1
$\begingroup$

When those authors state the following


$\bf{Theorem 6.}$ If $W_1$ and $W_2$ are finite-dimensional subspaces of a vector space $V$, then $W_1+W_2$ is finite-dimensional and \begin{eqnarray} \dim W_1 + \dim W_2 = \dim (W_1 \cap W_2) + \dim (W_1+W_2). \end{eqnarray} Proof. By Theorem 5 and its corollaries, $W_1 \cap W_2$ has a finite basis $\{\alpha_1,\cdots,\alpha_k\}$ which is part of a basis \begin{eqnarray} \{\alpha_1,\dots,\alpha_k,\beta_1,\dots,\beta_m\} \hspace{.3cm} \text{for $W_1$} \end{eqnarray} and part of a basis \begin{eqnarray} \{\alpha_1,\dots,\alpha_k,\gamma_1,\dots,\gamma_m\} \hspace{.3cm} \text{for $W_2$}. \end{eqnarray} Then subspace $W_1+W_2$ is spanned by the vectors \begin{eqnarray} \alpha_1,\dots,\alpha_k,\beta_1,\dots,\beta_m,\gamma_1,\dots,\gamma_n \end{eqnarray}


How is this last part so?

$\endgroup$
1
1
$\begingroup$

Anything in $W_1$ is a sum of the $\alpha$'s and $\beta$'s, and everything in $W_2$ is a sum of the $\alpha$'s and the $\gamma$'s. Therefore, the span of the $\alpha$'s, $\beta$'s and $\gamma$'s contains both $W_1$ and $W_2$... but since the span is closed under addition it contains $W_1+W_2$.

On the other hand, all those basis vectors are in $W_1+W_2$, so the span can't "escape" $W_1+W_2$. So, you have equality.


Added to address comment.

I'm not sure exactly what you mean by "do these data suffice," but it's certain that they are relevant steps for proving the proposition. The follow-up is to consider the mapping from $W_1\times W_2\to W_1+W_2$ given by $(x,y)\mapsto x+y$. You can check that the kernel of the map is $W_1\cap W_2$, and so the rank-nullity theorem gives you the equation you seek. (The dimension of $W_1\times W_2$ is $dim(W_1)+dim(W_2)$, and the image of the map is $W_1+W_2$.)

$\endgroup$
2
  • $\begingroup$ Do these data suffice to show the first equality? $\endgroup$ – Trancot Apr 16 '13 at 21:16
  • $\begingroup$ @Trancot Added to the answer to address your comment. Thanks! $\endgroup$ – rschwieb Apr 17 '13 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.