By Jorge Rebaza

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28 Let V be a vector space, and U C V a vector subspace of V. 32) c*;. This linear combination does not need to be unique. 36 Let V = R 2 . Then, the set S! = {[1 1] T ,[0 I f , [2 2f} spans V; that is, any vector in R 2 can be expressed as a linear combination of the vectors in Si. 33) is not unique. 33), we could also use —7, 7, —2 and the new combination would still gives us [3 — 4] T . 32) need not be unique. 37 Again, let V = R 2 , and let S2 = {[1 1] T , [2 2] T }. 36, but it also spans V.
Uniqueness) The system has at most one solution for every b if and only ifN(A) {o}. = Proof. Existence: We have already seen that for arbitrary x, the vector Ax G col(A). Then, for Ax — b to have a solution, b must lie in the same subspace. Uniqueness: IfN(A) ^ {0}, then besides x = 0, there is another solution to Ax = b, with 6 = 0, which is a contradiction. On the other hand, assuming N(A) = {0}, if there is a 6 for which Ax = b has more than one solution, that is Ax\ = b and Ax2 = b, with x\ 7^x2, then A(x\ — X2) = Ax\ — Ax2 — b — b = 0, which means that x\ — X2 — 0, or x\ = X2.
63 Let S be the subspace of R 3 spanned by u\ = [2 1  1] T and u2 = [0  1 1] T . Then the matrix 4 0 0 2 1 1 P = 2 3 3 is a projection matrix onto S. 62: P not a projection. The projection matrices that probably have more applications are those that are also orthogonal. 54 An orthogonal projection matrix is a projection matrix P for which PT — P. 64 The matrix P = 1/2 1/2 1/2 1/2 is an orthogonal projection. It clearly satisfies P2 = P and PT = P. 62). 55 It is important to note that an orthogonal projection matrix is not necessarily an orthogonal matrix.