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Linear Algebra II, Eigenvectors and values.

Author: Andre Sostar

Exercise: T is the linear operator on P2 (R), defined by

T (f (x)) = f (x) + (x + 1)f ' (x),

b is the standard ordered basis for P2 (R), and A = [T ]b .

Solution:

We have T : P2 (R) == P2 (R), with b = {1, x, x2 } and dim = 3. Then, we have to find

A = [T ]b = T @ b in terms of g.

T (1) = 1 + (x + 1) * 0

== 1 * 1 + 0 * x + *x2

= (1,0,0)

'

T (x) = x + (x + 1) * (x) = 2x + 1

== 1 * 1 + 2 * x + 0 * x2

= (1,2,0)

2 2 2 '

2 2

T (x ) = x + (x + 1) * (x ) = x + 2x(x + 1) = 3x + 2x

== 0 * 1 + 2 * x + 3 * x2

= (0,2,3)

Then, we transpose the result and get the answer?

1 1 0

A = [T ]b = ?0 2 2?

0 0 3

Now, we have to find the characteristic polynomial?

1-a

1 0 2-a

2 ? = (1 - a)(2 - a)(3 - a)

det(A - I3) = det ? 0 0 0 3-a

Therefore we can see that:?

?a1 = 1

a2 = 2?

a3 = 3

Hence is an eigenvalue of T iff = 1, 2, 3.

1

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