Prove that if $W_1$ is any subspace of a finite-dimensional vector space $V$, then there exists a subspace $W_2$ of $V$ such that $V = W_1 \oplus W_2$

What I have done so far is to note that since $V$ is finite and $W_1$ is a subspace of $V$, we have $\dim(W_1) \leq \dim(V)$. If we have equality, then let $W_2 = \{0\}$, so that we have $V= W_1 \oplus W_2$. So now I need to look at the case when $\dim(W_1) \lt \dim(V)$.

What I have tried for this case is to let $$\beta = \{v_1,v_2,..,v_n\}$$ be a basis for $V$ and $$\gamma=\{u_1,u_2,..,u_m\}$$ a basis for $W_1$. My idea was to extend $\gamma$ to a basis for $V$, so let $$\alpha=\{u_1,u_2,..,u_m,w_1,w_2,..,w_{n-m}\}$$ be the extension of $\gamma$ to $V$, where $w_1,w_2,..,w_{n-m}$ are basis vectors for $W_2$. If I'm doing this right then I would just have to show that $W_1 \cap W_2 = \{0\}$ and $W_1 + W_2= V$

Am I heading in the right direction? Any hints would be greatly appreciated.

  • $\begingroup$ You're heading in the right direction. Keep going. $\endgroup$ – Pedro Tamaroff Apr 19 '14 at 22:02

If you already know that you can compete a basis of a subspace to a basis for the whole space then you are practically done.

Hint: Note that $\alpha$ is a basis for $V$ (this should give you $W_{1}+W_{2}=V$, why ?) and that the $w_{i}$ are linearly independent of the $u_{i}$ (this should show that $W_{1}\cap W_{2}=\{0\}$, why ?)

Note: The way I see it, there is no use for $\beta$ or of the $v_{i}$ in the proof

  • $\begingroup$ since $\alpha$ is a basis for $V$ then all vectors in $V$ can be written as a linear combination of vectors in $\alpha$, which is a linear combination of vectors in $\gamma$ plus linear combination of the basis vectors for $W_2$. Also, $W_1,W_2$ are subspaces of $V$ so they both contain the zero vector. $\endgroup$ – Miguel Landeros Apr 19 '14 at 22:13
  • $\begingroup$ One more thing, would I also have to show that $W_2$ is a subspace of V? $\endgroup$ – Miguel Landeros Apr 19 '14 at 22:16
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    $\begingroup$ @MiguelLanderos - $W_2$ is defined as a span of a set of vectors and hence is a subspace. You still need to show that the only vector in the intersection is the zero vector, can you show it ? $\endgroup$ – Belgi Apr 19 '14 at 22:21
  • $\begingroup$ @Belgi-I see, I think I should be able to show that last part. thanks $\endgroup$ – Miguel Landeros Apr 19 '14 at 22:24

Let $\dim V=n.$ Say you have a basis $\mathcal B=\{v_1,\dots,v_k\}$ for $W_1\subseteq V$, where $k\leq n$. Extend to a basis $\mathcal B'=\{v_1,\dots,v_n\}$ for $V$. Consider the subspace, call it $W_2$, spanned by those vectors you added to $\mathcal B$. Namely, $\{v_{k+1},\dots,v_n\}$. Then what can you say about $W_1$ and $W_2$? How do they relate to $V$?

EDIT: You say that you know $V=W_1+W_2$, but aren't sure how to get $W_1\cap W_2=\{0\}$. Let's take a look! Let $w\in W_1\cap W_2$. Then we can write $$ w = \sum_{i=1}^k a_iv_i \quad\text{and}\quad w = \sum_{i=k+1}^n a_iv_i, $$ for some $a_i\in F$. If you subtract these two equations, what does linear indendence (i.e. that $\{v_i\}$ is a basis) tell you about the $a_i$? What does this tell you about $W_1\cap W_2$?

  • $\begingroup$ I see that the sum of those two subspaces equals $V$. I only have to show that their intersection is the zero vector. I know since they are both subspaces of $V$ then they both contain the zero vector of $V$. I only need to show that the intersection is only the zero vector, but does this follow from the fact that both$W_1$ and $W_2$ are linearly independent? $\endgroup$ – Miguel Landeros Apr 19 '14 at 23:00
  • $\begingroup$ @MiguelLanderos: See edit. $\endgroup$ – user59083 Apr 20 '14 at 1:26
  • $\begingroup$ @5space-the $a_i$ all equal zero so the intersection is a linearly independent set. So the only vector in both is the zero vector. $\endgroup$ – Miguel Landeros Apr 20 '14 at 1:34
  • $\begingroup$ @MiguelLanderos: Exactly!! I think you've got it ;-) $\endgroup$ – user59083 Apr 20 '14 at 1:35

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