To verify: The matrix equation x = P y for the matrices A = [ 3 4 4 − 3 ] and P = [ − 4 2 3 − 1 ] if x = P y is eigenvectors of A and y is an eigenvector of A ^ = P − 1 A P . Given that, the matrices are A = [ 3 4 4 − 3 ] and P = [ − 4 2 3 − 1 ] . In order to verify the given statement, first find the matrix A ^ as follows. The inverse matrix of P = [ − 4 2 3 − 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 ] . P − 1 = 1 ( 4 ) − 6 [ − 1 − 2 − 3 − 4 ] = − 1 2 [ − 1 − 2 − 3 − 4 ] = [ 1 2 1 3 2 2 ] Now compute the value of P − 1 A P as follows. P − 1 A P = [ 1 2 1 3 2 2 ] [ 3 4 4 − 3 ] [ − 4 2 3 − 1 ] = [ 1 2 1 3 2 2 ] [ − 12 + 12 6 − 4 − 16 − 9 8 + 3 ] = [ 1 2 1 3 2 2 ] [ 0 2 − 25 11 ] = [ − 25 12 − 50 25 ] Thus, the value of the matrix A ^ = [ − 25 12 − 50 25 ] . In order to obtain y , compute the eigen vectors for the matrix A ^ = [ − 25 12 − 50 25 ] as follows. The characteristic equation of A ^ = [ − 25 12 − 50 25 ] is | A ^ − λ I | = 0 . | [ − 25 12 − 50 25 ] − λ [ 1 0 0 1 ] | = 0 | [ − 25 12 − 50 25 ] − [ λ 0 0 λ ] | = 0 | − 25 − λ 12 − 50 25 − λ | = 0 Simplify further and obtain the values of λ as follows. ( − 25 − λ ) ( 25 − λ ) + 600 = 0 λ 2 − 25 = 0 λ = 5 or λ = − 5 Now obtain eigen vector for λ = 5 as follows. A ^ − λ I = [ − 25 12 − 50 25 ] − 5 [ 1 0 0 1 ] = [ − 25 12 − 50 25 ] − [ 5 0 0 5 ] = [ − 30 12 − 50 20 ] Use row reduction to obtain the eigen vectors as follows. A ^ − λ I = [ − 30 12 − 50 20 ] ∼ [ − 30 12 0 0 ] ( R 2 → R 2 − 3 5 R 1 ) ∼ [ 1 − 2 5 0 0 ] ( R 1 → − 1 30 R 1 ) From above matrix, it is observed that the system of equation is { x 1 − 2 5 x 2 = 0 0 x 1 + 0 x 2 = 0 . So, the value of x 1 = 2 5 x 2 . Let x 2 = 5 , then the value of x 1 = 2 . Thus, the eigen vector is [ 2 5 ] . Now obtain eigen vector for λ = − 5 as follows. A ^ − λ I = [ − 25 12 − 50 25 ] + 5 [ 1 0 0 1 ] = [ − 25 12 − 50 25 ] + [ 5 0 0 5 ] = [ − 20 12 − 50 30 ] Use row reduction to obtain the eigen vectors as follows. A ^ − λ I = [ − 20 12 − 50 30 ] ∼ [ − 20 12 0 0 ] ( R 2 → R 2 − 2 5 R 1 ) ∼ [ 1 − 3 5 0 0 ] ( R 1 → − 1 20 R 1 ) From above matrix, it is observed that the system of equation is { x 1 − 3 5 x 2 = 0 0 x 1 + 0 x 2 = 0 . So, the value of x 1 = 3 5 x 2 . Let x 2 = 5 , then the value of x 1 = 3 . Thus, the eigen vector is [ 3 5 ] . For the eigen value λ = 5 the vector Py is computed as follows. [ − 4 2 3 − 1 ] [ 2 5 ] = [ − 8 + 10 6 − 5 ] P y = [ 2 1 ] ⋯ ⋯ ( 1 ) For the eigen value λ = − 5 the vector Py is computed as follows. [ − 4 2 3 − 1 ] [ 3 5 ] = [ − 12 + 10 9 − 5 ] P y = [ − 2 4 ] ⋯ ⋯ ( 2 ) In order to compute the matrix x , compute the eigen vectors for the matrix A = [ 3 4 4 − 3 ] as follows. The characteristic equation of A = [ 3 4 4 − 3 ] is | A − λ I | = 0 . | [ 3 4 4 − 3 ] − λ [ 1 0 0 1 ] | = 0 | [ 3 4 4 − 3 ] − [ λ 0 0 λ ] | = 0 | 3 − λ 4 4 − 3 − λ | = 0 Simplify further and obtain the values of λ as follows. ( 3 − λ ) ( − 3 − λ ) − 16 = 0 λ 2 − 25 = 0 ( λ − 5 ) ( λ + 5 ) = 0 λ = 5 or λ = − 5 Now obtain eigen vector for λ = 5 as follows. A − λ I = [ 3 4 4 − 3 ] − 5 [ 1 0 0 1 ] = [ 3 4 4 − 3 ] − [ 5 0 0 5 ] = [ − 2 4 4 − 8 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 2 4 4 − 8 ] ∼ [ 4 − 8 − 2 4 ] ( R 2 ↔ R 1 ) ∼ [ 4 − 8 0 0 ] ( R 2 → R 1 + 2 R 2 ) ∼ [ 1 − 2 0 0 ] ( R 1 → 1 4 R 1 ) From above matrix, it is observed that the system of equation is { x 1 − 2 x 2 = 0 0 x 1 + 0 x 2 = 0 . So, the value of x 1 = 2 x 2 . Let x 2 = 1 , then the value of x 1 = 2 . Thus, the eigen vector is x = [ 2 1 ] ⋯ ⋯ ( 3 ) . Now obtain eigen vector for λ = − 5 as follows. A − λ I = [ 3 4 4 − 3 ] + 5 [ 1 0 0 1 ] = [ 3 4 4 − 3 ] + [ 5 0 0 5 ] = [ 8 4 4 2 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ 8 4 4 2 ] ∼ [ 8 4 0 0 ] ( R 2 → R 1 − 2 R 2 ) ∼ [ 1 1 2 0 0 ] ( R 1 → 1 8 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 = 4 , then the value of x 1 = − 2 . Thus, the eigen vector is x = [ − 2 4 ] ⋯ ⋯ ( 4 ) . From the equations (1) and (3), it is verified that x = P y and from the equations (2) and (4), it is verified that x = P y .

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