# Linear maps and diagonal matrices

• Apr 5th 2010, 06:03 AM
kevinlightman
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
• Apr 5th 2010, 11:23 AM
tonio
Quote:

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

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