Highlights Math 304 Linear Algebra From Wednesday: I basis and dimension I transition matrix for change of basis Harold P. Boas Today:
[email protected] I row space and column space June 12, 2006 I the rank-nullity property Example Example continued 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 Let A = 2 4 3 7 1 . R2−2R1 2 4 3 7 1 −−−−−→ 0 0 1 3 −1 4 8 5 11 3 R −4R 4 8 5 11 3 3 1 0 0 1 3 −1 Three-part problem. Find a basis for 1 2 1 2 1 1 2 0 −1 2 R3−R2 R1−R2 5 −−−−→ 0 0 1 3 −1 −−−−→ 0 0 1 3 −1 I the row space (the subspace of R spanned by the rows of the matrix) 0 0 0 0 0 0 0 0 0 0 3 I the column space (the subspace of R spanned by the columns of the matrix) Notice that in each step, the second column is twice the first column; also, the second column is the sum of the third and the nullspace (the set of vectors x such that Ax = 0) I fifth columns. We can analyze all three parts by Gaussian elimination, Row operations change the column space but preserve linear even though row operations change the column space. relations among the columns. The final row-echelon form shows that the column space has dimension equal to 2, and the first and third columns are linearly independent.