Let $V$ subspace of $W$ and both have same dimension and same basis. Then can we safely say that $V= W$ ?

I believe yes. For example there may be an element $x \in V$ written as a linear combination of the basis elements. This linear combination is unique.

Now let's take another vector $y \in W$. The $y$ is written as a linear combination of the basis vectors.

We equate the two linear combinations, and since the basis elements are linearly independent, we get that $x=y$.

Thus $V=W$.

Do you agree?

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    $\begingroup$ If you're talking about finite dimensions, then yes. A basis generates a vector space. The same basis could not generate two different vector spaces. I'm not really sure about your reasoning: you can't take arbitrary $x \in V$ and $y \in W$ and decide somehow that the two are equal. $\endgroup$
    – User8128
    Commented Jul 25, 2016 at 22:10
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    $\begingroup$ If $B$ is a basis for $W$ and $B\subset V$, then every element of $W$ is a linear combination of elements of $V$, so is in $V$. $\endgroup$ Commented Jul 25, 2016 at 22:29
  • $\begingroup$ yes, by the definition of the space basis ( generator , then generate the same set ) $\endgroup$
    – user354674
    Commented Jul 25, 2016 at 22:57
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    $\begingroup$ If $V$ and $W$ both have basis $B$, then both $V$ and $W$ are the linear span of $B$, and thus are equal. The statements that $V$ is a subspace of $W$ and the statement about dimension (which is an easy consequence of the "same basis") are not needed. $\endgroup$ Commented Jul 25, 2016 at 22:58

3 Answers 3


The answer is yes. But you don't need they have the same basis.

More precisely:

Let $V$ be a subspace of the finite dimensional space $W$; if $\dim V=\dim W$, then $V=W$.

Proof. Since $\dim V\le \dim W$ for any subspace $V$ of $W$ we can prove the equivalent statement that if $V\subsetneq W$ (proper subspace), then $\dim V<\dim W$. If $\{v_1,v_2,\dots,v_n\}$ is a basis for $V$ and $w\in W$ but $w\notin V$, it is easy to show that $\{v_1,v_2,\dots,v_n,w\}$ is linearly independent (prove it). However, any linearly independent subset of $W$ can be extended to a basis, implying $\dim W\ge n+1$.

Your attempt shows you have the right idea, but express it poorly.

Let $w\in W$. Since $w$ is a linear combination of the common basis of $V$ and $W$, we conclude that $w\in V$. Together with the assumption that $V\subseteq W$, this ends the proof.

  • $\begingroup$ So if V subspace of W and dimV=dimW, then V=W. In your proof, you say dimV=n. And we said dimV=dimW, so dimW=n. And you show that dimW >= n+1. But how does this tells us that V=W ? To show this, we need to show that V and W have the same basis. But W may have as its basis any n elements of {u1,...,un, w} . So the bases of W and V may have the same number of elements, but not be equal. So how V= W? $\endgroup$
    – spyimp
    Commented Jul 26, 2016 at 10:08
  • $\begingroup$ @spyimp I proved a slightly different statement, namely that if $V$ is a proper subspace of $W$, then $\dim V<\dim W$. Thus, if $\dim V=\dim W$, we must have $V=W$. $\endgroup$
    – egreg
    Commented Jul 26, 2016 at 10:19
  • $\begingroup$ Wouldn't it be enough to show that, since both $W$ and $V$ have the same dimension $n$ and $V$ is a subspace of $W$ it is trivially possible to find a shared basis (just pick any $n$ linearly independent vectors from $V$). Then they are both equal to the set of all linear combinations of that basis, and are therefore equal to each other? $\endgroup$
    – jan
    Commented Jan 12, 2022 at 19:56

Here is another proof. We know that $V$ is a subspace of $W$ and $\dim(V) = \dim(W)$. Now suppose that V is a proper subset of W. We know that V must have a complement that is a subspace of $W$. Let that complement be $U$. We have that:

$$U + V = W\\U\cap V = \varnothing$$

By Greenman's theorem, we have: $$\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$$ which, using the relations above simplifies to: $$\dim(W) = \dim(U) + \dim(V) $$

Now, it is obvious that $\dim(U) \ge 1$: Suppose $\dim(V) = 0$, then $V = \varnothing$. But this is impossible since $U + \varnothing = U $, which contradicts the fact that $U$ is a complement of $V$ in $W$.

It follows immediately that $\dim(V) < \dim(W)$. This is a contradiction. Therefore $V$ is not a proper subset of $W$, in other words: $$V = W$$


Note that saying $V,W$ as having the same basis and same dimension is somewhat redundant; you should think and phrase that sentence as '$V,W$ are both generated by a basis of the same dimension', having the same basis for instance will directly imply they have the same dimension.

If they have the same basis then clearly they are the same, since every element is generated by it.

So, if indeed $V \leq W$ and both are generated by a basis of the same dimension then $V = span\{v_{1},..,v_{n}\}$ for $v_{i} \in W$, and indeed $\exists \{v_{1},..,v_{n},v_{n+1},..v_{m}\} \subset W$ completed to a basis of $W$, assuming the dimensions are finite.

But since we assume $dim V = dim W$ then $m = n$ and any addition to $\{v_{1},..,v_{n}\}$ results in a linearly dependent set within $W$. Then $span\{v_{1},..,v_{n}\} = W = V$.

If $dim V = dim W \approx N$, for example, and $V \leq W$ take:

$V = span\{cos(nx) | n \in N\} < span\{sin(nx),cos(nx) | n \in N \} = W$ to see how this proposition doesn't hold

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    $\begingroup$ Granted that they are over the same field as well. $\endgroup$ Commented Jul 25, 2016 at 22:42
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    $\begingroup$ In the infinite-dimensional case, having the same basis is a stronger condition. $\endgroup$ Commented Jul 25, 2016 at 22:54

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