Determine whether A is diagonalizable A 1 1 0 1 0 1 0 1 1 Y

Determine whether A is diagonalizable. A = [1 1 0 1 0 1 0 1 1] Yes No Find an invertible matrix P and a diagonal matrix D such that P^-1AP = D. (Enter each matrix in the form [[row 1], [row 2], ellipsis], where each row is a comma-separated list. If A is not diagonalizable, enter NO SOLUTION.) (D, P) =

Solution

The characteristic equation of A is det (A- I₃) = 0 or, ³ -2² – +2 = 0 or, (-2)( +1)( -1) = 0. Hence the eigenvalues of A are ₁ = 2, ₂ = -1 and ₃ = 1. We know that the eigenvector(s) of A correspoding to the eigenvalue are solutions to the equation (A- I₃)X = 0. Thus, the eigenvectors of A corresponding to the eigenvalues _{1, 2}, ₃ are solutions to the equation (A- 2I₃)X = 0, (A+I₃)X = 0 and (A- I₃)X = 0 respectively.

We will reduce to its RREF, as under, the matrix A- 2I₃ =

-1

1

0

1

-2

1

0

1

-1

Multiply the 1st row by -1

Add -1 times the 1st row to the 2nd row

Multiply the 2nd row by -1

Add -1 times the 2nd row to the 3rd row

Add 1 times the 2nd row to the 1st row

Then the RREF of A- 2I₃ is

1

0

-1

0

1

-1

0

0

0

Now, if X = (x,y,z)^T, then the equation (A- 2I₃)X = 0 is equivalent to x-z = 0, y-z = 0 so that x = z and y = z. Hence X=(z,z,z)^T=z(1,1,1)^T.Thus,the eigenvector of A corresponding to the eigenvalues ₁=2 is v₁= (1,1,1)^T. Similarly, the eigenvectors of A corresponding to the eigenvalues ₂=-1and ₃ =1 are v₂ = (1,-2,1)^Tand     v₃ = (-1,0,1)^T respectively. Further, the RREF of the matrix with these eigenvectors as its columns is I₃. This impolies that v₁,v₂,v₃ are linearly independent. Also these 3 eigenvectors are distinct. Therefore, A is diagonalizable. Also, P =

1

1

-1

1

-2

0

1

1

1

and D =

2

0

0

0

-1

-0

0

0

1

Note: D is the diagonal matrix with the eigenvalues of A on its leading diagonal and P is the matrix with the eigenvectors of A as its columns ( in the same order

-1	1	0
1	-2	1
0	1	-1