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?

  • 3
    $\begingroup$ "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. $\endgroup$ Feb 5 at 22:18
  • $\begingroup$ 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? $\endgroup$
    – J.G.
    Feb 5 at 22:22
  • 1
    $\begingroup$ 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 $V$. $\endgroup$ Feb 5 at 22:24
  • $\begingroup$ @GerryMyerson that was my first thought as well since it uses the definition of dimension directly. $\endgroup$ Feb 5 at 22:33
  • $\begingroup$ Notice: you prove a result, but you solve a problem. $\endgroup$
    – Pedro
    Feb 5 at 22:38

2 Answers 2


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.

  • $\begingroup$ 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 $V$, $\dim(T) = \dim(V) \implies T = V$? $\endgroup$
    – lafinur
    Feb 6 at 16:10
  • $\begingroup$ @lafinur Yes, every subspace (not just any set) $T\subset V$, which satisfies $\mathrm{dim}(T)=\mathrm{dim}(V)$ must be equal to $V$. $\endgroup$
    – Ben S.
    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 $V=U\oplus \text{span}(v)$.


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