The eigen-basis of A = [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] and diagonalize it. The diagonal matrix is D = [ − 2 0 0 0 1 0 0 0 4 ] _ . Theorem used: Diagonalizatiom of a matrix: “If an n × n matrix A has a basis of eigenvectors, then D = X -1 AX is diagonal with the eigen values of A as the entries on the main diagonal. Here, X is the matrix with these eigenvectors as column vectors .” Calculation: Given that, the matrix is A = [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] . The characteristic equation of A = [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] is | A − λ I | = 0 . | [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] − λ [ 1 0 0 0 1 0 0 0 1 ] | = 0 | [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] − [ λ 0 0 0 λ 0 0 0 λ ] | = 0 | − 5 − λ − 6 6 − 9 − 8 − λ 12 − 12 − 12 16 − λ | = 0 Simplify further and obtain the values of λ as follows. [ ( − 5 − λ ) ( ( − 8 − λ ) ( 16 − λ ) + 144 ) − ( − 6 ) ( − 144 + 144 ) + 6 ( 108 − 96 ) ] = 0 [ ( − 5 − λ ) ( λ 2 − 8 λ + 16 ) − 0 + 72 ] = 0 − λ 3 + 3 λ 2 + 6 λ − 8 = 0 Use synthetic division to compute the values of λ as follows. − 1 3 6 − 8 − 2 0 2 − 10 8 − 1 5 − 4 0 So, the equation becomes ( λ + 2 ) ( − λ 2 + 5 λ − 4 ) = 0 . ( λ + 2 ) ( − λ 2 + 5 λ − 4 ) = 0 ( λ + 2 ) ( − λ 2 + 4 λ + λ − 4 ) = 0 ( λ + 2 ) ( λ − 1 ) ( λ − 4 ) = 0 λ = − 2 or λ = 1 or λ = 4 Now obtain eigen vector for λ = − 2 as follows. A − λ I = [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] + 2 [ 1 0 0 0 1 0 0 0 1 ] = [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] + [ 2 0 0 0 2 0 0 0 2 ] = [ − 3 − 6 6 − 9 − 6 12 − 12 − 12 18 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 3 − 6 6 − 9 − 6 12 − 12 − 12 18 ] ∼ [ − 12 − 12 18 − 9 − 6 12 − 3 − 6 6 ] ( R 1 ↔ R 3 ) ∼ [ − 12 − 12 18 − 6 0 6 − 6 0 6 ] ( R 3 → R 2 − R 3 R 2 → R 2 − R 3 ) ∼ [ − 12 − 12 18 0 − 12 6 0 0 0 ] ( R 3 → R 2 − R 3 R 2 → R 1 − 2 R 2 ) Simplify further, A − λ I ∼ [ 1 1 − 3 2 0 1 − 1 2 0 0 0 ] ( R 2 → − 1 12 R 2 R 1 → − 1 12 R 1 ) ∼ [ 1 0 − 1 0 1 − 1 2 0 0 0 ] ( R 1 → R 1 − R 2 ) From above matrix, it is observed that the system of equation is { x 1 − 0 x 2 − x 3 = 0 0 x 1 + x 2 − 1 2 x 3 = 0 0 x 1 + 0 x 2 + 0 x 3 = 0 . So, the value of x 1 = x 3 and the value of x 2 = 1 2 x 3 . Let x 3 = 2 , then the value of x 1 = 2 and x 2 = 1 . Thus, the eigen vector is [ 2 1 2 ] . Now obtain eigen vector for λ = 1 as follows. A − λ I = [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] − 1 [ 1 0 0 0 1 0 0 0 1 ] = [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] − [ 1 0 0 0 1 0 0 0 1 ] = [ − 6 − 6 6 − 9 − 9 12 − 12 − 12 15 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 6 − 6 6 − 9 − 9 12 − 12 − 12 15 ] ∼ [ − 12 − 12 15 − 9 − 9 12 − 6 − 6 6 ] ( R 1 ↔ R 3 ) ∼ [ − 12 − 12 15 0 0 3 4 0 0 − 3 2 ] ( R 2 → R 2 − 3 4 R 1 R 3 → R 3 − 1 2 R 1 ) ∼ [ − 12 − 12 15 0 0 3 4 0 0 1 ] ( R 3 → − 2 3 R 3 ) Simplify further, A − λ I ∼ [ − 12 − 12 0 0 0 0 0 0 1 ] ( R 2 → R 3 − 4 3 R 2 R 1 → R 1 − 15 R 3 ) ∼ [ 1 1 0 0 0 1 0 0 0 ] ( R 2 ↔ R 3 R 1 → − 1 12 R 1 ) From above matrix, it is observed that the system of equation is { x 1 + x 2 + 0 x 3 = 0 0 x 1 + 0 x 2 + x 3 = 0 0 x 1 + 0 x 2 + 0 x 3 = 0 . So, the value of x 1 = − x 2 and the value of x 3 = 0 . Let x 2 = 1 , then the value of x 1 = − 1 . Thus, the eigen vector is [ − 1 1 0 ] . Now obtain eigen vector for λ = 4 as follows. A − λ I = [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] − 4 [ 1 0 0 0 1 0 0 0 1 ] = [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] − [ 4 0 0 0 4 0 0 0 4 ] = [ − 9 − 6 6 − 9 − 12 12 − 12 − 12 12 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 9 − 6 6 − 9 − 12 12 − 12 − 12 12 ] ∼ [ − 12 − 12 12 − 9 − 12 12 − 9 − 6 6 ] ( R 1 ↔ R 3 ) ∼ [ − 12 − 12 12 0 − 6 6 0 − 6 6 ] ( R 3 → R 2 − R 3 R 2 → R 2 − R 3 ) ∼ [ − 12 0 0 0 − 6 6 0 0 0 ] ( R 3 → R 2 − R 3 R 1 → R 1 − 2 R 2 ) Simplify further, A − λ I ∼ [ 1 0 0 0 1 − 1 0 0 0 ] ( R 1 → − 1 12 R 1 R 2 → − 1 6 R 2 ) From above matrix, it is observed that the system of equation is { x 1 + 0 x 2 + 0 x 3 = 0 0 x 1 + x 2 − x 3 = 0 0 x 1 + 0 x 2 + 0 x 3 = 0 . So, the value of x 2 = x 3 and the value of x 1 = 0 . Let x 2 = 1 , then the value of x 3 = 1 . Thus, the eigen vector is [ 0 1 1 ] . From the theorem, the matrix X is X = [ 2 − 1 0 1 1 1 2 0 1 ] . The inverse matrix of X = [ 2 − 1 0 1 1 1 2 0 1 ] is computed as follows. X − 1 = 1 | X | a d j [ 2 − 1 0 1 1 1 2 0 1 ] = 1 1 a d j [ 2 − 1 0 1 1 1 2 0 1 ] = [ 1 1 − 1 1 2 − 2 − 2 − 2 3 ] Now obtain the diagonal matrix as follows. D = X -1 AX = [ 1 1 − 1 1 2 − 2 − 2 − 2 3 ] [ − 5 − 6 6 − 9 − 8 12 − 12 − 12 16 ] [ 2 − 1 0 1 1 1 2 0 1 ] = [ 1 0 0 − 2 1 0 1 − 2 1 ] [ − 4 − 1 0 − 2 1 4 − 4 0 4 ] = [ − 2 0 0 0 1 0 0 0 4 ] Thus, the diagonal matrix is D = [ − 2 0 0 0 1 0 0 0 4 ] _ .

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