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# 5.2 - 5.2 Characteristic Equation Recall If RE form...

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5.2 Characteristic Equation Recall: If RE form for (square) A is A U with r row interchanges, Then det A = ( - 1) r det U = ( - 1) r × product of pivots A is invertible there are no free variables det U = 0 det A = 0 Ax = 0 has only trivial solution x = 0 A has an ev 0 there is x = 0 such that Ax = 0 · x = 0 Invertible Matrix Theorem Equivalences Continued: r) A invertible s) det A = 0 t) 0 is not an eigenvalue of A Another viewpoint: λ is an ev. of A Ax = λx for some x = 0 ( A - λI ) x = 0 for x = 0 p ( λ ) = det( A - λI ) = 0 Characteristic equation . 1

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For triangular A = a 11 · · · · · · a 1 n 0 . . . . . . . . . . . . . . . 0 · · · 0 a nn det( A - λI ) = a 11 - λ · · · · · · a 1 n 0 . . . . . . . . . . . . . . . 0 · · · 0 a nn - λ = ( a 11 - λ )( a 22 - λ ) · · · ( a nn - λ ) = 0 λ = a ii some i evs of A are just diagonal entries! EX: A = 2 0 0 0 1 0 0 0 - 4 evs. λ 1 = 2 , λ 2 = 1 , λ 3 = - 4 S c = k 0 0 1 evs. λ 1 = k, λ 2 = 1 S h = 1 1 0 1 evs. λ 1 = 1 , λ 2 = 1 Is there more than 1 linear independent e-vector? (Hint: What is S h geometrically?)
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5.2 - 5.2 Characteristic Equation Recall If RE form...

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