Let $\mathcal{L}(\mathbb{C}^{n \times n})$ denote the algebra of all linear mappings from $\mathbb{C}^{n \times n}$ to $\mathbb{C}^{n \times n}$ and let $\mathcal{C} \subseteq \mathcal{L}(\mathbb{C}^{n \times n})$ denote the subalgebra of all $\phi \in \mathcal{L}(\mathbb{C}^{n \times n})$ which satisfy $$ \phi(U^*AU) = U^*\phi(A)U $$ for all $A \in \mathbb{C}^{n \times n}$ and all unitary $U \in \mathbb{C}^{n \times n}$ (in other words, $\mathcal{C}$ is the commutant of the set of all conjugations by unitary matrices).

Question. Is there an explicit description of $\mathcal{C}$?

Of course, there is some freedom for interpretation of the word "explicit"; I would be most happy with a set of mappings in $\mathcal{L}(\mathbb{C}^{n \times n})$ which spans $\mathcal{C}$.


  • Clearly, the identity $\operatorname{id}_{\mathbb{C}^{n \times n}}$ is an element of $\mathcal{C}$.

  • The operator $\tau: \mathbb{C}^{n \times n}\to \mathbb{C}^{n \times n}$ given by $$ \tau(A) = \operatorname{tr}(A) \cdot \operatorname{id}_{\mathbb{C}^{n \times n}} $$ is an element of $\mathcal{C}$ (where $\operatorname{tr}(A)$ denotes the trace of the matrix $A$).

  • The span of $\operatorname{id}_{\mathbb{C}^{n \times n}} $ and $\tau$ is a subalgebra of $\mathcal{C}$ (since $\tau^2 = n\tau$), but I don't know whether $\mathcal{C}$ is larger than this span.

  • 3
    $\begingroup$ The conjugation representation of $\mathrm U(n)$ on $\mathbb M_n(\mathbb C)$ decomposes into irreducible representations, and the algebra of $\mathrm U(n)$-equivariant maps is completely determined by this decomposition: it is a sum of matrix algebras, one for each irreducible component, of the size equal to the multiplicity of that irreducible component. Now, there is, for instance, $\mathfrak{su}(n)$ as an irreducible subspace (and $\mathbb C\cdot 1$, as well). I don't know the decomposition of $\mathfrak{su}(n)^\perp$ off the top of my head, but it should be in the literature. $\endgroup$ Nov 30, 2019 at 10:38
  • $\begingroup$ I just realised that $\mathfrak{su}(n)$ is a real, not complex, subrep'n. Its complex counterpart is $\mathfrak{sl}(n)$, and it's irreducible (see answer below). $\endgroup$ Nov 30, 2019 at 11:22

3 Answers 3


Building up on my comment, I can now give the complete answer. The space of matrices can be decomposed as follows: $$ \mathbb M_n(\mathbb C) = \mathbb C\cdot\mathrm{id}\oplus \mathfrak{sl}(n), $$ where $$ \mathfrak{sl}(n) = \{X\in\mathbb M_n(\mathbb C)\mid \mathrm{Tr}(X) = 0\}. $$

Thus, the conjugation representation of $\mathrm{U}(n)$ decomposes as the sum of a trivial representation and the conjugation repesentation on $\mathfrak{sl}(n)$. The latter is irreducible as a complex representation of $\mathrm{U}(n)$ because:

  • the complexification of $\mathfrak{u}(n)$ is $\mathfrak{gl}(n)$, and
  • the Lie algebra $\mathfrak{sl}(n)$ is a simple complex Lie algebra.

Therefore the algebra of linear $\mathrm U(n)$-equivariant maps is isomorphic to $\mathbb C\oplus \mathbb C$. The elements $(1,0)$ and $(0,1)$ are just orthogonal projections to $\mathbb C\cdot \mathrm{id}$ and its orthogonal complement $\mathfrak{sl}(n)$.

So, the space $\mathcal C$ from the question is indeed spanned by $\mathrm{id}_{\mathbb M_n}$ and $\tau$.

  • $\begingroup$ Thanks a lot for your answer! Could you specify what you mean by $\mathfrak{u}(n)$ (as opposed to $U(n)$)? $\endgroup$ Nov 30, 2019 at 21:20
  • 1
    $\begingroup$ The Lie algebra of $\mathrm{U}(n)$: $\mathfrak u(n) = \{X\in\mathbb M_n(\mathbb C)\mid X^* = -X\}$. $\endgroup$ Dec 1, 2019 at 9:18

$\mathcal{C}$ is simply the span of the two maps that you noted (the identity and the trace) -- there is nothing else in the commutant.

One (admittedly somewhat roundabout) way of seeing this is to notice that if you unpack $\phi$ into an $n^2 \times n^2$ matrix $\Phi$ in the "usual" way (i.e., instead of thinking of it as a linear transformation acting on matrices, think of it as a matrix acting on their vectorizations), then your commutation relation is equivalent to $$ (U \otimes \overline{U})\Phi(U \otimes \overline{U})^* = \Phi $$ for all unitary $U \in \mathbb{C}^{n\times n}$ (here $\overline{U}$ is the entrywise complex conjugate of $U$).

This is the defining property of something called an isotropic state from quantum information theory, and it is well-known (see this paper, for example) that all matrices with this property are linear combinations of the identity matrix and the "maximally entangled state" $\rho = \sum_{i,j=1}^n \mathbf{e}_i\mathbf{e}_j^* \otimes \mathbf{e}_i\mathbf{e}_j^*$ (where $\{\mathbf{e}_i\}$ is the standard basis of $\mathbb{C}^n$). These two matrices correspond to the trace linear map and the identity linear map, respectively, once you "un-vectorize" everything.

  • $\begingroup$ +1 Thank you very much! After I while of thinking I finally chose to accept Vadim Alekseev' answer (although your post answers my question as well) since he was a bit earlier and since his answer "forced" me to learn a few facts about Lie algebras ;-). Anyway, I'm very grateful to you for pointing out the connection to quantum information theory! $\endgroup$ Dec 3, 2019 at 16:45

I posted (nearly) this on MSE, so if it doen't belong here plese remove.

Let $\Phi\in\mathcal{L}(\mathbb{C}^{n \times n})$ and suppose that for any unitary conjugation operator $\mathcal{U}\in\mathcal{L}(\mathbb{C}^{n \times n})$, $\mathcal{U}\Phi=\Phi\mathcal{U}$. Let $\{e_1,e_2,\dots,e_n\}$ denote the standard orthonormal basis for $\mathbb{C}^n$, let $e_{ij}$ denote the $n\times n$ matrix with $1$ in the $i$th row $j$th column and zeros elswhere, and let $E_{ij}=\Phi(e_{ij})$. Claim, there exists $r,s\in\mathbb{C}$ such that, \begin{equation} E_{ij}=\begin{cases} re_{ij} & \text{ if } i\neq j, \text{ and }\\ % re_{ij}+sI & \text{ if } i= j, \end{cases}\tag{1} \end{equation} so that for $B\in\mathbb{C}^{n\times n}$, $\Phi(B)=rB+\mathrm{tr}(B)sI$. Accordingly, fix $1\leq i,j\leq n$ with $i\neq j$, let $K_{ij}$ denote the span of $\{e_i,e_j\}$, and let $P_{ij}$ denote the orthogonal projection onto $K_{ij}$. Notate the compressions to $K_{ij}$ by \begin{gather*}\epsilon_{ii}=P_{ij}E_{ii}\Bigm|_{K_{ij}}= \begin{bmatrix} a_{ii} & b_{ii} \\ c_{ii} & d_{ii} \end{bmatrix}\qquad % \epsilon_{ij}=P_{ij}E_{ij}\Bigm|_{K_{ij}}= \begin{bmatrix} a_{ij} & b_{ij} \\ c_{ij} & d_{ij} \end{bmatrix}\\ % \epsilon_{ji}=P_{ij}E_{ji}\Bigm|_{K_{ij}}= \begin{bmatrix} a_{ji} & b_{ji} \\ c_{ji} & d_{ji} \end{bmatrix}\qquad % \epsilon_{jj}=P_{ij}E_{jj}\Bigm|_{K_{ij}}= \begin{bmatrix} a_{jj} & b_{jj} \\ c_{jj} & d_{jj} \end{bmatrix} % \end{gather*}

Let $U_{1}$, $U_{2}$, and $U_{3}$ be the unitary matrices which fix the orthogonal complement of $K_{ij}$ with action on $K_{ij}$ given by $u_{1}=\begin{bmatrix} i & 0 \\ 0 & 1\end{bmatrix}$, $u_{2}=\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$, and $u_{3}=\frac{1}{\sqrt 2}\begin{bmatrix} 1 & 1 \\ -1 & 1\end{bmatrix}$ respectively. Note that for $A=\begin{bmatrix} a & b \\ c & d\end{bmatrix}\in\mathbb{C}\times\mathbb{C},$

$$u_1Au_1^\dagger=\begin{bmatrix} a & -ib \\ ic & d\end{bmatrix}\quad u_2Au_2^\dagger=\begin{bmatrix} d & c \\ b & a\end{bmatrix}\quad u_3Au_3^\dagger= \frac12\begin{bmatrix} a+b+c+d & -a+b-c+d \\ -a-b+c+d & a-b-c+d \end{bmatrix}$$

For $k=1,2,3$, $U_kP_{ij}=P_{ij}U_k$ so that $\mathcal{U}_{k}\Phi= \Phi\mathcal{U}_{k}$ implies for $\ell,m\in\{i,j\}$ , \begin{equation} \label{eq:compression} u_k\epsilon_{\ell m}u_k^\dagger= P_{ij}\Phi(U_ke_{\ell m}U_k^\dagger)\Bigm|_{K_{ij}}\tag{2} \end{equation} With $k=1$, equation (2) shows that the off-diagonal entries of $\epsilon_{ii}$ and $\epsilon_{jj}$ equal zero and that all entries of $\epsilon_{ij}$ and $\epsilon_{ji}$ except for $b_{ij}$ and $c_{ji}$ must equal zero. Since $\mathcal{U}_2(e_{ii})=e_{jj}$, $a_{ii}=d_{jj}$ and $d_{ii}=a_{jj}$, and since $\mathcal{U}_2(e_{ij})=e_{ji}$, $b_{ij}=c_{ji}$. With these identities, it follows that $2u_3\epsilon_{ij}u_3^\dagger= \begin{bmatrix}b_{ij} & b_{ij} \\ -b_{ij} & -b_{ij} \end{bmatrix}$. Further, since $2U_3e_{ij}U_3^\dagger=e_{ii}+e_{ij}-e_{ii}-e_{ii}$, $$\begin{bmatrix}b_{ij} & b_{ij} \\ -b_{ij} & -b_{ij} \end{bmatrix}= \begin{bmatrix} a_{ii}-d_{ii} & b_{ij} \\ -b_{ij} & d_{ii}-a_{ii}\end{bmatrix},$$ so that $a_{ii}-d_{ii} = b_{ij}$. Letting $r=b_{ij}$ and $s=d_{ii}$ one has, $$\epsilon_{ii}=\begin{bmatrix} s+r & 0 \\ 0 & s \end{bmatrix}\quad % \epsilon_{ij}=\begin{bmatrix} 0 & r \\ 0 & 0 \end{bmatrix}\qquad % \epsilon_{ji}=\begin{bmatrix} 0 & 0 \\ r & 0 \end{bmatrix}\qquad % \epsilon_{jj}=\begin{bmatrix} s & 0\\ 0 & s+r \end{bmatrix}\qquad % $$ Letting $i,j$ run through all unequal pairs yields equation (1).

Remark. Using similar techniques one can show the following. Let $\Phi\in\mathcal{L}(\mathbb{C}^{n\times n})$ and suppose that for any orthogonal conjugation operator $\mathcal{O}\in\mathcal{L}(\mathbb{C}^{n\times n})$, $\mathcal{O}\Phi=\Phi\mathcal{O}$. There exists $r,s,t\in\mathbb{C}$ such that, for $B\in\mathbb{C}^{n\times n}$, $\Phi(B)=rB+sB^\top+\mathrm{tr}(B)tI$.


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