Linear Algebra I [Lecture notes] by Peter M. Neumann

By Peter M. Neumann

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Xm ) , where xi := v, ui for i = 1, . . , m. Show that T is a linear mapping. (c) For T as in part (b), show that Ker T = U ⊥ and that rank T = m. (d) Deduce that dim U ⊥ = n − m and that V = U ⊕ U ⊥ . 11. 9, and let U be the subspace spanned by the polynomials 1 and x . 10 find polynomials p(x) ∈ U and q(x) ∈ U ⊥ such that x2 = p(x) + q(x) . Vectors v1 , v2 , . . , vn in an inner product space are said to form an 1 if i = j, orthonormal set if vi , vj = δi,j = 0 if i = j. 6. Note. If vectors v1 , v2 , .

Vr of V . Thus dim V = n + r . For 1 i r define wi := T (vi ). Our aim is to prove that w1 , . . , wr form a basis for Im T . 9(3) we know that the vectors T (u1 ), . . , T (un ), T (v1 ), . . , T (vr ) span Im T . Since u1 , . . , un ∈ Ker T , the first n of those vectors are 0W however, and do not contribute to the span. Therefore Im T is spanned by the remainder, that is, by w1 , . . , wr . We also need to show that w1 , . . , wr are linearly independent. So consider an arbitrary linear dependence relation α1 w1 + · · · + αr wr = 0W , where α1 , .

Us , v1 , . . vr for F n . Defining wi := LA (vi ), we then know that m. w1 , . . , wr form a basis of Im LA . Extend this to a basis w1 , . . , wr , wr+1 , . . , wm of Fcol Now consider the matrix of LA with respect to the ordered bases n and w , . . , w , w m v1 , . . vr , u1 , . . , us of Fcol 1 r r+1 , . . , wm of Fcol . Since LA (vi ) = wi for 1 i r and LA (uj ) = 0 for 1 j s, the matrix of coefficients Ir 0r×t and therefore the matrix of LA with that is seen on the page has the block form 0s×r 0s×t Ir 0r×s .

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