Question

Let $\phi$ define a $*$-automorphism from the matrix algebras $M_n(\mathbb{C})$ to $M_n(\mathbb{C})$ such that $\phi(I) = I$. Is it true that any such map $\phi$ can be represented as $\phi(x) = U x U^{\dagger}$ (where $U$ is a suitable unitary matrix)? If not, what is the most general expression?

Answer 1

Here is one generalization:

Every $*$-automorphism of the algebra of compact operators on a Hilbert space is conjugation by a unitary operator on that space.

Using the fact that the algebra of compact operators is irreducible, this can be seen as a special case of:

Every irreducible $*$-representation of the algebra of compact operators on a Hilbert space is unitarily equivalent to the identity representation.

A proof can be found for instance in Section 1.4 of Arveson's An invitation to C* algebras. Another proof of the first assertion that gives more information can be found in Proposition 1.6 of Raeburn and Williams's Morita equivalence and continuous trace C*-algebras.

The first part is still true if you take all bounded operators instead of only the compact ones. (And these are the same thing in the finite dimensional case.)

Answer 2

If $\phi$ is a $*$-automorphism then $\psi:A\mapsto\phi(\overline A)$ is a $\mathbb{C}$-automorphism. By the Skolem-Noether theorem every $\mathbb{C}$-automorphism of $M_n(\mathbb{C})$ is inner, that is of the form $\psi(A)=UAU^{-1}$. This must commute with the $*$-operation: $A\mapsto\overline{A}^t$. This leads to $UAU^{-1} =\overline{U^t}^{-1}A\overline{U^t}$ for all $A$. Thus implies that $U$ and $\overline{U^t}^{-1}$ are the same up to a constant multiple. By multiplying $U$ by a constant we may make $U$ unitary.

Answer 3

As an alternative to Robin Chapman's solution, I would like to state Exercise 7.8 from Rørdam's, Larsen's and Laustsen's "Introduction to the K-theory of C*-algebras":

For every unital AF-algebra $A$ there is a short exact sequence $$ 1\to\overline{\mathrm{Inn}}(A)\to\mathrm{Aut}(A)\to\mathrm{Aut}(K_0(A))\to 1, $$ where $\overline{\mathrm{Inn}}(A)$ denotes approximately inner automorphisms and $\mathrm{Aut}(K_0(A))$ denotes group automorphisms preserving the unit class and the positive cone in $K_0(A)$.

If $A$ is the matrix ring, then $\mathrm{Aut}(K_0(A))$ is trivial and hence every automorphism of $A$ is approximately inner. Since $A$ is separable, every approximately inner automorphism is the pointwise limit of a sequence of inner automorphisms. And I think the finite-dimensionality of $A$ implies that the pointwise limit of a sequence of inner automorphisms is again inner.

Using the statement above, one immediately sees that, for instance, $\mathbb C\oplus\mathbb C $ possesses an automorphism which is not approximately inner.

Answer 4

Another proof can be obtained using that $M_n(\mathbb{C})$ is singly generated (and finite-dimensional). So $M_n(\mathbb{C})=C^*(s)$ for some $s$ (the shift, for example). Now, of course, $\phi(s)$ is a generator for the image. And by Spetch's theorem, $\phi(s)$ and $s$ are unitarity equivalent (because $\phi$ is multiplicative and it preserves the trace). Then there exists a unitary $U\in M_n(\mathbb{C})$ with $\phi(s)=UsU^{-1}$. If you now take any $a\in M_n(\mathbb{C})$, we have $a=\sum_{j=0}^{n-1} \alpha_js^j+\sum_{j=1}^{n-1}\beta_j(s^*)^j$, for coefficients $\alpha_j,\beta_j$, and so $$ \phi(a)=\sum_{j=0}^{n-1} \alpha_j\phi(s)^j+\sum_{j=1}^{n-1}\beta_j\phi(s^*)^j $$ $$ =\sum_{j=0}^{n-1} \alpha_j((UsU)^{-1})^j+\sum_{j=1}^{n-1}\beta_j(Us^*U^{-1})^j=UaU^{-1} $$

Answer 5

Yet another proof would be to consider a system $(e_{kj})$ of matrix units in $M_n(\mathbb{C})$, coming from some orthonormal basis (the canonical one, say). It is then easy to check that $(\phi(e_{kj}))$ is another system of matrix units, and so it corresponds to another orthonormal basis. The unitary implementing the change of basis is the one implementing $\phi$.