# Thread: Linear maps and diagonal matrices

1. ## Linear maps and diagonal matrices

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.
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

2. 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 $A$ is diagonalizable there exist an invertible matrix $P$ and a diagonal matrix $D$ s.t. $P^{-1}AP=D$ .

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

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

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
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