Why is a *-surjective homomorphism between algebras of compact operators an irreducible representation?

Question

Let $\varphi:K(\mathcal{H})\to K(\mathcal{K})$ be an surjective * -homomorphism, where $K(X)$ is the C* algebra of compact operators on Hilbert space X. I want to conclude that $\varphi$ is an irreducible representation. It is clearly a representation.

I tried using the fact that a representation $\pi:A\to B(\mathcal{H})$ is irreducible iff $\pi(A)\xi=\{\pi(a)\xi:a\in A\}$ is dense in $\mathcal{H}$ for every $\xi\in \mathcal{H},\;\xi\neq0$.

So, for $\xi\in\mathcal{H}\setminus\{0\}$ we have $\varphi(K(\mathcal{H}))\xi=K(\mathcal{K})\xi$ and at this point I am stuck for I don't know why $\overline{K(\mathcal{K})\xi}=\mathcal{K}$.

Also, I have the same question but when $\varphi:B(\mathcal{H})\to B(\mathcal{K})$ is surjective *-homomorphism where I want to conclude $\overline{B(\mathcal{K})\xi}=\mathcal{K}$.

Answer 1

For all $\xi,\zeta\in\mathcal K\setminus\{0\}$ there exists a compact operator (and even a rank $1$ operator) sending $\xi$ to $\zeta.$

Answer 2

Without doing any computation, you could use that irreducibility is $\varphi(A)''=B(\mathcal K)$. In this case the surjectivity gives you $\varphi(A)=K(\mathcal K)$, which has trivial commutant, so you are done.

You can also do as Anne suggests. If you fix $\xi\in\mathcal K$, then for any $\eta\in\mathcal K$, then if $T\in B(\mathcal K)$ is the rank-one operator with $T\xi=\eta$, by the surjectivity there exists $S\in A$ with $\varphi(S)=T$, so $\varphi(S)\xi=\eta$.

Note that in both cases the domain is irrelevant. For any C$^*$-algebra $A$, if $\varphi:A\to B(\mathcal K)$ is surjective, then it is irreducible by the above.

As for your last question, again by the above any representation $\varphi:A\to B(\mathcal K)$ such that $\varphi(A)\supset K(\mathcal K)$ is irreducible.