The eigen-basis of A = [ − 19 7 − 42 16 ] and diagonalize it. The diagonal matrix is D = [ 2 0 0 − 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 = [ − 19 7 − 42 16 ] . The characteristic equation of A = [ − 19 7 − 42 16 ] is | A − λ I | = 0 . | [ − 19 7 − 42 16 ] − λ [ 1 0 0 1 ] | = 0 | [ − 19 7 − 42 16 ] − [ λ 0 0 λ ] | = 0 | − 19 − λ 7 − 42 16 − λ | = 0 Simplify further and obtain the values of λ as follows. ( − 19 − λ ) ( 16 − λ ) + 294 = 0 λ 2 + 3 λ − 304 + 294 = 0 λ 2 + 3 λ − 10 = 0 λ = 2 or λ = − 5 Now obtain eigen vector for λ = 2 as follows. A − λ I = [ − 19 7 − 42 16 ] − 2 [ 1 0 0 1 ] = [ − 19 7 − 42 16 ] − [ 2 0 0 2 ] = [ − 21 7 − 42 14 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 21 7 − 42 14 ] = [ − 42 14 − 21 7 ] ( R 1 ↔ R 2 ) = [ − 42 14 0 0 ] ( R 2 → R 1 − 2 R 2 ) = [ 1 − 1 3 0 0 ] ( R 1 → − 1 42 R 1 ) From above matrix, it is observed that the system of equation is { x 1 − 1 3 x 2 = 0 0 x 1 − 0 x 2 = 0 . So, the value of x 1 = 1 3 x 2 . Let x 2 = 3 , then the value of x 1 = 1 . Thus, the eigen vector is [ 1 3 ] . Now obtain eigen vector for λ = − 5 as follows. A − λ I = [ − 19 7 − 42 16 ] + 5 [ 1 0 0 1 ] = [ − 19 7 − 42 16 ] + [ 5 0 0 5 ] = [ − 14 7 − 42 21 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 14 7 − 42 21 ] = [ − 42 21 − 14 7 ] ( R 1 ↔ R 2 ) = [ − 42 21 0 0 ] ( R 2 → R 1 − 3 R 2 ) = [ 1 − 1 2 0 0 ] ( R 1 → − 1 42 R 1 ) From above matrix, it is observed that the system of equation is { x 1 − 1 2 x 2 = 0 0 x 1 − 0 x 2 = 0 . So, the value of x 1 = 1 2 x 2 . Let x 2 = 2 , then the value of x 1 = 1 . Thus, the eigen vector is [ 1 2 ] . From the theorem, the matrix X is X = [ 1 1 3 2 ] . The inverse matrix of X = [ 1 1 3 2 ] 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 2 − 3 [ 2 − 1 − 3 1 ] = − 1 [ 2 − 1 − 3 1 ] = [ − 2 1 3 − 1 ] Now obtain the diagonal matrix as follows. D = X -1 AX = [ − 2 1 3 − 1 ] [ − 19 7 − 42 16 ] [ 1 1 3 2 ] = [ − 2 1 3 − 1 ] [ 2 − 5 6 − 10 ] = [ 2 0 0 − 5 ] Thus, the diagonal matrix is D = [ 2 0 0 − 5 ] _ .

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