Given lambda_1 = 2, lambda_2 = -2, lambda_3 = 3 are the eigenvalues for matrix A

Question

Given lambda_1 = 2, lambda_2 = -2, lambda_3 = 3 are the eigenvalues for matrix A where A = [1 1 -3 -1 3 1 -1 1 -1] In addition, given the eigenvectors corresponding to the above eigenvalues respectively x_1 = (-1, 0, 1), x_1 = (1, -1, 4), x_3 = (-1, 1, 1), determine the matrix P such that P^-1AP = D where D = [2 0 0 0 -2 0 0 0 3] Matrix P: _____ Determine matrix A is diagonalizable given matrix A = [1 0 0 -2 0 0 1 1 -3]. Justify your answer. Matrix A is diagonalizable since _____ Matrix A is not diagonalizable since _____

Explanation / Answer

-1

1

-1

0

-1

1

1

4

1

2. The characteristic equation of A is det(A- I3 )= 0 or, (+3)( -1) = 0. Hence the eigenvalues of A are 1=-3, 2=1 and 3= 0. Further, the eigenvector of A corresponding to the eigenvalue is solution to the equation (A- I3)X = 0. Therefore, the eigenvector of A corresponding to the eigenvalue -3 is solution to the equation (A+3I3)X = 0. To solve this equation, we will reduce the matrix A+3I3 to its RREF as under:

Multiply the 1st row by ¼

Multiply the 2nd row by 1/3

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

Then the RREF of A+3I3 is

1

0

5/12

0

1

1/3

0

0

0

If X = (x,y,z)T, then the equation (A+3I3)X = 0 is equivalent to x+5z/12= 0 or, x = -5z/12 and y+z/3 = 0 or, y = -z/3. Then X = (-5z/12,-z/3,z)T = z/12(-5,-4,12)T. Thus, the eigenvector of A corresponding to the eigenvalue -3 is (-5,-4,12)T.

Similarly, the eigenvectors of A corresponding to the eigenvalues 1 and 0 are (1,0,0)T and (2,1,0)T respectively. Since the 3 eigenvectors of A are distinct and also apprently linearly independent, hence A is diagonalizable.

-1

1

-1

0

-1

1

1

4

1

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