4

I was solving problems of diagonalization of matrices and I wanted to know if a diagonalization of a matrix is always unique? but there's nothing about it in the books nor the net.

I was trying to look for counter examples but I found none.

Any hint would be much appreciated

Thanks!

user2021
  • 131

4 Answers4

4

The diagonal matrix is unique up to a permutation of the entries (assuming we use a similarity transformation to diagonalize). If we diagonalize a matrix $M = U\Lambda U^{-1}$, the $\Lambda$ are the eigenvalues of $M$, but they can appear in any order.

Aurey
  • 1,332
  • 8
  • 17
  • 2
    Also the matrix $U$ can vary as well, e.g. with different multiples of eigenvectors. – jdods May 25 '16 at 02:13
  • 3
    Or if there are several degenerate eigenvalues, you're allowed to take linear combinations of the eigenvectors. Like the identity matrix - any $UU^{-1}$ would work for that guy. – Aurey May 25 '16 at 02:21
4

Assume A is some diagnalizable matrix. Then, we can write A = P D$\ P^{-1}$. But, then, we can change the order of our eigenvalues along the diagnal in our matrix D, to produce some other matrix G. But, this corresponds to a change in the order of the eigenvectors in P, which again produces another matrix Q. So, in conclusion , we have A = Q G$\ Q^{-1}$. So, this implies we do NOT have uniqueness.

wesssg
  • 695
2

I assume you mean uniqueness up to similarity transformations (i.e. change of basis). Suppose that the linear transformation $T$ is represented by the matrix $\left(\begin{array}{cc}1 & 0 \\ 0 & 2 \end{array} \right)$ with respect to basis $\{e_1,e_2\}$. In other words, $T$ maps any vector $(a,b)=ae_1+be_2$ to $(a,2b)=ae_1+2be_2$. Then, with respect to basis $\{e_2,e_1\}$ (note that the order of the basis elements is different), the linear transformation $T$ is represented by the matrix $\left(\begin{array}{cc}2 & 0 \\ 0 & 1 \end{array} \right)$. Hence, there is a similarity transformation that takes one matrix to the other. So, the diagonal matrices that represent a linear transformation are not unique. But the diagonal matrices will be unique up to some permutation of the diagonal entries.

svsring
  • 1,222
0

Actually it is not unique. Even without changing the order of eigenvalues. For example, the matrix $$A=\begin{pmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0&0&1\end{pmatrix}=\begin{pmatrix}1&0& -1 \\ 1 & 0 & -1 \\ 0&1&0\end{pmatrix}\begin{pmatrix}1&0& 0 \\ 0 & 1 & 0 \\ 0&0&-1\end{pmatrix}\begin{pmatrix}1&1& 0 \\ 0 & 0 & 1 \\ -1&1&0\end{pmatrix}=\begin{pmatrix}\frac{1}{\sqrt{2}}&0& -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0&1&0\end{pmatrix}\begin{pmatrix}1&0& 0 \\ 0 & 1 & 0 \\ 0&0&-1\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}& 0 \\ 0 & 0 & 1 \\ -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\end{pmatrix}.$$

Quarantiner
  • 128