# Is matrix diagonalization unique?

From the following statement, it seems matrix diagonalization is just eigen decomposition.

Diagonalizing a matrix is also equivalent to finding the matrix's eigenvalues, which turn out to be precisely the entries of the diagonalized matrix. Similarly, the eigenvectors make up the new set of axes corresponding to the diagonal matrix.

http://mathworld.wolfram.com/MatrixDiagonalization.html

However, from what I have learned, Spectral Theorem is closest to this conclusion. But how the spectral theorem is related to it, or is there some other theorem grants this statement?

Spectral Theorem: Suppose that $V$ is a complex inner-product space and $T \in L(V)$. Then $V$ has an orthonormal basis consisting of eigenvectors of $T$ if and only if $T$ is normal.

• The operator defined by the matrix in some basis has a basis of orthonormal eigenvectors. The matrix of the same operator in this basis is the diagonalization. The change of basis gives you the equivalence.
– OR.
Commented Jul 25, 2013 at 22:22
• There are many different types of matrix factorization each of which finds use in different problem domains: en.wikipedia.org/wiki/Matrix_decomposition Commented Feb 13, 2018 at 22:35

Matrix diagonalization is more general than the spectral theorem. For instance, you may not be in an inner product space, and it still may be helpful to diagonalize a matrix. Not every matrix can be diagonalized, though; for instance,

$$\left[\begin{matrix} 1 & 1 \\ 0 & 1 \end {matrix}\right]$$

has eigenvalues 1 and 1, but cannot be diagonalized.

The spectral theorem tells you that in a certain situation, you are guaranteed to be able to diagonalize. Even better, the eigenvectors have some extra structure: they are orthogonal to each other.

If a matrix is diagonalized, its diagonal form is unique, up to a permutation of the diagonal entries. This is because the entries on the diagonal must be all the eigenvalues. For instance,

$$\left[\begin{matrix} 1 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 1 \end {matrix}\right] \text { and }\left[\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 2 \end {matrix}\right]$$

are examples of two different ways to diagonalize the same matrix.

• Thanks Eric. I am asking if a matrix can be diagonalized, then if that is unique.. Commented Jul 25, 2013 at 22:47
• I added to my answer. Commented Jul 25, 2013 at 22:54
• What does "up to a permutation of the diagonal entries" mean? You mean up to the $n$, where $n$ is the (maximum) number of permutations? Commented Jan 31, 2016 at 11:55
• @mavavilj it basically boils down to the fact you can interchange any two eigenvalues on the diagonals. Commented Feb 13, 2018 at 22:30

While the diagonal matrix $D$ is determined up to permutation, the change of basis $U$ is only determined up to scalar multiples of the columns. So we may indeed have diagonalizations $$A = U_1 D_1 U_1^{-1} = U_2 D_2 U_2^{-1}$$ where $D_1 \neq D_2$, $U_1 \neq U_2$. So diagonalization is not unique in general, but unique up to permutation of the diagonal entries in $D$, and multiples of the columns of $U$.

Eric's answer is spot on if by "diagonalization" you mean finding invertible $Q$ so that $QAQ^{-1}$ is diagonal. However we may also find unitary $U^\star, V$ so that $U^\star A V$ is diagonal, the singular value decomposition.

Another possibility is if you find $L, D, U$ where $L$ is lower triangular, $D$ is diagonal, $U$ is upper triangular, and $A=LDU$. The LDU decomposition is useful for solving linear systems. If $A$ is not invertible, this doesn't work, but if it is the diagonal $D$ is unique.

• And it is uniquely to be the eigenvalues, if the matrix is normal, right? Commented Jul 26, 2013 at 16:53
• Absolute values of the eigenvalues, yes. Commented Jul 26, 2013 at 17:28

Following up on Vadim's answer, there is another practically consequential sense in which diagonalization is not unique.

Consider the 2x2 positive-definite Hermitian matrix

$$H = \left( \begin{array}{cc} \omega & \lambda \\ \lambda & \omega \end{array} \right),$$

with $$\omega > 0$$, $$\lambda > 0$$, and $$\lambda < \omega$$. Since $$H$$ is Hermitian, it can be diagonalized by a unitary transformation,

$$D = U^\dagger H U,$$

with

$$D = \left( \begin{array}{cc} \omega + \lambda & 0 \\ 0 & \omega - \lambda \end{array} \right), \\ \\ U = \frac{1}{\sqrt{2}} \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right).$$

But here is an alternative way to diagonalize $$H$$. Let's write

$$D = T^\dagger H T,$$

with

$$D = \left( \begin{array}{cc} \epsilon & 0 \\ 0 & \epsilon \end{array} \right), \\ \\ T = \frac{1}{\sqrt{\lambda^2-(\omega-\epsilon)^2}} \left( \begin{array}{cc} -\lambda & \omega - \epsilon \\ \omega-\epsilon & -\lambda \end{array} \right),$$

and

$$\epsilon = \sqrt{\omega^2 - \lambda^2}.$$

You can check that this works. But you can also check that $$T$$ is not unitary. Consequently, it does not have the form $$D = P^{-1} H P$$ and is also not a singular value decomposition. It is also clearly not a $$LDU$$ decomposition.

The above example is the simplest example of symplectic diagonalization outlined by Williamson's theorem: any 2N x 2N positive-definite Hermitian matrix $$H$$ can be diagonalized as $$D = T^\dagger H T$$ by a symplectic transformation $$T$$. I will not go into the details of how to find $$T$$ here, but it involves the solution of a generalized eigenvalue problem.

In the language of quantum mechanics, this is also known as a bosonic Bogoliubov transformation. It is a transformation that is constructed by imposing the symplectic constraints of classical and quantum mechanics.