The eigen-basis of A = [ − 4.3 7.7 1.3 9.3 ] and diagonalize it. The diagonal matrix is D = [ 10 0 0.00005 − 5 ] _ . 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 = [ − 4.3 7.7 1.3 9.3 ] . The characteristic equation of A = [ − 4.3 7.7 1.3 9.3 ] is | A − λ I | = 0 . | [ − 4.3 7.7 1.3 9.3 ] − λ [ 1 0 0 1 ] | = 0 | [ − 4.3 7.7 1.3 9.3 ] − [ λ 0 0 λ ] | = 0 | − 4.3 − λ 7.7 1.3 9.3 − λ | = 0 Simplify further and obtain the values of λ as follows. ( − 4.3 − λ ) ( 9.3 − λ ) − 10.01 = 0 λ 2 − 5 λ − 39.99 − 10.01 = 0 λ 2 − 5 λ − 50 = 0 λ = 10 or λ = − 5 Now obtain eigen vector for λ = 10 as follows. A − λ I = [ − 4.3 7.7 1.3 9.3 ] − 10 [ 1 0 0 1 ] = [ − 4.3 7.7 1.3 9.3 ] − [ 10 0 0 10 ] = [ − 14.3 7.7 1.3 − 0.7 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 14.3 7.7 1.3 − 0.7 ] = [ − 143 77 13 − 7 ] ( R 1 → 10 R 1 R 2 → 10 R 2 ) = [ − 143 77 0 0 ] ( R 2 → R 1 + 11 R 2 ) = [ 1 − 0.5385 0 0 ] ( R 1 → − 1 143 R 1 ) From above matrix, it is observed that the system of equation is { x 1 − 0.5385 x 2 = 0 0 x 1 − 0 x 2 = 0 . So, the value of x 1 = 0.5385 x 2 . Let x 2 = 1 , then the value of x 1 = 0.5385 . Thus, the eigen vector is [ 0.5385 1 ] . Now obtain eigen vector for λ = − 5 as follows. A − λ I = [ − 4.3 7.7 1.3 9.3 ] + 10 [ 1 0 0 1 ] = [ − 4.3 7.7 1.3 9.3 ] + [ 5 0 0 5 ] = [ 0.7 7.7 1.3 14.3 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ 0.7 7.7 1.3 14.3 ] = [ 1.3 14.3 0.7 7.7 ] ( R 2 ↔ R 1 ) = [ 13 143 7 77 ] ( R 1 → 10 R 1 R 2 → 10 R 2 ) = [ 13 143 0 0 ] ( R 2 → R 1 − 143 77 R 2 ) = [ 1 11 0 0 ] ( R 1 → 1 13 R 1 ) From above matrix, it is observed that the system of equation is { x 1 + 11 x 2 = 0 0 x 1 − 0 x 2 = 0 . So, the value of x 1 = − 11 x 2 . Let x 2 = 1 , then the value of x 1 = − 11 . Thus, the eigen vector is [ − 11 1 ] . From the theorem, the matrix X is X = [ 0.5385 − 11 1 1 ] . The inverse matrix of X = [ 0.5385 − 11 1 1 ] is computed as follows. It is known that, the inverse matrix for 2 × 2 is [ a b c d ] − 1 = 1 a d − b c [ d − b − c a ] . X − 1 = 1 0.5835 + 11 [ 1 11 − 1 0.5385 ] = 1 11.5385 [ 1 11 − 1 0.5385 ] = [ 1 11.5385 11 11.5385 − 1 11.5385 0.5385 11.5385 ] Now obtain the diagonal matrix as follows. D = X -1 AX = [ 1 11.5385 11 11.5385 − 1 11.5385 0.5385 11.5385 ] [ − 4.3 7.7 1.3 9.3 ] [ 0.5385 − 11 1 1 ] = [ − 2 1 3 − 1 ] [ 5.38445 55 10.00005 − 5 ] = [ 10 0 0.00005 − 5 ] Thus, the diagonal matrix is D = [ 10 0 0.00005 − 5 ] _ .

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