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Let $V=\mathbb{R}^4$. $W_1$ is a subspace of $V$ spanned by vectors $a_1=(1, 2, 0, 1)$ and $a_2=(1,1,1,0)$. $W_2$ is a subspace of $V$ spanned by vectors $b_1=(1,0,1,0)$ and $b_2=(1,3,0,1)$. Determine $\dim(W_1+W_2)$ and $\dim(W_1 \cap W_2)$.

Attempt
The vectors $a_1$ and $a_2$ are linearly independent and span $W_1$, so they form a basis for $W_1$. Hence $\dim(W_1)=2$. The vectors $b_1$ and $b_2$ are linearly independent and span $W_2$, so they form a basis for $W_2$. Hence $\dim(W_2)=2$.

This is the step that I'm unsure of. I added vectors from $W_1$ and $W_2$ together, i.e. $a_1 + b_1, a_1 + b_2, a_2 + b_1, a_2 + b_2$ to form $$W_1+W_2=\left\{(2,2,1,1),(2,5,0,2),(2,1,2,0),(2,4,1,1) \right\}.$$

The first three vectors are linearly independent, but $$(2,4,1,1)= (2,5,0,2)+(2,1,2,0)-(2,2,1,1).$$ Hence there are only $3$ vectors that are linearly independent and span $W_1+W_2$, so $\dim(W_1+W_2)=3$.

We know that $\dim(W_1+W_2) = \dim(W_1) + \dim (W_2) - \dim(W_1 \cap W_2)$. Therefore $\dim(W_1 \cap W_2)=2+2-3=1$.

Question
Is my attempt correct?

Thank you for your time.

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    $\begingroup$ Everything is fine, although I would make it clear that $W_1+W_2$ is the set spanned by those four vectors, not just the set that is those four vectors. Also, again being picky, instead of saying "there are only $3$ vectors that are linearly independent and span $W_1+W_2$", you should say that "There is a basis of $W_1+W_2$ consisting of three vectors. You've certainly got the right method though. $\endgroup$ – Tom Oldfield May 21 '13 at 19:34
  • $\begingroup$ @TomOldfield Thank you for your comment. $\endgroup$ – user4167 May 21 '13 at 19:44
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    $\begingroup$ No worries, hope it helped! $\endgroup$ – Tom Oldfield May 21 '13 at 19:58
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Your attempt is correct, but note that your understanding of $W_1 + W_2$ might be wrong. By definition, $W_1 + W_2 = \{\vec{w}_1 + \vec{w}_2 | \vec{w}_i \in W_i, i=1, 2\}.$ However, This doesn't mean that the bases have to be added together. In other words, you can simply take $\{a_1,a_2,b_1,b_2\}$ as a spanning set for the space $W_1 + W_2$ and then toss out any linearly dependent vectors. The easiest way to do this is to row reduce the matrix whose columns are the corresponding basis vectors. For your example, the matrix will row reduce to (why? -- try it!) $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}.$$ This tells you that you have three linearly independent vectors and so the dimension of the $W_1 + W_2$ is 3.

Edit: another related question that is useful to ask yourself: if $a_1, a_2, a_3, a_4$ are linearly independent, are $a_1 + a_2, a_1 - a_2, a_3 + a_4, a_3 - a_4$? How would you check? (hint: checking independence can always be reduced to row reducing a matrix)

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  • $\begingroup$ Hi, thank you for your answer. I added the vectors from $W_1$ and $W_2$ because of the definition $W_1 + W_2 = \{\vec{w}_1 + \vec{w}_2 | \vec{w}_i \in W_i, i=1, 2\}$. Could you please explain why I can simply take $\{a_1,a_2,b_1,b_2\}$ as a spanning set for $W_1 + W_2$? $\endgroup$ – user4167 May 21 '13 at 19:35
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    $\begingroup$ Sure. I'll avoid the \vec signs. Any $w_1 \in W_1$ is a linear combination of $a_1$ and $a_2$, same for $b_1, b_2$. Thus, to say that you take any $w_1 \in W_1$ is to say that you can take $c_1a_1 + c_2 a_2$ for any scalars $c_1, c_2$, same for $W_2$. As such, $\{w_1 + w_2 | w_i \in W_i\} = \{c_1 a_1 + c_2 a_2 + d_1 b_1 + d_2 b_2 | c_i, d_i \in \mathbb{R}\} = \mathrm{span}(a_1,a_2,b_1,b_2)$. Intuitively, I just need to be able to get any vector in $W_1$ and any vector in $W_2$, then add them together. To do that, I just need basis vectors from both spaces, and those may overlap. Clear? $\endgroup$ – snar May 21 '13 at 19:41
  • $\begingroup$ Yes, thank you! $\endgroup$ – user4167 May 21 '13 at 19:44
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    $\begingroup$ @user4167: Here's another way to see it. Since $\vec 0\in W_1,W_2$, it follows that $W_1,W_2\subseteq W_1+W_2$. In fact, $W_1+W_2$ is the smallest subspace of $V$ containing $W_1,W_2$. (Why?) Thus, if we have given bases (or even just spanning sets) $B_1,B_2$ for $W_1,W_2$ respectively, then $B_1\cup B_2$ will be a spanning set (though not necessarily a basis) for $W_1+W_2$. $\endgroup$ – Cameron Buie May 21 '13 at 19:44
  • $\begingroup$ @CameronBuie Thank you for your explanation. It really helps! $\endgroup$ – user4167 May 21 '13 at 19:47

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