It is mentioned on [Wikipedia](https://en.wikipedia.org/wiki/Approximately_finite-dimensional_C*-algebra#Finite-dimensional_C*-algebras) that every unital \*-homomorphism $\Phi:M\_i\to M\_j$ is necessarily of the form $\Phi(a)=U^\*(a\otimes I\_r)U$ for some unitary $U$ and some $r$. (Here $M\_i$ are the $i\times i$ complex matrices and $I\_r$ is the $r\times r$ identity.)


No proof or reference is given.


How is this proven? (A reference to a textbook is also OK for me.)


I will need to formalize the proof in a computer-aided theorem prover, so an elementary proof would be preferred. (By elementary I mean not more than introductory graduate-level textbook level, maybe.)


**Notes:**


* This [answer](https://mathoverflow.net/a/33592/101775) seems to touch the question but I don't understand it. (It uses "Morita equivalence" which as far as I can tell is quite a power-tool in this context.)
* Information about whether something similar holds also for the bounded operators on a larger Hilbert space would be appreciated.



Unital *-homomorphisms between matrices
**Theorem:** For Hilbert spaces $H,K$, every normal unital \*-homomorphism $\Phi:\mathcal B(H)\to\mathcal B(K)$ is of the form $\Phi(a)=U(a\otimes 1\_{K\_0})U^∗$ for some Hilbert space $K\_0$ and some unitary $U$.


Here is a super down-to-earth proof, from a functional analysis / operator-algebras perspective. I'll start by working in the infinite-dimensional setting. Let $H,K$ be Hilbert spaces, write $\newcommand{\mc}{\mathcal}\mc B(H)$ for the algebra of all bounded operators on $H$, and $\mc K(H)$ for the compact operators. Let $\Phi:\mc B(H)\rightarrow\mc B(K)$ be a unital $\*$-homomorphism which is *normal* (aka weak$^\ast$-continuous). As $\mc K(H)$ is weak$^\ast$-dense in $\mc B(H)$, $\Phi$ is completely determined by its restricton to $\mc K(H)$, That $\Phi$ is unital corresponds to $\Phi:\mc K(H)\rightarrow\mc B(K)$ being *non-degenerate* meaning that $\{ \Phi(\theta)(\xi) : \theta\in\mc K(H), \xi\in K \}$ has a dense linear span in $K$.


If $H$ is finite-dimensional, of course $\mc B(H) = \mc K(H) \cong M\_i$ where $i$ is the dimension of $H$.


So, consider a non-degenerate $\*$-homomorphism $\Phi:\mc K(H)\rightarrow\mc B(K)$. For $\xi,\eta\in H$ write $\theta\_{\xi,\eta}$ for the rank-one operator $\alpha\mapsto (\alpha|\eta) \xi$. Then $\theta\_{\xi,\eta}^\* = \theta\_{\eta,\xi}$ and $\theta\_{\xi,\eta} \theta\_{\xi\_1,\eta\_1} = (\xi\_1|\eta) \theta\_{\xi,\eta\_1}$. Here I write $(\cdot|\cdot)$ for the inner-product on $H$. Fix a unit vector $\xi\_0\in H$, and consider $\theta\_{\xi\_0, \xi\_0}$ which is a projection (self-adjoint idempotent). So $p = \Phi(\theta\_{\xi\_0, \xi\_0})$ is also a projection. Let $K\_0\subseteq K$ be the (closed) subspace forming the image of $p$.


Define $U:H\odot K\_0 \rightarrow K$ by
$$ U(\xi\otimes\alpha) = \Phi(\theta\_{\xi,\xi\_0})(\alpha), $$
and extend by linearity. I write $\odot$ for the algebraic tensor product, and will write $\otimes$ for the (completed) Hilbert space tensor product. Then
\begin{align\*}
( U(\xi\otimes\alpha) | U(\eta\otimes\beta) )
&= (\Phi(\theta\_{\xi,\xi\_0})(\alpha) | \Phi(\theta\_{\eta,\xi\_0})(\beta)) \\
&= (\Phi(\theta\_{\xi\_0, \eta} \theta\_{\xi,\xi\_0})(\alpha) | \beta ) \\
&= (\xi|\eta) (\Phi(\theta\_{\xi\_0,\xi\_0})(\alpha) | \beta )
= (\xi|\eta) (p(\alpha) | \beta ) \\
&= (\xi|\eta) (\alpha | \beta ).
\end{align\*}
Thus $U$ is an isometry, and so extends to $H\otimes K\_0$. Now compute
\begin{align\*}
U^\* \Phi(\theta\_{\xi,\eta}) U(\xi\_1\otimes\alpha)
&= U^\* \Phi(\theta\_{\xi,\eta}) \Phi(\theta\_{\xi\_1,\xi\_0})(\alpha) \\
&= (\xi\_1|\eta) U^\*\Phi(\theta\_{\xi,\xi\_0})(\alpha) \\
&= (\xi\_1|\eta) U^\*U(\xi\otimes\alpha) \\
&= \theta\_{\xi,\eta}(\xi\_1) \otimes \alpha.
\end{align\*}
So $U^\*\Phi(\theta\_{\xi,\eta})U = \theta\_{\xi,\eta}\otimes 1$ and so by linearity and continuity, $U^\*\Phi(\theta)U = \theta\otimes 1$ for each $\theta\in\mc K(H)$.


If we can show that $U$ has dense range, it must be onto (as it's an isometry), and so will be a unitary, and so $UU^\*=1$ and so $\Phi(\theta) = U(\theta\otimes 1)U^\*$ as required.


If $\xi\_1$ is another vector, we see that
$$ \Phi(\theta\_{\xi,\xi\_1})(\alpha) = \Phi(\theta\_{\xi,\xi\_0})\Phi(\theta\_{\xi\_0,\xi\_1})(\alpha) = U(\xi \otimes \beta), $$
say, where $\beta = \Phi(\theta\_{\xi\_0,\xi\_1})(\alpha)$. Letting $\xi, \xi\_1,\alpha$ vary, taking linear span, and using non-degeneracy, we see that $U$ does indeed have dense range.


