Question

6. Let T P2 P be a linear transformation such that T P2P2 is still a linear trans formation such that T(1) 2r22 T(2-)=2 T(1) = 2r22 T(12 - )=2 T(x2x= 2r T(r2)2x (a) (6 points) Find the matrix for T in some basis B. Specify the basis that you use. (d) (4 points) Find a basis for the eigenspace E2. (b) (2 points) Find det(T) and tr(T') (e) (4 points) Find a basis = (f,9,h) for P2 such that T(f T(h)= ch for scalars a, b, e, or explain why such a basis do (c) (4 points) Given that 2 is an eigenvalue of T, find all eigenvalues and their algebraic multiplicities

Answer 1

6. (a). Since x² = (1/2)[ (x² +x)+( x² -x)] and x = (1/2)[ (x² +x)-( x² -x)] and since T is a linear transformation, hence T(x²) = (1/2)T[(x² +x)+( x² -x)] = (1/2)[T(x² +x)+T( x² -x)] = (1/2)[(2x+2)] = x+1 and T(x) = (1/2) T[ (x² +x)-( x² -x)] = (1/2)[T(x² +x)-T( x² -x)] = (1/2)[(2x-2)] = x -1.

Thus, the standard matrix of T with respect to the standard basis B = {x²,x,1} of P₂ is A(say) = [T(x²),T(x),T(1)] =

0	0	2
1	1	0
1	-1	2

It may be observed that the entries in the columns of A are the coefficients of x²,x and scalar multiples of 1 in T(x²),T(x), T(1) respectively.

(b). det(T)= det(A) = 2*[1*(-1)-1*1] = -4. Also tr(T) = trace(A) = 0+1+2 = 3.

( c). The eigenvalues of A are solutions to its characteristic equation det(A- λI₃)= 0 or, λ³-3λ²+4 = 0. We know that ʎ₁ = 2 is an eigenvalue of A. On divoding λ³-3λ²+4 by (ʎ-2), we get , λ²-λ-2 = (ʎ-2)(ʎ+1). Therefore, the other 2 eigenvalues of A are ʎ₂ = 2 and ʎ₃ = -1. Thus, the eigenvalues of T atre 2 ( of algebraic multiplicity 2) and -1 (of algebraic multiplicity 1 ).

(d). The eigenvectors of A corresponding to the eigenvalue 2 are solutions to the equation (A-2I₃)X= 0. To solve this equation, we have to reduce A-2I₃ to its RREF which is

1	0	-1
0	1	-1
0	0	0

Now, if X = (x,y,z)^T, then the equation (A-2I₃)X= 0 is equivalent tox-z = 0 or, x = z and y-z = 0 or, y = z. Then X = (z,z,z)^T = z(1,0,0)^T. This implies that every solution to the the equation (A-2I₃)X= 0 is a scalar multiple of the vector (1,1,1)^T. Hence, the only eigenvector of A corresponding to the eigenvalue 2 is (1,1,1)^T. Thus, the set {(1,1,1)^T } is a basis for E₂.