zero. Hence, \(c_{1}=\cdots=c_{n}=d_{1}=\cdots d_{k}=0\). Going back to the original equation, we conclude that \(a_{1}=\cdots=a_{k}=b_{1}=\cdots=b_{m}=0\). This implies, the set \(\{w_{1},\cdots,w_{k},u_{1},\cdots,u_{m},v_{1},\cdots,v_{n}\}\) is linearly independent. We therefore have shown that this is a basis for \(U+V\).

From this, the formula that relates the dimensions of the subspaces \(U,V,U\cap V\) and \(U+V\) follows:

\[ \boxed{\text{dim}(U+V)=\text{dim}(U)+\text{dim}(V)-\text{dim}(U\cap V)} \]