Suppose $W_1,W_2$ are subspaces of $\mathbb{R}^4$. $W_1$ is spanned by $(1,2,3,4), (2,1,1,2)$ and $W_2$ is spanned by $(1,0,1,0),(3,0,1,0)$. I have to find a basis for $W_1\cap W_2$.

I have calculated basis and dimension for $W_1+W_2$, by row-reducing the matrix whose rows are these four spanning vectors. I have found that $\dim(W_1+W_2)=3$. So it follows the dimension of $\dim(W_1\cap W_2)=1$. But how do I find a basis for it? My textbook provides a long method, by first finding a homogenous system and row-reducing the corresponding matrix. Is there a shorter way?



The quickest way is to close your eyes and write down a nonzero vector that you can show is in both vector spaces. Since you know the space is one-dimensional, this will be your basis. It's always better to be lucky than good.

This is not as ridiculous as it sounds; you know the vector must have $0$ in the second and fourth coordinates. Hence a reasonable guess is $(1,2,3,4)-2(2,1,1,2)=(-3,0,1,0)$. Now if this happens to be in the second vector space, you are done. And, in fact, $(-3,0,1,0)=3(1,0,1,0)-2(3,0,1,0)$.

  • $\begingroup$ Thanks for answer , but why you have subtracted one vector minus twice of other $\endgroup$ – godonichia Dec 31 '14 at 7:09
  • $\begingroup$ One answer is because I need a linear combination of the basis of $W_1$ that has a zero in the second and fourth coordinates. The second answer is that 2 is twice 1, and 4 is twice 2, so this particular linear combination will get me what I need. $\endgroup$ – vadim123 Dec 31 '14 at 7:11
  • $\begingroup$ The vector is inboth w1 and w2 but how can one be sure that it spans it , $\endgroup$ – godonichia Dec 31 '14 at 7:13
  • $\begingroup$ Because you have proven that this space ($W_1\cap W_2$) is one-dimensional. $\endgroup$ – vadim123 Dec 31 '14 at 7:15
  • $\begingroup$ It can span vectors which areintegral multiples of 3 in 1st coordinate , what about other vectors say ( 2,0,1,0 ) $\endgroup$ – godonichia Dec 31 '14 at 7:18

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