My Math Forum

My Math Forum (http://mymathforum.com/math-forums.php)
-   Linear Algebra (http://mymathforum.com/linear-algebra/)
-   -   intersecting two matrices problem (http://mymathforum.com/linear-algebra/344575-intersecting-two-matrices-problem.html)

zollen July 18th, 2018 04:17 PM

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?

Country Boy July 19th, 2018 09:26 AM

Quote:

Originally Posted by zollen (Post 596751)
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?

skipjack July 19th, 2018 11:03 PM

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.

zollen July 21st, 2018 11:07 AM

So.. what is the total number of basis for U ∩ V??

romsek July 21st, 2018 11:14 AM

Quote:

Originally Posted by skipjack (Post 596782)
<snip>
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 (Post 596861)
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.

This is a plane embedded in 7 dimensional space.


All times are GMT -8. The time now is 12:42 AM.

Copyright © 2019 My Math Forum. All rights reserved.