If A is a real diagonalizable matrix and B is real symmetric and positive definite, is the matrix B − 1 A real diagonalizable?Let G be the square root of B and G = G T , then B − 1 A = G − 1 ( G − 1 A G − T ) G is similar to G − 1 A G − T , which is again symmetric and thus diagonalizable.However, this method fails when A is just real diagonalizable. Are there any tricks to modify the proof, or is there a counterexample?

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If   A is a real diagonalizable matrix and   B is real symmetric and positive definite, is the matrix       B          −      1        A real diagonalizable?Let G be the square root of B and   G  =      G    T  , then       B          −      1        A  =      G          −      1        (      G          −      1        A      G          −      T        )  G is similar to       G          −      1        A      G          −      T      , which is again symmetric and thus diagonalizable.However, this method fails when A is just real diagonalizable. Are there any tricks to modify the proof, or is there a counterexample?

Damion Ray · Accepted Answer

The statement is not true. Take                    A        =                  (                                                    0                                            1                                                                    0                                            −                1                                  /                                2                                              )                ,                B        =                  (                                                    1                                            −                1                                  /                                2                                                                    −                1                                  /                                2                                            1                                              )                &amp;gt;        0        ,            then                              B                      −            1                          A        =                  (                                                    0                                            1                                                                    0                                            0                                              )                    is not diagonalizable. But A is clearly diagonalizable because it has two distinct eigenvalues.

Zachery Mckee · Accepted Answer

Hint: for a counterexample, start with a matrix   C that is not diagonalizable, and see if you can find a symmetric positive definite matrix   B so that   A  =  B  C is diagonalizable. Almost any one will do.

If A is a real diagonalizable matrix and B is real symmetric and positive definite, is the matrix B^(−1) A real diagonalizable?

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