Linear Algebra Solutions 69

Linear Algebra Solutions 69 - (ii) Let B A I2 . Then B 2 =...

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(ii) Let B - A - αI 2 . Then B 2 = ( A - αI 2 ) 2 = A 2 - 2 αA + α 2 I 2 = A 2 - ( a + d ) A + ( ad - bc ) I 2 , But by problem 3, chapter 2.4, A 2 - ( a + d ) A + ( ad - bc ) I 2 = 0, so B 2 = 0. (iii) Now suppose that B 6 = 0. Then BE 1 6 = 0 or BE 2 6 = 0, as BE i is the i –th column of B . Hence BX 2 6 = 0, where X 2 = E 1 or X 2 = E 2 . (iv) Let X 1 = BX 2 and P = [ X 1 | X 2 ]. We prove P is non–singular by demonstrating that X 1 and X 2 are linearly independent. Assume xX 1 + yX 2 = 0. Then xBX 2 + yX 2 = 0 B ( xBX 2 + yX 2 ) = B 0 = 0 xB 2 X 2 + yBX 2 = 0 x 0 X 2 + yBX 2 = 0 yBX 2 = 0 . Hence y = 0 as BX 2 6 = 0. Hence xBX 2 = 0 and so x = 0. Finally, BX 1 = B ( BX 2 ) = B 2 X 2 = 0, so (
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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