By Jorge Rebaza

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M, h c^^fcj, which is the same as r a^ a2j _ ^77lj _ = ^lj Cn Ci 2 Clfc C21 C22 C2fc + ^2j +•*+ . Cm2 _ ^777,1 ^j I _ Cmk J This tells us that the columns of A are linear combinations of the k vectors c. A). By applying the ideas above to AT, we can also show that dim row(A) < dim col(A), and therefore conclude that indeed dim col(A) = dim TOW(A). n We next state an important result on the rank of product of matrices. 44 Let A and B be two arbitrary matrices whose product AB is well defined.
And since these first two columns are linearly independent, this implies that co\(A) is 2dimensional. Geometrically then, col(^4) is the plane — x + 2y — z = 0 passing through the origin, with normal vector [—1 2 — 1] T . 48, the null space N(A) is onedimensional. In fact, by solving the system Ax = 0, we observe that all solutions are of the form [xi X2 X3]7\ with X2 = — 2#i, xs = x\. That is, N(A) is a line spanned by the vector [—1 2 — 1] T . 47). 59 "2 0 1 3 0" 0 3 1 2 0 In this case the columns 1, 2, and 5 are linearly indeLet A = 2 3 2 5 1 0 0 0 0 0 pendent and they form a basis of col( A).
More formally, we can characterize them through the following definition. 27 Let V be a vector space and U a nonempty subset ofV. Then, we say U is a vector subspace of V if (i) u + v G U, for all u,v G U. (ii) cu G U, for all scalars c and all u G U. 3 The line y = x is a subspace U of V = R2. This definition simply says that operations performed with elements of U will generate vectors that will stay within U. 29 Let V = E 2 . 26), and let U = {(#, y) G i? 2 : y = #}• In other words, U is the line y = x.