The eigen-basis of A = [ 1 0 2 − 1 ] and diagonalize it. The diagonal matrix is D = [ 1 0 1 − 1 ] _ . 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 = [ 1 0 2 − 1 ] . The characteristic equation of A = [ 1 0 2 − 1 ] is | A − λ I | = 0 . | [ 1 0 2 − 1 ] − λ [ 1 0 0 1 ] | = 0 | [ 1 0 2 − 1 ] − [ λ 0 0 λ ] | = 0 | 1 − λ 0 2 − 1 − λ | = 0 Simplify further and obtain the values of λ as follows. ( 1 − λ ) ( − 1 − λ ) = 0 λ 2 − 1 = 0 λ = 1 or λ = − 1 Now obtain eigen vector for λ = 1 as follows. A − λ I = [ 1 0 2 − 1 ] − 1 [ 1 0 0 1 ] = [ 0 0 2 − 2 ] From above matrix, it is observed that the system of equation is { 0 x 1 + 0 x 2 = 0 2 x 1 − 2 x 2 = 0 . So, the value of x 2 = x 1 . Let x 1 = 1 , then the value of x 2 = 1 . Thus, the eigen vector is [ 1 1 ] . Now obtain eigen vector for λ = − 1 as follows. A − λ I = [ 1 0 2 − 1 ] + 1 [ 1 0 0 1 ] = [ 1 0 2 0 ] From above matrix, it is observed that the system of equation is { x 1 + 0 x 2 = 0 2 x 1 + 0 x 2 = 0 . So, the value of x 1 = 0 . Consider the value of x 2 = 1 . Thus, the eigen vector is [ 0 1 ] . From the theorem, the matrix X is X = [ 1 0 1 1 ] . The inverse matrix of X = [ 1 0 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 1 − 0 [ 1 0 − 1 1 ] = [ 1 0 − 1 1 ] Now obtain the diagonal matrix as follows. D = X -1 AX = [ 1 0 − 1 1 ] [ 1 0 2 − 1 ] [ 1 0 1 1 ] = [ 1 0 − 1 1 ] [ 1 0 1 − 1 ] = [ 1 0 1 − 1 ] Thus, the diagonal matrix is D = [ 1 0 1 − 1 ] _ .

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