Let A 3 1 0 1 3 0 0 0 5 a Find the eigenvalues of A b For e

Let A = (3 1 0 1 3 0 0 0 5). (a) Find the eigenvalues of A. (b) For each eigenvalue of A, find a basis for the corresponding eigenspace. (c) Find an orthonormal basis for R^3 consisting of eigenvectors of A. Hence find a matrix P such that P^T AP is a diagonal matrix. (d) Consider the linear transformation T: R^3 rightarrow R defined by T(x) = Ax for every x elementof R^3. Is there a basis B of R^3 such that the matrix representation of T with respect to B is a diagonal matrix.

Solution

(a) The eigenvalues of A are solutions to the characteristic equation det(A- I₃)= 0 or, ³-11²+38-40 = 0 or, (-5)( -4)( -2)=0. Thus, the eigenvalues of A are ₁=5, ₂=4 and ₃=2.

(b) The eigenvector of A corresponding to the eigenvalue 5 is solution to the equation (A-5I₃)X= 0. To solve this equation, we will reduce A-5I₃ to its RREF as under:

Multiply the 1st row by -1/2

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

Multiply the 2nd row by -2/3

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

Then the RREF of A is

1

0

0

0

1

0

0

0

0

Now, if X = (x,y,z)^T, then the equation (A-5I₃)X= 0 is equivalent to x=0 and y=0 so that X=(0 ,0,z)^T = z(0,0,1)^T. Thus, the eigenvector of A corresponding to the eigenvalue 5 is (0,0,1)^T. Similarly, the eigenvector of A corresponding to the eigenvalue 4 is solution to the equation (A-4I₃)X= 0. The RREF of A-4I₃ is

1

-1

0

0

0

1

0

0

0

Now, if X = (x,y,z)^T, then the equation (A-4I₃)X= 0 is equivalent to x-y =0 or, x = y and z = 0 so that X= (y,y,0)^T = y(1,1,0)^T. Thus, the eigenvector of A corresponding to the eigenvalue 4 is (1,1,0)^T. Also, the eigenvector of A corresponding to the eigenvalue 2 is solution to the equation (A-2I₃)X= 0. The RREF of A-2I₃ is

1

1

0

0

0

1

0

0

0

Now, if X = (x,y,z)^T, then the equation (A-2I₃)X= 0 is equivalent to x+y =0 or, x = -y and z = 0 so that X= (-y,y,0)^T = y(-1,1,0)^T. Thus, the eigenvector of A corresponding to the eigenvalue 2 is (-1,1,0)^T.

A basis of the eigenspace of A corresponding to the eigenvalue 5 is {(0,0,1)^T}. Also, bases of the eigenspaces of A corresponding to the eigenvalues 4 and 2 are {(1,1,0)^T} and {(-1,1,0)^T} respectively.

(c) The eigenvectors of A are v₁ = (0,0,1)^T, v₂ = (1,1,0)^T and v₃ = (-1,1,0)^T. Since v₁.v₂ = v₂.v₃ = v₃.v₁ = 0, the eigenvectors of A are already orthogonal. We only need to convert v₂,and v₃ into unit vectors as v₁ is a unit vector Let u₂ = v₂/||v₂||= [1/(1+1+0)]v₂ = (1/2) (1,1,0)^T= (1/2,1/2,0)^T and u₃ = v₃/||v₃||= [1/(1+1+0)]v₂ = (1/2) (-1,1,0)^T= (-1/2,1/2,0)^T. Then {u₁,u₂,u₃} = {(0,0,1)^T, (1/2,1/2,0)^T, (-1/2,1/2,0)^T} is an orthonormal basis for R³, comprising the eigenvectors of A.

Further, we know that an orthogonal matrix P (say) has orthonormal columns and for such a matrix P, we have P^T = P^-1. Now, since A has 3 distinct and linearly indfependent eigenvectors, hence , as per the diagonalization theorem, we have D = P^-1AP = P^T AP where, the columns of P are u₁,u₂,u₃, and D has the eigenvalues of A on its leading diagonal, in the same order. Then D =

5

0

0

0

4

0

0

0

2

and P =

0

1/2

-1/2

0

1/2

1/2

1

0

0

(d) We know that if M is the B- matrix , for T, then A is similar to M. Now, A can be similar to a diagonalmatrix if and only if A has 3 linearly independent eigenvectors. Also, the eigenvectors

corresponding to different eigenvalues are linearly independent. Here, the characteristic equation of A has degree 3. Therefore, A will be diagonalizable if and only if either A has 3 distinct eigenvalues or if the eigenspace corresponding to the eigenvalue with multiplicity 2 has dimension 2.

1	0	0
0	1	0
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