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.
Best Answer
I think is false:
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.