Find a "canonical form" for the linear operators that are both self-adjoint and unitary in a finite-dimensional complex inner product space.

What's the meaning of "find a canonical form" in this problem? How to find it? I know that self-adjoint operators are diagonalizable and there is an orthonormal basis of eigenvectors. So, a self-adjoint operator can be represented by a diagonal matrix with real entries. Also, the modulus of eigenvalues of a unitary operator is $1$.

So, can we conclude that there is an orthonormal basis such that such an operator (self-adjoint and unitary) can be represented by the identity matrix? Is the identity matrix the "canonical form" of such an operator?

  • $\begingroup$ By "canonical form" one often means a representation for matrices of a certain class which allows us to decide effectively whether any two such matrices are similar. So while "identity matrix" is not quite right here, it is of course easy to detect whether a self-adjoint, unitary matrix is similar to the identity matrix. $\endgroup$
    – hardmath
    Commented Nov 16, 2018 at 0:56
  • $\begingroup$ en.wikipedia.org/wiki/Involutory_matrix might be of further interest $\endgroup$
    – Milten
    Commented Apr 19 at 17:33

2 Answers 2


Not too sure just what "canonical form" means in this context, but with respect to general matrix terminology, the phrase is usually used to refer to some sort of diagonal, or "near diagonal", representation, e.g., "Jordan canonical form", "rational canonical form", and so forth. In most of these "forms", the diagonal of a matrix plays a central role. Bearing these thoughts in mind, I offer the following words on the matter:

We have a unitary $U$:

$UU^\dagger = U^\dagger U = I, \tag 1$

which is also self-adjoint:

$U = U^\dagger; \tag 2$

(2) implies $U$ may be diagonalized by some unitary $W$:

$W^\dagger U W = \Lambda = \text{diag}(\mu_1, \mu_2, \ldots, \mu_n), \tag 3$


$W^\dagger W = WW^\dagger = I; \tag 4$

the $\mu_i$ are the eigenvalues of $U$; from (2), they are all real; therefore

$\Lambda^\dagger = \Lambda; \tag 5$

from (3),

$\Lambda^\dagger = W^\dagger U^\dagger W; \tag 6$

and thus,

$\Lambda^2 = \Lambda^\dagger \Lambda = W^\dagger U^\dagger W W^\dagger U W; \tag 7$

since we have (2) and (4),

$\Lambda^2 = W^\dagger U^\dagger IU W = W^\dagger U^\dagger U W =W^\dagger I W =W^\dagger W = I; \tag 8$


$\Lambda^2 = \text{diag}(\mu_1^2, \mu_2^2, \ldots, \mu_n^2); \tag 9$


$\mu_i^2 = 1 \Longrightarrow \mu_i = \pm 1, \; 1 \le i \le n; \tag{10}$

thus, $\Lambda$ must be a matrix of the form

$\Lambda = \text{diag}(\pm 1, \pm 1, \ldots, \pm 1); \tag{11}$

and hence $U$ is of the general form

$U = W\Lambda W^\dagger = W \text{diag}(\pm 1, \pm 1, \ldots, \pm 1)W^\dagger, \tag{12}$

for some choice of $\pm 1$ on the diagonal of $\Lambda$, and for some appropriate choice of unitary $W$.

It should be clear that any matrix given by (12) satisfies (1) and (2).

This representation is as "canonical" as I know how to make it!


You can be a little more efficient in finding a "canonical form" here: if $\alpha\colon V \to V$ is self-adjoint and unitary, then considering it first as a self-adjoint linear map there must an orthonormal basis of $V$ consisting of eigenvectors for $\alpha$, with each eigenvalues of $\alpha$ real. But as $\alpha$ is unitary, $\alpha\alpha^* = I_V$ so that as $\alpha = \alpha^*$ we see that $\alpha^2 = I_V$. In particular, any eigenvalue $\lambda$ of $\alpha$ must satisfy $\lambda^2=1$, and hence the eigenvalues of $\alpha$ lie in $\{\pm 1\}$, and $\alpha$ is determined up to conjugacy by the dimension of its $+1$ eigenspace.

If $\dim(V)=n$ and we set $r = \dim(\ker(\alpha-I_V))$ then a ``canonical form'' for $\alpha$ is $$ \Lambda_r :=\left(\begin{array}{cc} I_r & 0 \\ 0 & -I_{n-r}\end{array}\right) $$


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