Please explain how to find basis for set of 2x3 matrices whose null space contains (2, 1, 1). This is a question from Recitation video in http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/matrix-spaces-rank-1-small-world-graphs/ It's not very well explained in the video.

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Please explain how to find basis for set of 2x3 matrices whose null space contains (2, 1, 1). This is a question from Recitation video in http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/matrix-spaces-rank-1-small-world-graphs/ It's not very well explained in the video.

MIT 18.06 Linear Algebra, Spring 2010
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Lokeshh, if A is a (2,3) matrix with u=(2,1,1) in its null space then it must be that both rows of A are orthogonal to u. Any two independent vectors v and w orthogonal to u will span a plane perpendicular to u, and the recitation lecturer suggests the two vectors v=(1,0,-2) and w=(0,1,-1) are natural choices. As long as the rows of A are linear combinations of v and w then we will have Au=0; and because v and w span the whole plane, linear combinations of v and w will generate all vectors orthogonal to u. So we can make a basis for all (2,3) matrices satisfying Au=0 by using the basis [1,0,-2; 0,0,0], [0,1,-1; 0,0,0], [0,0,0; 1,0,-2], [0,0,0; 0,1,-1]. Does that help? Josh.

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