Is there a linear map $f: \mathbb{R^{4}} \rightarrow \mathbb{R^{4}} $ and a base vector a, with (f: a ,a ) matrix to be diagonal?
I've found this theorem:
Let L be the linear transformation
L(v) = Av
then A is diagonalizable with n linearly independent eigenvectors S = {v1, ... >,vn} if and only if the matrix of L with respect to S is diagonal.
Do you know if it is true the other way around?
What I mean is, if A is diagonalizable with n linearly independent eigenvectors $S = {v_1, ... ,v_n}$ then the matrix of L is diagonal. Is that true?
I've found this:
If n characteristic vectors correspond with all different eigenvalues, then these characteristic vectors are linear independent. The characteristic vectors can be used as a basis of V. The matrix of t relative to that basis is diagonal and the eigenvalues are the diagonal elements of the matrix.
Which states that if all the eigenvectors of a matrix A are linearly independent then the matrix of the linear map, is diagonal.
So if we have a matrix A that is diagonalizable, then the linear map, would be f(x) = A*x ?
Thank you for your time!