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Let $W_1=\operatorname{span}\{(1,-1,-3),(3,0,-3),(1,2,3),(4,5,6)\}$ and $W_2=\operatorname{span}\{(6,3,7),(0,0,11),(-4,-2,0)\}$. Find dimension and one basis of $W_1,W_2,W_1+W_2,W_1\cap W_2$.

$$ A= \begin{bmatrix} 1 & -1 & -3 \\ 3 & 0 & -3 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}\sim \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\Rightarrow \operatorname{rank}(A)=2\Rightarrow \dim(W_1)=2$$

How to find one basis for $W_1$?

$$ B= \begin{bmatrix} 6 & 3 & 7 \\ 0 & 0 & 11 \\ -4 & -2 & 0 \\ \end{bmatrix}\sim \begin{bmatrix} 1 & 1/2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}\Rightarrow \operatorname{rank}(B)=2\Rightarrow \dim(W_2)=2$$

How to find one basis for $W_2$?

For $W_1+W_2,$ $$\begin{bmatrix} 1 & -1 & -3 \\ 3 & 0 & -3 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \\ 6 & 3 & 7 \\ 0 & 0 & 11 \\ -4 & -2 & 0 \\ \end{bmatrix}\sim \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\Rightarrow \dim(W_1+W_2)=3$$

How to find one basis of $W_1+W_2$?

From previous, $\dim(W_1\cap W_2)=1$

How to find one basis of $W_1\cap W_2$?

Could someone show the procedure for finding at least one basis for either $W_1,W_2,W_1+W_2$ or $W_1\cap W_2$?

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  • $\begingroup$ the basis for $W_1$ are the non-sero rows, that is, $W_1$ have dimension $2$. Same for the other cases. $\endgroup$ – Masacroso Apr 23 '17 at 8:51
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For $W_1$, $W_2$ and $W_1 + W_2$ you can take the non-zero rows as a basis of the vector spaces. To find the elements in the intersection it must be written as a linear combination of the basis of both vector spaces, so it reduces to solving the equation:

$$a(1,0,-1) + b(0,1,2) = c(1,\frac 12,0) + d(0,0,1)$$

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  • $\begingroup$ $b=\frac{1}{2}a,c=a,d=0$. One basis of $W_1\cap W_2$ is $\{ a\begin{bmatrix} 1 \\ 1/2 \\ 0 \\ \end{bmatrix}: a\in\mathbb R\}$ If $\dim(W_1\cap W_2)>1$, how would we know be basis? $\endgroup$ – user300048 Apr 23 '17 at 10:23
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    $\begingroup$ In a similar manner, but you instead of one free variable you will have more. $\endgroup$ – Stefan4024 Apr 23 '17 at 10:35
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Just thik it geometrically.You have made my answer easier by calculating the rank of each of the subspaces.Rank of W1 is 2 implies that W1 represents a plane in 3 dim. space.That spanning set which you have given...any two elements are independent...so you may take any two of them to form the basis. ie. {(1,-1,-3);(3,0,-3)} is a basis (look you may take the basis in 4C2 ways). From the similar idea you can find out the basis of W2. Now dimension of W1+W2 is 3(as you calculated).Look both W1and W2 are subspace of R3(3 dimensional euclidean space).So you have the easiest answer!Just take your standard basis {(1,0,0);(0,1,0);(0,0,1)}. Now look intersection of W1 and W1 has dimension 1 which represents a straight line.You are to find out those planes that represents W1 and W2 and obviously their intersecting part is that straight line.Just calculate it

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