Welcome to our community

Be a part of something great, join today!

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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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 $\dim (P_1+P_2)=2$ wich 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.
 
Last edited: