By Jorge Rebaza
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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\. That is, x is the first column of H. 24 Let x = Take u = x — e\ = 6 -2 0 -4 2 VECTOR SPACES 21 Then, the matrix H = I 2uu2 4 0 - 4 2 ] 0 6 0 0 -4 0 - 2 4 2 0 4 4J is orthogonal, with x as its first column.
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. This means that geometrically, col(A) is a 3-dimensional hyperplane (in M4) spanned by those three columns, and therefore N(A) is a two-dimensional subspaceof M5.
12 For an arbitrary matrix A, the matrices A + AT and A AT are symmetric. is complex, we say A is Hermitian if A = A*. 15 IfAnXn following matrix is Hermitian: 1 4-2z 4 + 2% - 9 In the set of real numbers, we can define a number a to be positive if a x2 > 0, for all numbers x ^ 0. The generalization of this definition for matrices leads to the so-called positive definite matrices. 23) for all vectors x ^ 0. If instead we have xT A x > 0, then the matrix is said to be positive semidefinite. In several applications, especially optimization, we will encounter matrices that are both, symmetric and positive definite.