Show that maximal ideal space of $M_n(\mathbb{C})$ is empty? Show that maximal ideal space of $M_n(\mathbb{C})$ is empty?
how to approach this problem ?? what i know is that it is non commutative Banach algebra with unity.any hint ??
 A: You can play the game to show that $E_{ij}AE_{kl}=a_{jk}E_{il}$, where $E_{ij}$ is the matrix whose $(i,j)$ entry is $1$ and all other entries are $0$.
With such, one can show that every two-sided ideal of $M_{n}(R)$ has the form $M_{n}(I)$ for some two-sided ideal $I$ of $R$.
And a consequence is that, if $D$ is a division ring, then $M_{n}(D)$ is simple, which means that $M_{n}(D)$ has no submodule $M\ne 0,S$.
${\bf{C}}$ is a division ring, so the result follows.
A: If $n\geq2$ and $\alpha:M_n(\mathbb C)\to\mathbb C$ is an algebra homomorphism, $$\alpha(E_{kk})=\alpha(E_{k1}E_{1k})=\alpha(E_{k1})\alpha(E_{1k})=\alpha(E_{1k})\alpha(E_{k1})=\alpha(E_{1k}E_{k1})=\alpha(E_{11}).$$ Then 
$$
0=\alpha(0)=\alpha(E_{11}E_{22})=\alpha(E_{11})\alpha(E_{22})=\alpha(E_{11})^2. 
$$
Thus $\alpha(E_{11})=0$, and 
$$
\alpha(I)=\sum_k\alpha(E_{kk})=0,
$$
so $\alpha=0$. 
A: Of course we assume that $n>1$.
Say $\phi$ is a (nonzero) complex homomorphism of $M_n(\Bbb C)$. Let $A$ be the subalgebra consisting of the diagonal matrices. Then $A$ is canonically isomorphic to $C(K)$, where $K=\{1,2,\dots,n\}$. Since every complex homomorphism of $C(K)$ is given by evaluation at a point of $K$ we obtain
$\newcommand\diag{\text{diag}}$


There exists $j$ such that $\phi(\diag(d_1,\dots,d_n))=d_j$ for every $\diag(d_1,\dots, d_n)\in A$.


But if $\sigma$ is a permutation of $\{1,\dots,n\}$ then $y=\diag(d_{\sigma(1)},\dots,d_{\sigma(n)})$ is similar to $x=\diag(d_1,\dots,d_n)$. Hence $\phi(y)=\phi(x)$. With the above this shows that $d_j=d_k$ for every $d_j,d_k\in\Bbb C$.
