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Let $V$ be a vector space over a field $k$ and let $U$, $W$ be finite-dimensional subspaces of $V$.

Prove that $\dim(U+W) + \dim(U\cap W) = \dim U + \dim W$

I'm given that to begin this problem I can find the bases:

$\{v_1,\dots,v_p\}$ for $U\cap W$

$\{v_1,\dots,v_p, u_1,\dots,u_q\}$ for $U$

and $\{v_1,\dots,v_p, w_1,\dots,w_r\}$ for $W$

and then I just need to show that $\{v_1,\dots,v_p, u_1,\dots,u_q, w_1,\dots,w_r\}$ is a basis for $U+W$.

My question is: how does one go about showing that it is a basis for $U+W$ and then use that to prove the above question?

Side note: This question has already been asked here: Given two subspaces $U,W$ of vector space $V$, how to show that $\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)$

However, the first answer given does not apply to solving it the way I want to with finding the bases. The second answer simply gives me what I already knew to start with. Thus, I am asking this question again since I'm asking how to solve it a particular way instead of just any general hints towards solving it.

Community
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Lindsey G
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2 Answers2

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Just do the computations; the fact that the set spans $U+W$ should be clear, so we prove linear independence.

Suppose $$ \alpha_1v_1+\dots+\alpha_pv_p+ \beta_1u_1+\dots+\beta_qu_q+ \gamma_1w_1+\dots+\gamma_rw_r=0 $$ Then $$ x=\underbrace{\alpha_1v_1+\dots+\alpha_pv_p+ \beta_1u_1+\dots+\beta_qu_q}_{\in U}= -(\underbrace{\gamma_1w_1+\dots+\gamma_rw_r}_{\in W}) $$ belongs to $U\cap W$. Thus $$ x=\delta_1v_1+\dots+\delta_pv_p $$ and therefore $$ \delta_1v_1+\dots+\delta_pv_p=-(\gamma_1w_1+\dots+\gamma_rw_r) $$ so that $$ \delta_1v_1+\dots+\delta_pv_p+\gamma_1w_1+\dots+\gamma_rw_r=0 $$ Since the set $\{v_1,\dots,v_p,w_1,\dots,w_r\}$ is linearly independent, we conclude $$ \delta_1=0,\quad\dots,\quad\delta_p=0,\quad \gamma_1=0,\quad\dots,\quad\gamma_r=0 $$ and also that $$ \alpha_1v_1+\dots+\alpha_pv_p+\beta_1u_1+\dots+\beta_qu_q=0 $$ so, from linear independence of $\{v_1,\dots,v_p,u_1,\dots,u_q\}$ we get $$ \alpha_1=0,\quad\dots,\quad\alpha_p=0,\quad \beta_1=0,\quad\dots,\quad\beta_q=0 $$

egreg
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4

A shorter proof: consider $T:U \times W \to U + W$ by $T(u, w) = u - w,$ then $\ker T = U \cap W$ and the theorem of dimension $\dim \ker T + \dim \ \mathrm{image}\ T = \dim\ \mathrm{domain}\ T$ gives the result at once (since $T(U \times W) = U + W$ and $\dim U \times W = \dim U + \dim W$).

William M.
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  • How can $\ker T = U \cap W$? $U \cup W$ isn't even a subspace of $U \times W$. – user193319 Jun 24 '19 at 14:34
  • Whoops, I meant "$U \cap W$ isn't even a subspace of $U \times W$." I think the proof can be fixed, however; it would seem that $\ker T$ is isomorphic to $U \cap W$. – user193319 Jun 24 '19 at 15:15
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    @user193319 i know this is an old post but you are correct. The kernel of $T$ is the set of all ordered pairs of the form $(v,v)$ where $v \in U \cap W$. There is a known map called the diagonal map defined as $\Delta: V \to V\times V$ via $v \mapsto (v,v)$. So, ker $T = \Delta(U \cap W)$. – Important_man74 Jan 13 '24 at 03:24
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    As you noted, $\Delta(U \cap W)$ has the same dimension as $U \cap W$. – Important_man74 Jan 13 '24 at 03:25