I am to prove the following corollary.

Let $V_1$ and $V_2$ be subspaces of a $n$-dimensional vector space $V$. If the sum of the dimensions of $V_1$ and $V_2$ is greater than $n$, then $V_1$ and $V_2$ contain a common nonzero vector.

What we know:

  • Since, $V_1$ is a subspace of $V$, then $V_1 $has a finite basis $B_1$ with $n_1$ linearly independent vectors. Similarly for $V_2$.
  • We know span$B_1$ = span$\{\alpha_1, ..., \alpha_{n1}\} = V_1$
  • We know span$B_2$ = span$\{\gamma_1, ..., \gamma_{n2}\} = V_2$
  • I know the intersection of any two subspaces is a subspace, so if I can show $V_1 \cap V_2$ contains another vector besides $0$, I am done.
  • Now the dimensional formula says: dim($V_1$) + dim($V_2$) = dim($V_1 + V_2$) + dim($V_1 \cap V_2$).
  • By hypothesis, we know dim($V_1$) + dim($V_2$) $ = n_1 + n_2>$ dim($V$) $= n$
  • I need to prove $V_1 \cap V_2 \neq \{0\}$, i.e, not disjoint.
  • I could try to prove the corollary by contradiction. I could claim, $V_1 \cap V_2$ $= \{0\} \Rightarrow$ \begin{eqnarray} \text{dim}(V_1) + \text{dim}(V_2) = \text{dim}(V_1 + V_2). \end{eqnarray}
  • Thus dim($V_1 + V_2$) = $n_1 + n_2$. This is where I need to show that \begin{eqnarray} n_1 + n_2 &\neq& \text{dim}(V_1 + V_2). \end{eqnarray}
  • Since we are claiming $V_1 \cap V_2 = \{0\}$. I can say span $D$ = span$\{\alpha_1, ..., \alpha_{n1},\gamma_1, ..., \gamma_{n2}\} = V_1 + V_2$. Since $V$ is a $n$-dimensional vector space, there can only be at most $n$ linearly independent vectors in the span of $V_1 + V_2$. Thus dim$(V_1 + V_2) \leq n$. However, we said dim$(V_1 + V_2)$ = $n_1 + n_2$, which is a contradiction. This would mean, in order for the dimensional formula to hold true, $V_1\cap V_2$ can not be disjoint.
  • I am not sure if I would need to show a few of these vectors are linear combinations of the other, i.e, the span is linearly dependent. This would in turn show that $V_1 \cap V_2$ does have at least on vector in common, thus not disjoint. However, I am not sure how to go about this.

This is where I am stuck. I feel like I am missing a crucial piece of information to properly execute a proof. Any suggestions? Are any of my claims incorrect or am I going about this the wrong way?

Thank You for your time. I appreciate all the helpful advice and guidance you have given me. Take care and have a wonderful day.


You're overthinking. The dimension formula $$ \dim(V_1\cap V_2)=\dim V_1+\dim V_2-\dim(V_1+V_2) $$ already tells you that $\dim(V_1\cap V_2)>0$, because $$ \dim V_1+\dim V_2>\dim V\ge\dim(V_1+V_2) $$ A subspace with dimension greater than $0$ contains a nonzero vector.

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  • $\begingroup$ wow. Looking at what you wrote, I am overthinking it. $\endgroup$ – Kevin_H Feb 2 '16 at 21:57
  • $\begingroup$ @Kevin_H You could also do by contrapositive: if $V_1\cap V_2=\{0\}$, then $\dim V_1+\dim V_2=\dim(V_1+V_2)\le n$. $\endgroup$ – egreg Feb 2 '16 at 22:09

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