Similarly, the eigenvectors with eigenvalue λ = 8 are solutions of Av= 8v, so (A−8I)v= 0 =⇒ −4 6 2 −3 x y = 0 0 =⇒ 2x−3y = 0 =⇒ x = 3y/2 and every eigenvector with eigenvalue λ = 8 must have the form v= 3y/2 y = y 3/2 1 , y 6= 0 . This ends up being a cubic equation, but just looking at it here we see one of the roots is 2 (because of 2−λ), and the part inside the square brackets is Quadratic, with roots of −1 and 8. Eigenvalues and Eigenvectors Po-Ning Chen, Professor Department of Electrical and Computer Engineering National Chiao Tung University Hsin Chu, Taiwan 30010, R.O.C. Therefore, λ 2 is an eigenvalue of A 2, and x is the corresponding eigenvector. 2. Since there are three distinct eigenvalues, they have algebraic and geometric multiplicity one, so the block diagonalization theorem applies to A. detQ(A,λ)has degree less than or equal to mnand degQ(A,λ)

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