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MIT 18.06 Linear Algebra, Spring 2010 8 Online
OpenStudy (anonymous):

so if you have a 0 in the pivot position such as the following matrix: 0 1 2 3 4 0 0 0 1 2 0 0 0 0 0 how would you go about finding the basis for all four subspaces?

OpenStudy (anonymous):

If all elements below pivot that is diagonal element is zero than matrix becomes zero

OpenStudy (anonymous):

Let A be the original 3 x 5 matrix. The column space has is a subset of R^3 spanned by the second and fourth columns of A. The left null space of A the subset of R^3 that is spanned by any vector that is perpendicular to both of those columns. The row space of A is a subset of R^5 that is spanned by the first two rows of A, and the null space of A is the subset of R^5 that is spanned by (0, 0, 1, 0, 0)', (-2, 0, 0, 1, 0)', and (-4, -2, 0, 0, 1). I got that basis from the free variables you gave in your example.

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