1. Aug 15, 2013

### Bipolarity

Suppose that $A$ is a diagonalizable $n \times n$ matrix. Then it is similar to a diagonal matrix $B$. My question is, is $B$ the only diagonal matrix to which $A$ is similar?

I have thought about this, but am unsure if my answer is correct. My claim is that $B$ is the only diagonal matrix to which $A$ is similar, up to a rearrangement (or permutation) of the eigenvectors of $A$ in an ordered eigenbasis for $ℝ^{n}$. Thus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for $[L_{A}]_{β}$ but this will be merely a rearrangement of the diagonal entries of B.

Is this true? If so, how might I go about proving this conjecture of mine?

BiP

2. Aug 15, 2013

### micromass

Staff Emeritus
Yes, it's true.

A hint for the proof: the diagonal entries are the eigenvalues of $A$.

3. Aug 15, 2013

### CompuChip

If $A = X^{-1}BX = Y^{-1}B'Y$ then $B = (Y X^{-1})^{-1} B' (Y X^{-1})$.
So if A is similar to two diagonal matrices, then these matrices are similar to each other.
I guess a starting point would be to check which similarity transformations keep diagonal matrices diagonal, and see what you can say about X and Y that way.

4. Aug 15, 2013

### HallsofIvy

Staff Emeritus
Not quite. A matrix can be similar to two different diagonal matrices if they have the same numbers on the diagonal in different orders.

5. Aug 17, 2013

### micromass

Staff Emeritus
That's exactly what the OP said:

6. Aug 20, 2013

### mathwonk

you might look at the characteristic polynomial.