If $W_1$ and $W_2$ are any two subspaces of a finite dimensional vector space $V$, then $\dim(W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1\cap W_2)$

Question

This is a proof verification question.

Proof:

By theorem 4.4, $W_1+W_2=\langle W_1\cup W_2\rangle$ s.t. $W_1$, $W_2$, and $W_1+W_2$ are subspaces. Let $\langle X\rangle=W_1$ and $\langle Y\rangle=W_2$ s.t. $|X|=m$ and $|Y|=n$.

$\Rightarrow$ Since $W_1+W_2:=\{w=\alpha + \beta$ $\mid$ $\alpha\in W_1$ and $\beta\in W_2\}$, then $\langle X\cup Y\rangle=W_1+W_2$ $\Rightarrow$

$\mathbf{Case 1}$ ($X\cap Y=\phi$): $\dim(W_1+W_2)=|X\cup Y|=|X|+|Y|+0=\dim W_1 +\dim W_2 +\dim(W_1\cap W_2)$.

$\mathbf{Case 2}$ ($X\cap Y\neq\phi$):\begin{align} |X\cup Y| &=|(X\cup Y)\setminus(X\cap Y)|\\ &=|X\cup Y|-|X\cap Y|\\&=|X|+|Y|-|X\cap Y|\\&=\dim W_1+\dim W_2-\dim(W_1\cap W_2). \Bbb{QED} \end{align}

$\mathbf{Theorem 4.4.}$ $W_1+W_2$ is the smallest subspace containing both $W_1$ and $W_2$; that is $W_1+W_2=\langle W_1\cup W_2\rangle$.

Answer 1

Your proof is not correct. You cannot start with arbitrary bases for $W_1$ and $W_2$. $X$ and $Y$ might not intersect, but $W_1\cap W_2$ may still have positive dimension. For example, $X = \{e_1, e_2\}$ and $Y = \{e_1 + e_2\}$.

You should start with a base for $W_1\cap W_2$ and extend that to a base of $W_1$ and a base of $W_2$.

Answer 2

This is not correct. In case 2, since $X\cap Y\neq\emptyset$, it is obviously false that$$|X\cup Y|=|X\cup Y|-|X \cap Y|.$$