# Finding eigen values and eigen space

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Thank-you so much for your expertise.

Let A = . Find each of the following (by hand, showing your work).

(a) The characteristic polynomial of A

(b) The eigenvalues of A.

(c) The algebraic multiplicity for each eigenvalue.

(d) A basis for each eigenspace.

(e) Is "A" diagonalizable? Why or why not?

https://brainmass.com/math/linear-algebra/finding-eigen-values-eigen-space-204062

## SOLUTION This solution is **FREE** courtesy of BrainMass!

HI,

Please find the explanations/solution attached herewith.

Let A = . Find each of the following (by hand, showing your work).

(a) The characteristic polynomial of A

(b) The eigenvalues of A.

(c) The algebraic multiplicity for each eigenvalue.

(d) A basis for each eigenspace.

(e) Is "A" diagonalizable? Why or why not?

Solution:

(a)

The characteristic equation is

(b)

Eigen values are 3 and 2.

(c)

The multiplicity of eigen value 2 is 2.

The multiplicity of eigen value 3 is 2.

(d)

The Eigen vector corresponding to =3, will be

Substitute =3 into the gives

Which gives y = z = w = 0

Therefore, the eigenvectors of A associated with the eigenvalue Î» = 3 are all vectors of the form ( x, 0, 0, 0)T = x(1, 0, 0, 0)T for xâ‰ 0. Removing the restriction that the scalar multiple be nonzero includes the zero vector and gives the full eigenspace:

Basis for this eigenspace is .

The Eigen vector corresponding to =2, will be

Substitute =3 into the gives

Which gives z = w = 0 , x = -2y

Therefore, the eigenvectors of A associated with the eigenvalue Î» = 2 are all vectors of the form ( -2y, y, 0, 0)T = y(-2, 1, 0, 0)T for yâ‰ 0. Removing the restriction that the scalar multiple be nonzero includes the zero vector and gives the full eigenspace:

Basis for this Eigen space is .

(e)

A is not diagonalizable as the its all eigen values are not different.

https://brainmass.com/math/linear-algebra/finding-eigen-values-eigen-space-204062