If A is a Banach algebra, we say that $\Phi: A \longrightarrow \mathbb{C}$ is a multiplicative linear functional if $\Phi$ is nontrivial, linear and $\Phi(xy)=\Phi(x)\Phi(y)$.
It is easy to see that there is no multiplicative linear functional on $M_n(\mathbb{C})$. Is this statement true for B(H)? if yes, how can i prove it?
On
Here's a direct argument. Since $\Phi$ is onto and multiplicative, $\phi(I)=1$. Write $I=P+Q$ for two orthogonal equivalent projections. Then, as $PQ=0$, we have $\Phi(P)\Phi(Q)=0$, so at least one of them is zero. But as $P=V^*V$, $Q=VV^*$, we get $\Phi(P)=\Phi(Q)$, so both are zero. Then $1=\Phi(P+Q)=\Phi(P)+\Phi(Q)=0$, a contradiction.
It is true for $B(H)$ also. Another way to see it for $M_n$ when $n>1$ that may be extended in spirit to $B(H)$ is to note that $M_n$ is simple, and the kernel of a multiplicative linear functional would be an ideal of codimension $1$. If $H$ is infinite dimensional then $B(H)$ is not simple, but it does have a unique maximal ideal, and the codimension is not $1$. E.g., in the separable case the maximal ideal is the set $K(H)$ of compact operators on $H$, and the Calkin algebra $B(H)/K(H)$ is infinite dimensional.