By Jorge Rebaza

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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). That is, any vector in col(^4) (in particular, the third and fourth columns of ^4) can be expressed as a unique combination of columns 1, 2, and 5.
The idea is to get an orthogonal matrix H, with its first column being the vector x, normalized, if necessary. , when proving several theorems in Chapter 3. Let x £ R n be a vector with llxll2 = xTx — 1. Define the vector u = x — e\. T T First observe that u x = x x — efx — 1 — ejx and that uTu = xTx — xTe\ — ejx + ejei = 2(1 — ejx), so that uTu = 2uTx. Then, Hx y uTu J X — X 2u(uTx) _ _ (2uTx)u — x — u = e\. The columns of this Householder matrix H are of course orthonormal, but according to the equality Hx = e\, we have that x — HTe\ = He\.
42) A direct sum of subspaces can be obtained by using the perp of a vector subspace. In fact, we have the following result. 36 Let V be a vector space, and let U be an arbitrary vector subspace ofV. Then, v = u®u±. Proof. , um} be an orthonormal basis of U. 40), let U = (vTUi)ui \ 1 {^Furr^Um Then, the vector w = v — u is orthogonal to each Ui, i — 1 , . . , m, and therefore, it is in U1. Hence, v = u + w, where u G U and w G U^. Also, it is clear that the only intersection ofU and U1 is the zero vector, for ifueU and u G U^, then \\u\\2 =uTu = 0.