Let A be a 3 times 3 diagonalizable matrix whose eigenvalues

Let A be a 3 times 3 diagonalizable matrix whose eigenvalues are lambda_1 = -2, lambda_2 = 1, and lambda_3 = 3. If v_1 = [l 0 0], v_2 - [1 1 0], v_3 = [0 1 1] are eigenvectors of A corresponding to lambda_1, lambda_2, and lambda_3 respectively, then factor A into a product XDX^-1 with D diagonal, and use this factorization to find A^5. A^5 = [_________ _____________ ____________ __________ ____________ _____________ __________ ___________ ______________]

Solution

The RREF of the matrix with v₁,v₂,v₃ as columns is I3. Hence v₁,v₂,v₃ are linearly independent. Since these are distinct vectors also, hence A is diagonalizable. Then there exists an invertible matrix X and a diagonal matrix D such that A = XDX^-1. Further, D has eigenvalues of A on its leading diagonal and P is the matrix with the eigenvectors of A as its columns, in the same order. Thus, D =

-2

0

0

0

1

0

0

0

3

and X =

1

1

0

0

1

1

0

0

1

Then X^-1 =

1

-1

1

0

1

-1

0

0

1

Also, A⁵ = XD⁵X^-1. Further, since (-2)⁵=-32 and 3⁵ = 243, D⁵=

-32

0

0

0

1

0

0

0

243

Hence, A⁵=

-32

33

-33

0

1

242

0

0

243

-2	0	0
0	1	0
0	0	3