8 71 Define T P2 rightarrow P2 by Ta0a1xa2x2 3a15a2 4a04a1

8) 7.1

Define T: P_2 rightarrow P_2 by T(a_0+a_1x+a_2x^2) = (-3a_1+5a_2) + (-4a_0+4a_1-10a_2) x+ 3a_2x^2. Find the eigenvalues. (Enter your answers from smallest to largest.) (lambda_1, lambda_2, lambda_3) = (________) Find the corresponding coordinate eigenvectors of T relative to the standard basis {1, x, x^2}. x_1 = ______ x_ 2=_____ x_3 = _________

Solution

We have T (a₀+a₁x+a₂x²) = (-3a₁ +5a₂)+ (-4a₀ +4a₁-10a₂)x +3a₂x². Then T(1) = 0+(-4+0-0)x +0x² = -4x, T(x) = (-3+0)+(0+4-0)x +0x²= -3+4x and T(x²) = (0+5)+(0+0-10)x +3x² = 5-10x +3x². Hence, the standard matrix of T is A =

0

-3

5

-4

4

-10

0

0

3

The eigenvalues of T, being the eigenvalues of A, are the solutions to its characteristic equation det(A- I₃) = 0 or, ³-7²+36 = 0 or (-3)( -6)( +2)= 0. Thus, the eigenvalues of A (also T) are ₁= -2, ₂ =2 and ₃ = 6. Further, the coordinate eigenvector of A, relative to the standard basis {1,x,x²}, corresponding to the eigenvalue -2, is solution to the equation (A+2I₃)X= 0. To solve this equation, we will reduce A+2I₃ to its RREF as under:

Multiply the 1st row by ½

Add 4 times the 1st row to the 2nd row

Interchange the 2nd row and the 3rd row

Multiply the 2nd row by 1/5       

Add -5/2 times the 2nd row to the 1st row

Then, the RREF of A+2I₃ is

1

-3/2

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-3y/2 = 0 or, x = 3y/2 and z= 0. Then X = (3y/2, y,0)^T = y/2(3,2,0)^T. Hence the eigenvector of A corresponding to the eigenvalue -2 is (3,2,0)^T. Similarly, the eigenvector of A corresponding to the eigenvalue 3 is solution to the equation (A-3I₃)X= 0. The RREF of A-3I₃ is

1

0

5/3

0

1

-10/3

0

0

0

Now, if X = (x,y,z)^T, then the equation (A-3I₃)X= 0 is equivalent to x+5z/3 = 0 or, x = -5z/3 and y -10 z/3= 0 or, y= 10z/3 . Then X = (-5z/3, 10z/3, z)^T = z/3(-5,10,3)^T. Hence the eigenvector of A corresponding to the eigenvalue 3 is (-5, 10,3)^T. Further, the eigenvector of A corresponding to the eigenvalue 6 is solution to the equation (A-6I₃)X= 0. The RREF of A-6I₃ is

1

½

0

0

0

1

0

0

0

Now, if X = (x,y,z)^T, then the equation (A-6I₃)X= 0 is equivalent tox +y/2 = 0 or, x = -y/2 and z = 0. Then X = (-y/2, y, 0)^T = y/2(-1,2,0)^T. Hence the eigenvector of A corresponding to the eigenvalue 6 is (-1, 2,0)^T.

Thus, the eigenvalues of T are, ₁= -2, ₂ =2 and ₃ = 6.

The corresponding coordinate eigenvectors of T, relative to the standard basis {1,x,x²} are x₁ = (3,2,0)^T,x₂ = (-5, 10,3)^T and x₃ = (-1, 2,0)^T .

0	-3	5
-4	4	-10
0	0	3