For the following linear operator T, I have to test T for diagonalizability, and if T is diagonalizable, I have to find a basis B for V such that is a diagonal matrix
and T is defined by T(f(x))=f(0) +f(1)(x+x^2)
Evaluate the characteristic polynomial of the operator, perhaps by first finding its matrix wrt the basis . You'll get , so the matrix (and thus T) are diagonalizable iff the geometric multiplicity of the eigenvalue is 2.
Check that this indeed is the case and then you're done.
If you had three distinct eigenvalues, since eigenvectors corresponding to distinct eigenvalues are independent, you would know immediately that this matrix has three independent eigenvectors. Here, since there is a double eigenvalue, you will have to check if there really exist two independent eigenvector corresponding to that eigenvalue.
Please show us what you did so we can tell you what you did wrong. Do you know the definition of the characteristic polynomial? You should evaluate , where is the matrix of the transformation with respect to your basis. There should be no such things as or in your result.
ok so I found that the second eigenvalue, 2 has multiplicity 2, so T is indeed diagonalizable. Now I have to find eigenspaces for each eigenvalue right? Im not sure where to proceed...
Coincidentally the matrix I got for where B is the standard basis for
Is this correct?
Good! Now check that you have the good characteristic polynomial. Once you have the eigenvalues, to find the eigenspace associated to the eigenvalue , solve the matrix equation for ; this will give you a subspace of which consists of the vectors whose entries are the coordinates, in the basis , of the eigenvectors of
ok now I am completely boggled. How did you get the char. polynomial to be x(x-2)^2?????
I got x(x-1)(x-2)
Am I missing something here? If the matrix A is :
then A-xI is:
So det(A-xI) = -x(1-x)(1-x)
How you possibly have gotten x(x-2)^2???
You are right. Tonio got the wrong characteristic equation and eigenvalues. (Well, it had to happen sooner or later!) They are, as you say, 0, 1, and 2. Since those are all distinct, there are three indpendent eigenvectors and, using those as basis vectors gives you a diagonal matrix.