# Can someone resolve my confusion about uniqueness of diagonalization?

I am a bit confused about diagonalization. I have $A$ which I know is diagonalizable. I want to find $P$ such that $A = P \Sigma P^{-1}$ where $\Sigma$ is diagonal. Under what circumstances is $P$ unique, if ever? If it is not unique, is it at least unique up to some operation?

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If $P$ diagonalizes $A$, and $B$ is any nonsingular matrix which commutes with $\Sigma$ ($B\Sigma = \Sigma B$), then $PB$ also diagonalizes $A$. –  p.s. Jul 19 '12 at 0:25

For any diagonalizible matrix $n\times n$, for $n\geq 2$, $P$ is not unique. If the eigenvalues of $A$ are not all equal, then $\Sigma$ is not unique as well. If you change the columns of $P$ that will correspond to changing the appropriate columns in $\Sigma$. Even if, suppose, you fix $\Sigma$, even then you can change the columns in $P$ that correspond to eigenvectors of the same eigenvalue, or you can multiply them by scalar or make linear combinations. In short, in this case (when you fix $\Sigma$), $P$ will be unique up to elementary matrix column operations, but only between columns that correspond to the same eigenvalue.

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In general, $P$ won't be unique. You can always:

1. Change the order of different eigenvalues in $\Sigma$; that is, the values along the main diagonal. This will produce changes in the order of the corresponding eigenvectors; that is, the columns of $P$.
2. Even keeping $\Sigma$ untouched, if you have eigenvalues with multiplicity greater than 1, this will produce subspaces of eigenvectors with dimension greater than one. Any basis of which can be used for the columns of $P$.

For instance, take $A$ to be the diagonal matrix

$$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$$

Then you can obviously take $\Sigma = A$ and $P$ to be the unit matrix, but any matrix of the form

$$P = \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & e \end{pmatrix}$$

will do (as long as $ad - bc \neq 0$ and $e \neq 0$).

Or, for the same $A$, you could also take

$$\Sigma = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

and any

$$P = \begin{pmatrix} e & 0 & 0 \\ 0 & a & b \\ 0 & c & d \end{pmatrix}$$

with the same restrictions as before is ok.

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P is never unique - if D is any diagonal matrix then PD also works to diagonalise A. P is not even unique up to multiplication by a diagonal matrix though - consider a matrix which permutes the basis.

In general, P is unique up to transformation by any matrix which takes each eigenvector of A to another eigenvector.

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This is to diagonalise A generally, not to a specific sigma. If we wish sigma to be fixed, P is unique up to any transformation that fixes the eigenspaces of A –  Tom Hutchcroft Jul 19 '12 at 0:25
Strictly speaking $PD$ only works if $D$ is invertible (no diagonal entries equal to $0$) since $P$ has to be invertible. But even if $D$ is not invertible $P$ is not unique. It is unique, assuming $A$ is a $n\times n$ matrix with entriesin a field $F$, precisely if either $n=0$ or $n=1$ and $F=\mathbb F_2$ –  Marc van Leeuwen Jul 19 '12 at 1:28