By Jorge Rebaza

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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. We will see some more applications of Householder matrices in Chapter 3. 5 VECTOR SPACES Although vector spaces can be very general, we restrict our attention to vector spaces in finite dimensions, and unless otherwise stated, we will always consider the scalars to be real numbers. We know that if we add, subtract, multiply, or divide (except division by zero) two real numbers, we get a real number again.
Rank(23). Abn]. ) < rank(^4). The second inequality is proved similarly using rows. (b) If B is nonsingular, we have rank(A) = mnkiABB'1) < rank(AjB). 38 BASICS OF LINEAR ALGEBRA By combining this with the first inequality in (a), we conclude that ra,nk(AB) = rank(A). The proof of the last statement is similar. □ We now define the null space of A, denoted as N(A). 45 The null space of a matrix Am x n is defined as the vector subspace of R n : N(A) = {x e Rn : Ax = 0 } . 57 1 2 From the definition, we see that the null space is the set 2 4 3 6 {[xi X2 xs]T : x\ + 2x2 + 3^3 = 0}.
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 3dimensional hyperplane (in M4) spanned by those three columns, and therefore N(A) is a twodimensional subspaceof M5. As stated before, one of the applications of column spaces, row spaces, and null spaces is in the solution of linear systems. We can now put together the links between the solution of Ax = b and the vector spaces col(^4) and N(A) in a theorem.