Originally Posted by

**kevinlightman** Let T:V->V be a linear map on a finite-dimensional real vector space.

Suppose A is the matrix representing T with respect to basis e.

Show that if A is diagonalizable then every matrix representation of T is diagonalizable.

With this question there is a hint that you may assume that any other matrix representation of T is of the form (X^-1)AX where X is an invertible matrix.

My thoughts for this question start with that if A is diagonalizable then there exists an invertible matrix X such that (X^-1)AX is diagonal. But I think I need to show that for all X (X^-1)AX is diagonalizable.

No: since $\displaystyle A$ is diagonalizable there exist an invertible matrix $\displaystyle P$ and a diagonal matrix $\displaystyle D$ s.t. $\displaystyle P^{-1}AP=D$ .

Now, if $\displaystyle B$ is ANY other matrix representation of the map $\displaystyle T$ then there exists an invertible matrix $\displaystyle Q$ s.t. $\displaystyle A=Q^{-1}BQ$ , but then:

$\displaystyle D=P^{-1}AP=P^{-1}(Q^{-1}BQ)P=(PQ)^{-1}B(PQ)$ ...and this means not only that $\displaystyle B$ is also diagonalizable but it will have the very diagonal form as $\displaystyle A$

Tonio

So I think that if (X^-1)AX is a matrix representation of T with respect to another basis then X must consist of the coefficients of the vectors in this basis as linear combinations of the vectors in e.

I think after this I have to show that (X^-1)AX has n linearly independent eigenvectors as this would imply diagonalizability but I'm not sure how to get there. Any help would be much appreciated