FinalTestReviewSpring2011

Because dim na 3 1 2 2 3 those corresponding

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Unformatted text preview: nd eigenvectors of A det 5 !6 ! ! 2 ! ! ! 2 ! 0 , ! ! !1, 2 3 !4 ! ! !1 : 6 !6 A ! !I ! 1 ,v ! " 3 !3 1 !!2: 3 !6 A ! !I ! 2 ,v! " 3 !6 1 b) Write A in diagonalized form, A ! SDS !1 where D is a diagonal matrix. Then find a formula for A k . As k # ", every column of 1k A k is a multiple of what vector - what is the 2 significance of this vector? A ! V $V Ak ! 1 Ak ! 2k !1 12 0 !1 12 11 ! !1 0 11 2 12 !!1 " k 0 !1 2 11 0 2k 1 !1 2 % 2 !k !!1 " k%1 !2 % 2 1!k !!1 " k 12 ! k # !1 1 00 !1 2 11 01 1 !1 2 0 1 !1 2 2 k%1 % !!1 " k%1 !2 k%1 % 2!!1 " k 2 k % 2!!1 " k%1 2 !2 ! 2 k % 4! ! 1 " k !1 12 !1 2 1 !1 0 11 ! k%1 1!k 2!k 1 !1 1 % 2 !!1 " !1 % 2 !!1 " the eigenvector with the largest eigenvalue, ! ! 2. , each column is a multiple of You can also calculate: 1 Ak ! 2k 12 !!1 " k 2 !k 0 11 # 0 ! 2 !2 1 !1 1 3 !1 6 4 !2 !3 !2 2. (12 pts) If A ! 2 1 The rank of A is __1___ so one eigenvalue is ! !_0___ . (the null space of A ! 0I is not just " 0) There must be _1 / 2 / 3_____ eigenvectors cor...
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