linear operator and diagonalizable

[lldk]

How to solve this problem? I try to transform it into a matrix with respect to a basis [1,x,x^2], but it seems doesn't work

 

[blamocur]

Can you post the resulting matrix?

 

[lldk]

Can you post the resulting matrix?

T(1)=k*1+l*0
T(x)=k*x+l*1
T(x^2)=k*x^2+2lx
Then we have
k l 0
0 k 2l
0 0 k

 

[blamocur]

I got the same. Which criteria of diagonalizability are you familiar with? (E.g., https://en.wikipedia.org/wiki/Diagonalizable_matrix).

 

[lldk]

by calculating the characteristic polynomial, we can only get one eigenvalue, hence can only get one eigenvector and implies that it is not diagonalizable, so I think this method will fail

 

[blamocur]

by calculating the characteristic polynomial, we can only get one eigenvalue, hence can only get one eigenvector and implies that it is not diagonalizable, so I think this method will fail

I agree with you about single eigenvalue, but not about single eigenvector. E.g., an identity matrix has more than one eigenvector.

 

[lldk]

I agree with you about single eigenvalue, but not about single eigenvector. E.g., an identity matrix has more than one eigenvector.

is it I need to find a matrix that is similar to the original matrix?

 

[blamocur]

According to https://en.wikipedia.org/wiki/Diagonalizable_matrix, the matrix is diagonizable iff its eigenvalues from a basis of the whole space ([imath]\mathbb R^3[/imath] in your case). But since there is only one eigenvector every vector is an eigenvector. Can you finish the rest ?