I want to prove that:

If $W_1$ and $W_2$ are finite-dimensional subspaces of a vector space $V$, then the subspace $W_1 + W_2$ is finite-dimensional.

I have written a skeleton of a proof but I don't know how to complete it. Could you guys look it over and tell me what's wrong about it? Thanks:

We know the intersection of two subspaces is also a subspace. So, $W_1 \cap W_2$ is a subspace. Since $W_1$ and $W_2$ are both finite-dimensional, their intersection must also be finite-dimensional. Thus, $W_1 \cap W_2$ is finite-dimensional. Because subspaces are, themselves, vector spaces we know that a finite-dimensional vector space includes a basis consisting of a finite number of vectors. So, $W_1 \cap W_2$ has a basis $\beta = \{u_1, \cdots, u_n\}$ with $n < \infty$. We know $W_1 \cap W_2 \subseteq W_1$ So, we can extend $\beta$ to a basis for $W_1$ which we will call $\gamma = \{u_1, \cdots , u_n, p_1, \cdots , p_m \}$. Similarly, since $W_1 \cap W_2 \subseteq W_2$ we can extend $\beta$ to a basis of $W_2$ which we will call $\alpha = \{u_1, \cdots , u_n, v_1, \cdots , v_k \}$. By definition of a basis, we know $\gamma$ spans $W_1$ and $\alpha$ spans $W_2$. Therefore, $\text{span}(W_1\cup W_2) = \text{span}(\gamma) + \text{span}(\alpha)$. But, $W_1 + W_2 = \text{span}(W_1 \cup W_2)$ by definition implying $W_1 + W_2 = \text{span}(\gamma) + \text{span}(\alpha)$. Since $\text{span}(\gamma)$ and $\text{span}(\alpha)$ are both finite-dimensional, $W_1 + W_2$ must also be finite-dimensional.

That is my proof, but I am still a little worried about the last few parts. Is it true that the span of the union of two subspaces is exactly the sum of the spans of the two subspaces? Thank you!

  • 1
    $\begingroup$ Is it easier to see that the intersection of two finite-dimensional subspaces is finite-dimensional (which you use without justification) than it is to see that their sum is? I think not. And at the end of your proof (last sentence of the first paragraph, ending "must also be finite dimensional") you seem to be using exactly what you are trying to prove. $\endgroup$ Apr 3, 2016 at 19:51

2 Answers 2


Since this is such basic stuff, you must really be clear about what definitions you are using. In the courses I know (and give) "finite dimensional" is actually defined before dimension is, and it means "finitely generated", in other words a space is finite dimensional if (and only if) it is the span of a finite set. Span of a set of vectors is formed by the set of their linear combinations, which forms the smallest subspace containing that set. From this it is easy to show that the span of $A\cup B$ is the sum of the span of $A$ and the span of $B$ (which sum can be defined as the span of their union). Once this is done, it is clear that the sum of two finite dimensional subspaces (spans of finite $A,B$ respectively) is finite dimensional (the span of the finite set $A\cup B$).

On the other hand it is not so obvious (though true) that any subspace of a finite dimensional space is finite dimensional, and in particular that the intersection of a finite dimensional subspace with another subspace (which may or may not be finite dimensional) is finite dimensional.


Someone will probably read through your argument. Let me instead point a direct of of proving what you want.

A subspace is finite dimensional if and only if it can be generated by a finite number of elements.

If $U$ and $V$ are finite dimensional subspaces of a vector space $W$, and $A$ and $B$ are finite subsets of $U$ and $V$, respectively, that generate them, then you should be able to check that $A\cup B$, which is also a finite set, generates the sum $U+V$. It follows from this that $U+V$ is finite dimensional.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .