# A more elegant solution to this linear algebra problem?

Consider the following problem.

Let $$V$$ be a $$\mathbb{K}$$-vector space of dimension $$n$$ and $$U \subset V$$ a subspace of dimension $$n - 1$$. Prove the following propositions:

If $$v \not\in U$$ then $$V = U \oplus \langle v\rangle$$, where $$\langle v \rangle$$ denotes the span of $$v$$.

The way I set out to show this is the following. Assume $$v \not\in U$$. Then the $$n$$th component of $$v$$ is non-null. Then $$cv \not\in U$$ for every non-null $$c \in \mathbb{K}$$. Then $$U \cap \langle v\rangle = \{0\}$$.

We must now only show $$U + \langle v\rangle = V$$. Every vector $$u \in U$$ may be expressed as $$(x_1, \ldots, x_{n-1}, 0)$$; this is, as an $$n$$-dimensional vector with the $$n$$th component null. Now observe $$v = (y_1, \ldots, y_{n-1}, y_n)$$. Then

\begin{align*} U + \langle v\rangle &= \begin{pmatrix} u_1 \\ \vdots \\ u_{n-1} \\ 0 \end{pmatrix} + \begin{pmatrix} v_1 \\ \vdots \\ v_{n-1} \\ v_n \end{pmatrix} \\ &= \begin{pmatrix} u_1 + v_1 \\ \vdots \\ u_{n-1} + v_{n-1} \\ v_n \end{pmatrix} \\ &= (u_1 + v_1)b_1 + \ldots + (u_{n-1}+ v_{n-1})b_{n-1} + v_nb_n .\end{align*}

where $$b_i$$ is the $$i$$th vector in the basis of $$V$$. Since $$u_i + v_i$$ might be any element in $$\mathbb{K}$$ (for we are taking $$u_i, v_i$$ to be arbitrary scalars in $$\mathbb{K}$$ and any scalar can be represented as the sum of other two), then we have found $$U + \langle v\rangle = V$$. $$\blacksquare$$

Fair enough. However, in attempting my proof I had to appeal to the interpretation of $$U$$ as the subspace of $$n$$th dimensional vectors $$v \in V$$ whose $$n$$th component is $$0$$. This allowed me to compute the sum $$U + \langle v\rangle$$ explicitly, as shown above. From the sum it readily follows that the resulting set of vectors was only $$V$$.

However, I find it rather unpleasant to perform this explicit sum. My intuition tells me that there must be some formal property that allows me to show $$U + \langle v\rangle = V$$ without actually observing the result of $$(u_1, ..., u_{n-1}, 0) + (y_1, ..., y_{n-1}, y_n)$$ and so on. What would be a more abstract or formal way to address this problem?

• "Then the $n$th component of $v$ is nonzero". First, vector don't necessarily have components. Second, if they do, it's with respect to some fixed basis, which you haven't provided. And third, there's no reason why the $n$th component of $v$ should be nonzero, unless you are very careful about the basis you have chosen. Feb 5 at 22:18
• I've approved a recent edit suggestion to replace each <v> with \langle v \rangle, but it won't take effect until someone else approves it. Any takers?
– J.G.
Feb 5 at 22:22
• I'd let $\{\,b_1,b_2,\dots,b_{n-1}\,\}$ be a basis for $U$, then show $\{\,b_1,b_2,\dots,b_{n-1},v\,\}$ is a linearly independent set, hence, a basis for $V$. Feb 5 at 22:24
• @GerryMyerson that was my first thought as well since it uses the definition of dimension directly. Feb 5 at 22:33
• Notice: you prove a result, but you solve a problem.
– Pedro
Feb 5 at 22:38

Just use dimensions: For two subvector spaces $$U,W \subset V$$, we have

$$\mathrm{dim}(U+V)=\mathrm{dim}(U)+\mathrm{dim}(W)-\mathrm{dim}(U\cap W).$$

Setting $$W:=\langle v \rangle,$$ we can easily see that $$W\cap U= 0$$ and therefore $$\dim(U+\langle v)\rangle= \mathrm{dim}(U)+\mathrm{dim}(\langle v \rangle)=n-1+1=n.$$ Therefore we must have $$U+\langle v \rangle =U \oplus \langle v \rangle =V$$.

The takeaway should be that when dealing with finite-dimensional vector spaces, the dimension is an important tool, which often can simplify arguments.

• This is quite clear, thanks! To be sure, the conclusion follows from the fact that $U + W \subseteq V$ and that for any subset $T$ of $V$, $\dim(T) = \dim(V) \implies T = V$? Feb 6 at 16:10
• @lafinur Yes, every subspace (not just any set) $T\subset V$, which satisfies $\mathrm{dim}(T)=\mathrm{dim}(V)$ must be equal to $V$. Feb 6 at 17:32

Your proof is vague and most likely wrong (maybe it can be fix). You’re assigning component to vectors without defining a basis for $$V$$. If $$\{b_1,…,b_n\}$$ is arbitrary basis of $$V$$. Then $$u\in U$$ may not be expressed as $$(x_1,…,x_{n-1},0)$$.

Proof: Since $$v\notin U$$, we have $$U\cap \text{span}(v)=\{0\}$$. So $$U, \text{span}(v)$$ are independent. Let $$B’=\{\alpha_1,…,\alpha_{n-1}\}$$ be basis of $$U$$. We show $$B=\{\alpha_1,…,\alpha_{n-1},v\}$$ is basis of $$V$$. If $$c_1 \alpha_1+…+c_{n-1}\alpha_{n-1}+c_nv=0$$, for some $$c_i\in F$$. Then $$c_1 \alpha_1+…+c_{n-1}\alpha_{n-1}=-c_nv\in U\cap \text{span}(v)$$. Since $$v\neq 0$$, we have $$c_n=0$$. By linear independence of $$B’$$, $$c_i=0$$, $$\forall i\in J_{n-1}$$. So $$B$$ is independent. Since $$|B|=n=\dim (V)$$, we have $$B$$ is basis of $$V$$. In particular, $$\text{span}(B)=V$$. Thus $$V=U+\text{span}(v)$$. Hence $$V=U\oplus \text{span}(v)$$.