Let $A$ be a unital Banach algebra, $n$ a positive integer, and $w : A \rightarrow \mathbb{C}^{n \times n}$ be a homomorphism of complex algebras such that $w(A) = \mathbb{C}^{n \times n}$ ($\mathbb{C}^{n \times n}$ is the algebra of complex $n \times n$ matrices).
Show that $w$ is continuous.
Also deduce that any linear functional $f : A \rightarrow \mathbb{C}$ satisfying $f(xy) = f(x)f(y)$ is continuous. HINT: Use that $\mathfrak{m} \subseteq A$ is a maximal ideal if and only if the quotient algebra $A / \mathfrak{m}$ is simple.
For the first part we have to show that for any open set $U_{n \times n}$ of $\mathbb{C}^{n \times n}$ we have that the pre-image $w^{-1}(U_{n \times n})$ is open in $A$; or equivalently for any closed set $C_{n \times n}$ of $\mathbb{C}^{n \times n}$, the pre-image $w^{-1}(C_{n \times n})$ is closed in $A$. I am kind of at a loss here.
I suppose the second question, namely, showing that any linear functional $f$ is continuous then follows from the first part by taking $n = 1$? Since the condition $f(xy) = f(x)f(y)$ makes $f$ into a homomorphism?
I will denote $\mathbb{C}^{n\times n}$ by $M_n(\mathbb{C})$. Since $w$ is an algebra homomorphism, we have that $\ker(w)$ is an ideal in $A$ and from the first isomorphism theorem one easily sees that $A/\ker(w)\cong M_n(\mathbb{C})$. Now any matrix algebra is simple, so $A/\ker(w)$ is simple. As the hint suggests, this implies that $\ker(w)$ is a maximal ideal. But this means that $\ker(w)$ is closed, since $\overline{\ker(w)}$ is also an ideal containing $\ker(w)$ (and of course if $\overline{\ker(w)}=A$ then $w=0$, a contradiction: indeed, if $\ker(w)$ is dense in $A$, then there exists an element $z\in B(1_A,1)$ with $w(z)=0$. But then $z$ is invertible, so $w(1)=w(zz^{-1})=w(z)w(z^{-1})=0$). But if the kernel is closed then $w$ is continuous.