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 July 18th, 2018, 04:17 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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? July 19th, 2018, 09:26 AM   #2
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Quote:
 Originally Posted by zollen U = span ( [ 1, 2, 3, 4, 3, 2, 1 ], [ 4, 0, 2, 0, -2, 0, -4 ], [ 3, 1, 2, 0, 1, -1, 0 ] )
So any vector in U is of the form of a[ 1, 2, 3, 4, 3, 2, 1 ]+ b[ 4, 0, 2, 0, -2, 0, -4 ]+ c[ 3, 1, 2, 0, 1, -1, 0 ]= [a+ 4b+ 3c, 2a+ c, 3a+ 2b+ 2c, 4a, 3a- 2b+ c, 2a- c, a- 4b]

Quote:
 V = span( [ 2, 1, 3, 4, 0, 3, -3 ], [ 4, 3, 5, 4, 4, 1, 1 ], [ 1, 1, 1, 1, 1, 1, 2 ] )
So any vector in V is of the form p[ 2, 1, 3, 4, 0, 3, -3 ]+ q[ 4, 3, 5, 4, 4, 1, 1 ]+ r[ 1, 1, 1, 1, 1, 1, 2 ]= [2p+ 4q+ r, p+ 3q+ r, 3p+ 5q+ r, 4p+ 4q+ r, 4q+ r, 3p+ q+ r, -3p+ q+ 2r]

Quote:
 1. Find the basis of U ∩ V
A vector is in both U and V if there exist a, b, c, p, q, r such that
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, over-determined. 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:
 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?

Last edited by skipjack; July 19th, 2018 at 11:09 PM. July 19th, 2018, 11:03 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 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. Thanks from zollen July 21st, 2018, 11:07 AM #4 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 So.. what is the total number of basis for U ∩ V?? July 21st, 2018, 11:14 AM   #5
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Quote:
 Originally Posted by skipjack 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.
Quote:
 Originally Posted by zollen So.. what is the total number of basis for U ∩ V??
As noted by Skipjack there are 2 degrees of freedom represented by the free variables p and q.

Thus the dimension of $U \cap V$ is 2.

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