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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, and $\dim (W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$.

Proof: let $\{a_1,a_2,...,a_i,v_1,v_2...v_n\}$ be basis for $W_1$, then $\dim(W_1)= i+n$ and $\{b_1,b_2,...,b_j,v_1,v_2...v_n\}$ be basis for $W_2$, then $\dim(W_2)= j+n$ . Since both of them are finite-dimensional, and the definition of sum of two subspaces tells us that the basis of the sum is a combination of those two subspaces, which is $\dim(W_1+W_2)= i +j+n$. Hence we can arrive that $W_1+W_2$ is finite-dimensional.

To prove $\dim (W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$. Since the basis of the sum of two subspaces is a combination of both subspaces, $\dim(W_1+W_2) = i +j+n$. Since the both subspaces have n elements in common, so $\dim(W_1 \cap W_2)= n$.

Hence $\dim (W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$ is $ i +j+n = (i +n) + (j+n)- n$. Q.E.D

Suggestions for improving?

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    $\begingroup$ It looks like you are assuming the results you want to prove. For the first part, a better argument is every vector in $W_1 + W_2$ can be written as a linear combination of vectors in the multiset $\{a_1,a_2,...,a_i,v_1,v_2...v_n, b_1,b_2,...,b_j,v_1,v_2...v_n\}\}$ (apply definition of $W_1 + W_2$) so $\text{dim} (W_1 + W_2) \leq \text{dim} (W_1) + \text{dim} (W_2) \lt \infty$. For the second part: I don't see why any $b_k$ can't be written as a linear combination of $a_r$'s or how you've proven inclusion-exclusion. $\endgroup$ Jan 23, 2020 at 4:40
  • $\begingroup$ @user8675309 Thank you for the feedback, can you be more specific with the second part? (how you would write?) $\endgroup$
    – xyz12354
    Jan 23, 2020 at 5:09
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    $\begingroup$ It depends on what you know. Do you know the 'basis creation algorithm'? I.e. if you have a set of linearly independent vectors, how can you extend to a basis for your space. So e.g. if my space is all of $\mathbb F^n$ and I have $\{a_1, a_2\}$ as a linearly independent set, I can take standard basis vectors $\{e_1, ..., e_n\}$ (which I know generate my space) and one at a time try taking one from there and augment $\{a_1, a_2\}$ with $e_1$ then $e_2$, etc. unless $ e_k$ is already in the span of my augmented set (then ignore that vector and continue). Same idea applies to your problem. $\endgroup$ Jan 23, 2020 at 5:37

1 Answer 1

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Some problematic points in your proof:

Proof: let $\{a_1,a_2,...,a_i,v_1,v_2...v_n\}$ be basis for $W_1$, then $\dim(W_1)= i+n$ and $\{b_1,b_2,...,b_j,v_1,v_2...v_n\}$ be basis for $W_2$, then $\dim(W_2)= j+n$ .

You are missing some fundamental information. What are $i,j,n$? Why do both bases contain the same vectors $v_1,\dots,v_n$?

the definition of sum of two subspaces tells us that the basis of the sum is a combination of those two subspaces,

Presumably, you mean "a combination of those two bases". In any case, the term "a combination of" is too vague for this statement to be correct.

which is $\dim(W_1+W_2)= i +j+n$. Hence we can arrive that $W_1+W_2$ is finite-dimensional.

Since the both subspaces have n elements in common, so $\dim(W_1 \cap W_2)= n$.

It is not true that the two subspace have $n$ elements in common. If we're talking about vector spaces over $\Bbb R$ or $\Bbb C$, then the subspaces should have either infinitely many elements or one element in common.


A correct proof, in which I have attempted to parallel yours as much as possible.

Let $v_1,\dots,v_n$ be a basis of $W_1 \cap W_2$. Since $W_1 \cap W_2 \subseteq W_1$, we can extend this to a basis $v_1,\dots,v_n,a_1,\dots,a_i$ of $W_1$. Similarly, let $v_1,\dots,v_n,b_1,\dots,b_j$ be a basis of $W_2$. It is clear that the union of these bases, $$ \mathcal B = \{v_1,\dots,v_n,a_1,\dots,a_i,b_1,\dots,b_j\} $$ is a spanning set of $W_1 + W_2$. In order to show that this is a basis, we must also show that $\mathcal B$ is linearly independent.

One we have proven the claim that $\mathcal B$ is indeed a basis, we may simply count the elements of each basis to find $$ \dim(W_1 \cap W_2) = n, \quad \dim(W_1) = n+i, \quad \\ \dim(W_2) = n+j, \quad \dim(W_1 + W_2) = n+i+j. $$ We can then verify the desired result by plugging these in to the desired equation.


Proof that $\mathcal B$ is linearly independent: let coefficients $c_k,p_k,q_k$ be such that $$ c_1v_1 + \cdots + c_nv_n + p_1 a_1 + \cdots + p_ia_i + q_1 b_1 + \cdots + q_j b_j = 0\tag{1} $$ Rearrange the above to get $$ c_1v_1 + \cdots + c_nv_n + p_1 a_1 + \cdots + p_ia_i = -q_1 b_1 - \cdots - q_j b_j. \tag{2} $$ Let $w$ denote the vector $w = -q_1 b_1 - \cdots - q_j b_j$. From the left-hand side of Equation (2), we know that $v \in W_1$. From the right-hand side of the equation, we know that $v \in W_2$. Thus, $v \in W_1 \cap W_2$.

Now, revisiting the left-hand side, we have $$ c_1 v_1 + \cdots + c_n v_n + p_1 a_1 + \cdots + p_ia_i = v \in W_1 \cap W_2. $$ Because $\{v_1,\dots,v_n\}$ is a basis of $W_1 \cap W_2$, there exist $c_1',\dots,c_n'$ such that $c_1'v_1 + \cdots + c_n'v_n = v$. Thus, we have $$ c_1 v_1 + \cdots + c_n v_n + p_1 a_1 + \cdots + p_ia_i = c_1' v_1 + \cdots + c_n' v_n \implies\\ (c_1-c_1') v_1 + \cdots + (c_n-c_n') v_n + p_1 a_1 + \cdots + p_ia_i = 0. $$ But $\{v_1,\dots,v_n,a_1,\dots,a_i\}$ is linearly independent, so we can conclude that $p_1 = \cdots = p_i = 0$ (and that $c_1 - c_1' = \cdots = c_n - c_n' = 0$, which we don't use).

With that, we can rewrite Equation (1) as $$ c_1v_1 + \cdots + c_nv_n + q_1 b_1 + \cdots + q_j b_j = 0. $$ The set $\{v_1,\dots,v_n,b_1,\dots,b_j\}$ is linearly independent, so we can conclude that $c_1 = \cdots = c_n = 0$ and $q_1 = \cdots = q_j = 0$. Thus, we began with Equation (1) and concluded that all coefficients must be zero. It follows that $\mathcal B$ is linearly independent.


An alternative proof, with the help of quotient spaces:

Denote $K = W_1 \cap W_2$. Consider the canonical projection map $\pi:V \to V/K$. For any subspace $W \subset V$, let $\pi(W)$ denote the image of $W$ under $\pi$, i.e. the image of $\pi|_{W}$. Note that the rank-nullity theorem tells us that $$ \dim(\pi(W)) = \dim(W) - \dim (\ker \pi|_W) $$ In the case that $K \subset W$, we have $\ker \pi|_W = K$.

Verify that $\pi(W_1) \cap \pi(W_2) = \{0\}$. Thus, the image of $W_1 + W_2$ can be decomposed into a (internal) direct sum: $$ \pi(W_1 + W_2) = \pi(W_1) \oplus \pi(W_2). $$ It follows that $\dim \pi(W_1 + W_2) = \dim\pi(W_1) + \dim(\pi(W_2))$. Applying the rank-nullity theorem allows us to rewrite this as $$ \dim(W_1 + W_2) - \dim (K) = [\dim(W_1) - \dim(K)] + [\dim(W_2) - \dim(K)]. $$ Adding $\dim (K)$ to both sides in the above yields the desired result.

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    $\begingroup$ I've added a corrected version of your proof to my post if you're interested. $\endgroup$ Jan 23, 2020 at 6:23
  • $\begingroup$ If I may ask, how do I prove that B is linearly independent? $\endgroup$ Mar 12 at 23:50
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    $\begingroup$ @A.Srivastava That's a fair question. See my latest edit. $\endgroup$ Mar 13 at 13:28

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