Question about diagonalizable matrices

  • Thread starter Bipolarity
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  • #1
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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
 

Answers and Replies

  • #2
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Yes, it's true.

A hint for the proof: the diagonal entries are the eigenvalues of ##A##.
 
  • #3
CompuChip
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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
HallsofIvy
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Yes, it's true.

A hint for the proof: the diagonal entries are the eigenvalues of ##A##.
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
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Not quite. A matrix can be similar to two different diagonal matrices if they have the same numbers on the diagonal in different orders.
That's exactly what the OP said:

Thus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for ##[L_A]_\beta## but this will be merely a rearrangement of the diagonal entries of B.
 
  • #6
mathwonk
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you might look at the characteristic polynomial.
 

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