Given that $W$ is a subspace of $V$, prove: $\dim W=\dim V \implies W=V$

Suppose $W$ is a subspace of the finite-dimensional vector space $V$ and $\dim W = \dim V=n$

Let $\{\overrightarrow w_1,\ldots,\overrightarrow w_n\}$ be a basis for $W$ and $\{\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ be a basis for $V$.

The set $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ still spans $W$. Since $W$ is a subspace of $V$, $W\subset V$ and all the $w_i$'s can be written in terms of $\{ \overrightarrow v_{1},\ldots,\overrightarrow v_n\}$.

$\therefore$ We can remove one by one all the $w_i$'s from the set $\{\overrightarrow w_{1},\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ while still having it span $W$, which leaves us with the set $\{\overrightarrow v_1,\ldots,\overrightarrow v_n\}$. We know it to be linearly independent since it is a basis for $V$. Therefore the basis of $V$ is also a basis of $W$. From there on, it is trivial to prove double inclusion $W\subset V$ and $V\subset W$ which gives us $W=V$.

Is my approach correct?

• Yes, this is correct! – Jon Warneke Mar 27 '16 at 23:22
• Easier: every linearly independent set in $V$ can be extended to a basis. A basis of $W$ is linearly independent. Can you add vectors to it getting again a linearly independent set? – egreg Mar 27 '16 at 23:23
• I think it is. You could also go with this route: There exists a linear mapping that is a bijection between the basis of $W$ and $V$. This mapping is an isomorphism, and then you can use it to prove that $V\subset W$. – Marra Mar 27 '16 at 23:24
• Please consider using the LaTeX command \dim to get $\dim$ instead $dim$. – Fly by Night Mar 27 '16 at 23:39
• I do not think this argument is correct, since you seem to be assuming that the $\vec{v_i}$ are in W when you claim that the set of $\vec{w_j}$ and $\vec{v_i}$ span W. The fact that the $\vec{v_i}$ are a basis for V implies that every vector in W can be written as a linear combination of the $\vec{v_i}$, which seems to be what you are showing. – user84413 Mar 27 '16 at 23:45

Your approach is incorrect: the set $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ spans $V$ and saying that it spans $W$ is the same as assuming $W=V$.

You should rather consider a basis $\{\overrightarrow w_1,\ldots,\overrightarrow w_n\}$ of $W$ and assume, by contradiction, that $W\ne V$. Then there exists $\vec{v}\in V$ with $\vec{v}\notin W$. It's easy to prove that $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v\}$ is linearly independent and now this is a contradiction, because we found a linearly independent set in $V$ with more elements than $\dim V$.

• Why am I assuming that $W=V$? I just added vectors to the basis of $W$, the spanning of W should not be affected. We can then remove the dependent vectors one by one. – Omrane Mar 28 '16 at 12:10
• @Sadem You also add vectors that span $V$, so the span is $V$, not $W$. Your argument would work exactly the same if $w\in\mathbb{R}^2$ and $\{v_1,v_2\}$ is a basis of $\mathbb{R}^2$: from the set $\{w,v_1,v_2\}$ you can surely remove $w$. – egreg Mar 28 '16 at 12:14

Your approach is fine and correct.

The easiest way to prove this is to notice that $W$ is a subspace of $V$, thus it is a subset of $V$. Thus, if $dimV = n$, and we have $n$ linearly independent vectors in $W$ that span $W$, these must also be linearly independent vectors in $V$. But then we have $n$ of them so these must span $V$ as well! So the vectors span $W$ AND $V$, so they are equal.

• The approach is not correct. While is is true that $\{w_1, \ldots, w_n, v_1, \ldots, v_n\}$ spans $W$, this is only because $W$ and $V$ are equal. Sadem's explanation on the other hand is completely wrong. – Dominik Mar 28 '16 at 8:26
• @Dominik, {$\overrightarrow w_{1},...,\overrightarrow w_{n}$} spans W, therefore adding any vector to the set will not affect the spanning of W, if anything it will make it larger. – Omrane Mar 28 '16 at 12:08
• @Sadem: Your definition of spanning is colloquial, not the mathematical definition. A set of vectors is said to span a vector space, if the span of these vectors is precisely the space - not if it is bigger than the space. Note that your argumentation doesn't even require at any point that the dimensions of $W$ and $V$ are equal. – Dominik Mar 28 '16 at 12:12
• It actually does, suppose if $W$ had a lower dimension than $V$, we would end up with a linearly independent set of $n$ elements for $W$ while its dimension is less than $n$. We could therefore not declare it a basis of $W$ which does not allow us to solve the problem. You are right about how I see spanning though, I don't know how to phrase it in another way though. – Omrane Mar 28 '16 at 12:16
• This doesn't correct your argumentation. In your comment, you've just shown that $\operatorname{dim} W \ne \operatorname{dim} V$ implies $W \ne V$. At no point does your argumentation use the fact that the dimensions are the same [cf. the comments under egreg's answer]. Your proof is quite simply false. – Dominik Mar 28 '16 at 15:19

Your proof is incorrect. You first choose a colloquial understanding of the word "spanning" and at a later point the mathematically correct understanding [which changes the meaning of the word!].

You say that $\{w_1, \ldots, w_n, v_1, \ldots, v_n\}$ spans $W$, but you can't deduce this directly. What you mean is that $W$ is a subset of $\operatorname{span}(\{w_1, \ldots, w_n, v_1, \ldots, v_n\})$. Since the $w_i$ can be represented as linear combinations of the $v_i$, you can succesively remove the $w_i$ without changing the span, i.e. $$\operatorname{span}(\{w_1, \ldots, w_n, v_1, \ldots, v_n\}) = \operatorname{span}(\{v_1, \ldots, v_n\})$$

Now at this point in your argumentation, you make the switch from the colloquial term "spanning" to the mathematically rigorous by saying that $\{v_1, \ldots, v_n\}$ spans $W$ and is therefore a basis for $W$. But this is incorrect, as you've only shown that $\{v_1, \ldots, v_n\}$ "overspans" $W$, i.e. $W \subset \operatorname{span}(\{v_1, \ldots, v_n\})$. But this inclusion is trivial, since the span of the $v_i$ is the whole of $V$.

Note that you've at no point in your argumentation actually used the fact that the dimensions of $V$ and $W$ are the same. Simply go throgh the proof and try to point at the part where you've used this assumption.