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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 $W_1+W_2=\langle W_1\cup W_2\rangle$.

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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 $W_2$.

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  • $\begingroup$ understood. i've considered it while writing the proof but managed to disregard it nevertheless. thanks! $\endgroup$ – TheLast Cipher Jan 26 '18 at 7:17
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This is not correct. In case 2, since $X\cap Y\neq\emptyset$, it is obviously false that$$|X\cup Y|=|X\cup Y|-|X \cap Y|.$$

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  • $\begingroup$ my case 2 is $X\cap Y\neq\phi$ not $X\cap Y=\phi$ $\endgroup$ – TheLast Cipher Jan 26 '18 at 7:10
  • $\begingroup$ @TheLastCipher That's what I meant. Sorry. I've edited my answer. $\endgroup$ – José Carlos Santos Jan 26 '18 at 7:13
  • $\begingroup$ sorry I thought i understand, but I do not. $\endgroup$ – TheLast Cipher Jan 26 '18 at 7:20
  • $\begingroup$ @TheLastCipher When you write that $|X\cup Y|=|X\cup Y|-|X \cap Y|$, you are claiming that a number is equal to itslef minus a number greater than $0$. $\endgroup$ – José Carlos Santos Jan 26 '18 at 7:21
  • $\begingroup$ ohh right. thank you! $\endgroup$ – TheLast Cipher Jan 26 '18 at 7:22

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