intersecting two matrices problem U = span ( [ 1, 2, 3, 4, 3, 2, 1 ], [ 4, 0, 2, 0, 2, 0, 4 ], [ 3, 1, 2, 0, 1, 1, 0 ] ) V = span( [ 2, 1, 3, 4, 0, 3, 3 ], [ 4, 3, 5, 4, 4, 1, 1 ], [ 1, 1, 1, 1, 1, 1, 2 ] ) 1. Find the basis of U ∩ V My Answer: dim(U) + dim(V) = dim(U + V) + dim(U ∩ V). dim(U) = 3 dim(V) = 3 dim(U + V) = 3 Therefore: dim(U ∩ V) = 3 U ∩ V = span( [ 2, 1, 3, 4, 0, 3, 3 ], [ 4, 3, 5, 4, 4, 1, 1 ], [ ?, ?, ?, ?, ?, ?, ? ] ) With much calculations, I managed to find the first two basis.. Any idea how to find the last basis? 
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a+ 4b+ 3c= 2p+ 4q+ r 2a+ c= p+ 3q+ r 3a+ 2b+ 2c= 3p+ 5q+ r 4a= 4p+ 4q+ r 3a 2b+ c= 4q+ r 2a c= 3p+ q+ r a 4b= 3p+ q+ 2r So we have 7 equations in 6 unknowns. That is, of course, overdetermined. What you can do is solve any six equations for the 6 unknowns, then see if those values satisfy the seventh equation. If they do, you can get put the values of a, b, c into the first three equations to find a basis for the intersection, else the intersection is only the 0 vector. (How did you determine that "dim(U+ V)= 3"? Without doing any calculation, it could be as large as 6.) Quote:

Adding 3a + 2b + 2c = 3p + 5q + r to 3a  2b + c = 4q + r gives 6a + 3c = 3p + 9q + 2r, but multiplying 2a + c = p + 3q + r by 3 gives 6a + 3c = 3p + 9q + 3r, so r = 0. Hence 3a  2b + c = 4q and 2a  c = 3p + q. Adding those gives 5a  2b = 3p + 5q, but 4a = 4p + 4q + r leads to 5a = 5p + 5q, so b = p. In summary, a = p + q, b = p, c = 2(p + q)  3p  q = q  p and r = 0. For any values of p and q, all of the 7 equations are now satisfied. 
So.. what is the total number of basis for U ∩ V?? 
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Thus the dimension of $U \cap V$ is 2. This is a plane embedded in 7 dimensional space. 
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