Hi guys,

I am stuck with this questions, and final is coming... I would be appreciate if anyone can help me to solve these questions.

Thanks

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- December 4th 2009, 07:50 AMdolphinOrthogonal basic && basic for the orthogonal complement
Hi guys,

I am stuck with this questions, and final is coming... I would be appreciate if anyone can help me to solve these questions.

Thanks

http://i188.photobucket.com/albums/z...n/Untitled.jpg - December 5th 2009, 07:33 AMHallsofIvy
For

"null space" is, of course, the set of all vector so that Av= 0:

.

That matrix row-reduces to .

The fourth row tells us nothing. The third row tells us that f= 0. The second row tells us that e= 0. The first row tells us that a- b- 2c= 0 or that a= b+ 2c. Notice that d does not appear in any of those. A general vector in the null space is of the form (b+ 2c, b, c, d, 0, 0)= b(1, 1, 0, 0, 0, 0)+ c(2, 0, 1, 0, 0, 0)+ d(0, 0, 0, 1, 0, 0). Those three vectors, (1, 1, 0, 0, 0, 0), (2, 0, 1, 0, 0, 0), and (0, 0, 0, 1, 0, 0) form a basis for the null space.

But "the null space of A containing (1, 1, 0, 0, 0, 0)" doesn't really make sense! Yes, (1, 1, 0, 0, 0, 0) is in the null space- "the null space containing (1, 1, 0, 0, 0, 0)" is just the null space itself. If the problem had said "the smallest subspace of the null space containing (1, 1, 0, 0, 0, 0)" then we can just take that smallest subspace to be the multiples of (1, 1, 0, 0, 0, 0)

To find a basis for "the orthogonal complement of the null space" of the given matrix, just repeat what I did to find a basis for the null space. Then look for vectors (a, b, c, d, e, f) that have dot product with each of those basis vectors equal to 0. Depending upon how many vectors you get in the basis (the dimension of the null space) you will have up to 6 equations for a, b, c, d, e, and f.