Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm a chemistry major and I haven't taken much math, but this came up in a discussion of quantum chemistry and my professor said (not very confidently) that if a matrix is diagonalizable, then you should be able to diagonalize it to the identity matrix. I suspect this is true for symmetrical matrices, but not all matrices. Is that correct?

share|cite|improve this question
I'm not sure what he means by "diagonalize" it to the identity matrix it. If you mean that if for some matrix $A$ you have an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^{-1}$, then yes. But we need not have $D = \alpha I$ for some scalar $\alpha \in \mathbb{C}$. In other words, $\begin{pmatrix} 1 & 0 \\ 0 & 2\end{pmatrix}$ is already diagonal, but cannot be expressed as a multiple of the identity. The fact about symmetric matrices is that if $A = A^T$ then $P$ can be taken so that $P^T = P^{-1}$ with $D$ real. – snarski Jan 30 '13 at 5:19
yes, that's what he meant. Thanks! – twilightisnotaprincess Jan 30 '13 at 5:21
Anyone who makes such a statement has no idea of what diagonalisation is, or at least a very different idea than everybody else in the scientific community. – Marc van Leeuwen Jan 30 '13 at 5:23
up vote 8 down vote accepted

No. If $PAP^{-1} = I$ where $I$ is the identity then $A = P^{-1}IP = P^{-1}P = I$. So in fact only the identity matrix can be diagonalized to the identity matrix.

share|cite|improve this answer

Take the $0$ $n\times n$ matrix. It's already diagonal (and symmetrical) but certainly can't be diagonalized to the identity matrix.

share|cite|improve this answer
Hm... but what about a nxn matrix with non-zero elements? – twilightisnotaprincess Jan 30 '13 at 5:22
There are many reasons why an arbitrary diagonalize matrix need not be diagonalizable to the identity. One is the reason given by @Jim above that shows that the only matrix diagonalizable to the identity is the identity itself. – Ittay Weiss Jan 30 '13 at 5:35
maybe I should restate my question: can a symmetric matrix be diagonalized into a matrix D=αI where α is a scalar. – twilightisnotaprincess Jan 30 '13 at 5:39
the same argument as above will show that the only matrix diagonalizable to $\alpha I$ is $\alpha I$. – Ittay Weiss Jan 30 '13 at 5:45
I see... thanks :) – twilightisnotaprincess Jan 30 '13 at 5:53

The usual meaning of "diagonalization" is diagonalization by similarity transform, which takes the form of $PAP^{-1}=D$. As the others have shown, the only matrix that can be diagonalized into the identity matrix is the identity matrix itself.

Yet, depending on the context, your professor may refer to diagonalization by congruence, which takes the form of $P^\ast AP=D$. If he implicitly assumes that $A$ is positive definite, then his assertion is true: $P^\ast AP=D\,\Rightarrow\,(PD^{-1/2})^\ast A(PD^{-1/2})=I$. However, since I know absolutely nothing about quantum chemistry, I can't say if he really meant this.

share|cite|improve this answer

I know that the question has already been answered, but I hope that the following line of thought will provide further insight nonetheless:

The trace of an n-dimensional Hermitian (because we want a guarantee that it has n eigenvalues) matrix is a basis invariant given by

a) the sum of the eigenvalues
b) the sum of the diagonal elements.

Consequently, if for every Hermitian matrix, there existed an orthogonal transformation capable of diagonalizing it to the identity matrix, we would equivalently have that

a) the sum of the the eigenvalues of every n-dimensional Hermitian matrix is n
b) the trace of every n-dimensional Hermitian matrix is n,

both of which are trivially disproven by counter-example. Even more disturbingly, since we are searching for a transformation that diagonalizes the matrix, we know that the we would have, in addition, that

c) every n-dimensional Hermitian matrix has only the n-fold degenerate eigenvalue 1.

and computational quantum chemistry would be very dull indeed.

share|cite|improve this answer
All-unity eigenvalues make Jack a dull matrix. – Brian May 27 '15 at 2:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.