# Theorem 6, Section 2.3 of Hoffman and Kunze’s Linear Algebra

If $$W_1$$ and $$W_2$$ are finite-dimensional subspaces of a vector space $$V$$, then $$W_1+W_2$$ is finite-dimensional and $$\mathrm{dim}(W_1)+ \mathrm{dim}(W_2)= \mathrm{dim}(W_1\cap W_2)+ \mathrm{dim}(W_1+W_2)$$.

For seek of completeness, first I will show Hoffman and Kunze’s proof after that I will show my potential approach.[1]

Hoffman and Kunze’s Proof: If $$W_1\cap W_2=\{0_V\}$$. $$\emptyset$$ is the basis of $$W_1\cap W_2=\{0_V\}$$, here is the proof. So $$\mathrm{dim}(W_1\cap W_2)=0$$. Since $$W_1,W_2$$ is finite-dimensional subspaces of $$V$$, we have $$\exists B_1=\{u_1,…,u_n\}$$ finite basis of $$W_1$$, $$\exists B_2=\{v_1,…,v_m\}$$ finite basis of $$W_2$$. Claim: $$B=B_1\cup B_2=\{u_1,…,u_n,v_1,…,v_m\}$$ is basis of $$W_1+W_2$$. Proof: let $$x\in W_1+W_2$$. Then $$\exists w_1\in W_1$$, $$\exists w_2\in W_2$$ such that $$x=w_1+w_2$$. Since $$W_1=\mathrm{span}(B_1)$$, $$W_2= \mathrm{span}(B_2)$$, we have $$w_1=\sum_{i=1}^n a_i\cdot u_i$$ and $$w_2=\sum_{i=1}^m b_i\cdot v_i$$. So $$x= \sum_{i=1}^n a_i\cdot u_i+ \sum_{i=1}^m b_i\cdot v_i \in \mathrm{span}(B)$$. Thus $$W_1+W_2\subseteq \mathrm{span}(B)$$. Conversely, let $$x\in \mathrm{span}(B)$$. Then $$x= \sum_{i=1}^n p_i\cdot u_i + \sum_{i=1}^m q_i\cdot v_i$$. Let $$u= \sum_{i=1}^n p_i\cdot u_i$$ and $$v=\sum_{i=1}^m q_i\cdot v_i$$. So $$u\in \mathrm{span}(B_1)=W_1$$ and $$v\in \mathrm{span}(B_2)=W_2$$. So $$x=u+v\in W_1+W_2$$. Thus $$\mathrm{span}(B)\subseteq W_1+W_2$$. Another equivalent to show $$\mathrm{span}(B)\subseteq W_1+W_2$$. It follows from a more general fact, if $$S\subseteq W$$ and $$W\leq V$$, then $$\mathrm{span}(S)\subseteq W$$. Hence $$W_1+W_2=\mathrm{span}(B)$$. If $$\sum_{i=1}^n r_i\cdot u_i + \sum_{i=1}^m s_i\cdot v_i =0_V$$. Let $$u= \sum_{i=1}^n r_i\cdot u_i$$ and $$v= \sum_{i=1}^m s_i\cdot v_i$$. Then $$u=-v$$. $$u\in W_1$$. Since $$v\in W_2$$, we have $$-v\in W_2$$. So $$u=-v\in W_1\cap W_2=\{0_V\}$$. Thus $$u=-v=0_V$$. Since $$B_1$$ is independent, $$u= \sum_{i=1}^n r_i\cdot u_i=0_V$$ implies $$r_i=0_F$$, $$\forall i\in J_n$$. Since $$B_2$$ is independent, $$-v= \sum_{i=1}^m (-s_i)\cdot v_i=0_V$$ implies $$-s_i=-1_F \cdot s_i=0_F$$, $$\forall i\in J_m$$. So $$s_i=0_F$$, $$\forall i\in J_m$$. Hence $$B=B_1\cup B_2$$ is linearly independent. $$B$$ is finite basis of $$W_1+W_2$$. By theorem 4 corollary 1 section 2.3, $$\mathrm{dim}(W_1+W_2)=|B|=n+m$$. So $$\mathrm{dim}(W_1)+ \mathrm{dim}(W_2)=n+m=0+(n+m)=\mathrm{dim}(W_1\cap W_2)+ \mathrm{dim}(W_1+W_2)$$.

If $$W_1\cap W_2\neq \{0_V\}$$, then $$\exists \alpha_1 \in W_1\cap W_2$$ such that $$\alpha_1 \neq 0_V$$. $$\{\alpha_1\}$$ is linearly independent. By theorem 5 section 2.3, $$\exists B’\subseteq W_1\cap W_2$$ such that $$B’$$ is finite basis of $$W_1\cap W_2$$ and $$\alpha_1 \subseteq B’$$. Let $$B’=\{\alpha_1,…,\alpha_k\}$$. Since $$B’$$ is independent and $$B’\subseteq W_1\cap W_2\subseteq W_1, W_2$$, by theorem 5 section 2.3, $$\exists B_1\subseteq W_1$$ such that $$B_1$$ is finite basis of $$W_1$$ and $$B’\subseteq B_1$$, $$\exists B_2\subseteq W_2$$ such that $$B_2$$ is finite basis of $$W_2$$ and $$B’\subseteq B_2$$. Let $$B_1=\{\alpha_1,…,\alpha_k,\beta_1,…,\beta_m\}$$ and $$B_2=\{\alpha_1,…,\alpha_k, \gamma_1,…,\gamma_n\}$$. Claim: $$B=B_1\cup B_2=\{\alpha_1,…,\alpha_k, \beta_1,…,\beta_m, \gamma_1,…,\gamma_n\}$$ is basis of $$W_1+W_2$$. Proof: let $$x\in W_1+W_2$$. Then $$\exists w_1\in W_1$$, $$\exists w_2\in W_2$$ such that $$x=w_1+w_2$$. Since $$W_1=\mathrm{span}(B_1)$$, $$W_2= \mathrm{span}(B_2)$$, we have $$w_1=\sum_{i=1}^k a_i\cdot \alpha_i+\sum_{i=1}^m b_i\cdot \beta_i$$ and $$w_2=\sum_{i=1}^k c_i\cdot \alpha_i+\sum_{i=1}^n d_1\cdot \gamma_i$$. So $$x=w_1+w_2= \sum_{i=1}^k (a_i+c_i)\cdot \alpha_i+\sum_{i=1}^m b_i\cdot \beta_i+ \sum_{i=1}^n d_1\cdot \gamma_i$$. Thus $$x=w_1+w_2\in \mathrm{span}(B)$$. Hence $$W_1+W_2=\mathrm{span}(B)$$. If $$\sum_{i\in J_k} a_i\cdot \alpha_i + \sum_{i\in J_m} b_i\cdot \beta_i + \sum_{i\in J_n} d_1\cdot \gamma_i=0_V$$. Let $$x= \sum_{i\in J_k} a_i\cdot \alpha_i$$ , $$y= \sum_{i\in J_m} b_i\cdot \beta_i$$ and $$z= \sum_{i\in J_n} d_1\cdot \gamma_i=0_V$$. Then $$x+y=-z$$. Since $$x\in W_1\cap W_2$$, $$y\in W_1$$, we have $$x+y\in W_1$$. Since $$z\in W_2$$, we have $$-z\in W_2$$. So $$x+y=-z\in W_1\cap W_2=\mathrm{span}(B’)$$. $$-z=\sum_{i\in J_k}d_i\cdot \alpha_i$$ implies $$\sum_{i\in J_k}d_i\cdot \alpha_i +z= \sum_{i\in J_k}d_i\cdot \alpha_i + \sum_{i\in J_n}c_i\cdot \gamma_i=0_V$$. Since $$B_2=\{\alpha_1,…,\alpha_k,\gamma_1,…,\gamma_n\}$$ is independent, $$d_i=0_F$$, $$\forall i\in J_k$$ And $$c_i=0_F$$, $$\forall i\in J_n$$. So $$z=0_V$$. We have $$x+y= \sum_{i\in J_k} a_i\cdot \alpha_i + \sum_{i\in J_m} b_i\cdot \beta_i=0_V$$. Since $$B_1=\{\alpha_1,…,\alpha_k,\beta_1,…,\beta_m\}$$ is independent, $$a_i=0_F$$, $$\forall i\in J_k$$ and $$b_i=0_F$$, $$\forall i\in J_m$$. Thus $$a_i=0_F$$, $$\forall i\in J_k$$, $$b_i=0_F$$, $$\forall i\in J_m$$ and $$c_i=0_F$$, $$\forall i\in J_n$$. Hence $$B$$ is linearly independent. $$B$$ is finite basis of $$W_1+W_2$$. $$\mathrm{dim}(W_1)+ \mathrm{dim}(W_2)=(k+m)+(k+n)=k+(k+m+n)= \mathrm{dim}(W_1\cap W_2)+ \mathrm{dim}(W_1+W_2)$$.

My attempt: In Hoffman and Kunze’s proof, they start digging from basis of $$W_1\cap W_2$$. Which I think is really clever. If we start digging from basis of $$W_1+W_2$$, that don’t really help much, because we can’t construct basis of $$W_1$$ & $$W_2$$, in general. I would have approach this theorem from basis of $$W_1$$ & $$W_2$$, since we are given $$W_1$$ & $$W_2$$ is finite-dimensional subspaces of $$V$$. Let $$B_1$$ and $$B_2$$ be finite basis of $$W_1$$ and $$W_2$$, respectively. Question: Can we claim, $$(1)$$ $$B_1\cap B_2$$ is basis of $$W_1\cap W_2$$, and $$(2)$$ $$B_1\cup B_2$$ is basis of $$W_1+W_2$$?

[1] Hoffman, K.; Kunze, R., Linear algebra, Englewood Cliffs, N. J.: Prentice-Hall, Inc. VIII, 407 p. (1971). ZBL0212.36601.

• @XanderHenderson why did you delete all comments? Isn’t that abuse of power. I can’t even delete my own post! Jun 28 at 18:23
• Counterexample: $W_1=\Bbb{R}^2$, $W_2=\{(x,y)|x=y\}$. $B_1=\{(1,0),(0,1)\}$ is basis of $W_1$ and $B_2=\{(1,1)\}$ is basis of $W_2$. $(1)$ $B_1\cap B_2=\emptyset$ is not basis of $W_1\cap W_2=W_2$. (2) $B_1\cup B_2=\{(1,0),(0,1), (1,1)\}$ is not basis of $W_1+W_2= \Bbb{R}^2$. Jun 28 at 20:45

In $${\Bbb R}^3$$ with canonical basis let $$e_1+e_2,e_2$$ be a basis for $$U$$ and $$e_1+e_3,e_3$$ be a basis for $$W$$. Then $$U\cap W$$ is generated by $$e_1$$ which is not among the given basis vectors.
• Thank you for the answer. Those two claims are false. What is $U$ and $W$? Jun 28 at 8:16
• The way I denoted your subspaces $U_1$ and $U_2$. Is a counterexample about your claim Jun 28 at 8:20