Let V be the three-dimensional space of real-valued polynomials of a real variab
ID: 3037353 • Letter: L
Question
Let V be the three-dimensional space of real-valued polynomials of a real variable of degree lessthanorequalto 2. Let T: V - V be the translation operator that acts on a polynomial f as (Tf)(t) = f(t+2). (a) Construct the matrices [T]V of T with respect to the basis V = {1, t, t^2}. (b) Show that this matrix has only one eigenvalue (the multiplicity of which is 3). Show that the dimension of the corresponding eigenspace is 1 and find a non-zero vector belonging to this space. (c) Use the above to define the eigenvalue and the eigenspace of T.Explanation / Answer
(a) Since T (f(t)) = f(t+2), we have T(1)=0+1= 1=1*1+0*t+0*t2 ( t is 0 here so that f(t) becomes a constant function), T(t) = t+2 = 2*1+1*t +0*t2 and T(t2) = (t+2)2 = t2+4t+4= 4*1+4*t +1*t2. Hence the matrix of T with respect to the basis V = {1,t,t2 } is [T]V = A (say)=
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(b)The characteristic polynomial of the matrix [T]V = A is det(A- I3) = 0 or, 3-32 +3 -1 = 0 or, (-1)3 = 0. Thus, the eigenvalues of [T]V = A are =1 ( of multiplicity 3). We know that the eigenvectors of A corresponding to the eigenvalue are solutions to the equation (A-I3)X = 0. Hence, the eigenvectors of A corresponding to the eigenvalue 1 are solutions to the equation (A-I3)X = 0. To solve this equation, we have to reduce the matrix A-I3 to its RREF which is
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Now, if X = (x,y,z)T , then the equation (A-I3)X = 0 is equivalent to y = 0 and z = 0 so that X = (x,0,0)T= x (1,0,0)T. Thus, the eigenvector of A corresponding to the eigenvalue 1 is (1,0,0)T. Therefore, the eigenspace of A corresponding to the eigenvalue 1 is span {(1,0,0)T}. Its dimension is 1.
(c ) Since T is defined as T (f(t)) = f(t+2), and since [T]V = A is the standard matrix of T, the eigenvalue and the eigenspace of T are the same as those of [T]V = A. Thus, the eigenvalue and the eigenspace of T are 1 and span {(1,0,0)T} respectively.
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