MHB Jae 's question at Yahoo Answers (Intersection of subspaces)

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The discussion centers on finding the intersection of two subspaces, P1 and P2, defined by their spanning vectors. The user has set up a system of equations using these vectors but is unsure how to proceed. A response clarifies that the four vectors span the sum of the subspaces, leading to a dimension calculation that shows the intersection is equal to P1. The basis for the intersection is identified as the vectors spanning P1. The original question was deleted, leaving the responder frustrated.
Fernando Revilla
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Here is the question:

P1 = span( (1,2,2) , (0,1,1) )
P2 = span( (2,1,1) , (1,0,0) )

What I currently did:

a[1 2 2] + b[0 1 1] - c[2 1 1] - d[1 0 0] = 0
[1 0 -2 -1
2 1 -1 0
2 1 -1 0]

From this matrix, I get a = 2c + d and b = -3c -2d

I'm not sure where I go from here. I know you have to chug a,b,c,d back in but not sure on how that works.

Any help would be greatly appreciated.
Thanks in advance!

Here is a link to the question:

Intersection of subspaces P1 and P2? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Jae,

Those four vectors span $P_1+P_2$. Besides
$$\begin{bmatrix}1&2&2\\0&1&1\\2&1&1\\1&0&0 \end{bmatrix} \sim\ldots \sim \begin{bmatrix}1&2&2\\0&1&1\\0&0&0\\0&0&0 \end{bmatrix}$$ This means that $\dim (P_1+P_2)=2$ which implies $$\dim (P_1\cap P_2)=\dim P_1+\dim P_2-\dim (P_1+P_2)=2+2-2=2$$
But $P_1\cap P_2\subset P_1$ (this happens in general) and $\dim (P_1\cap P_2)=\dim P_2$ (in this case), hence $P_1\cap P_2=P_1$, so $B=\{(1,2,2),(0,1,1)\}$ (for example) is a basis of $P_1\cap P_2$.

P.D. The question has just been deleted, so I need to cry at least one minute.
 
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