As is known (see [Kadison-Ringrose](http://bookstore.ams.org/gsm-15), 3.4.1) each closed ideal $I$ in the $C^\*$-algebra $C(X)$ of continuous functions on a compact space $X$ has the form
$$
I=\{f\in C(X): \ \forall x\in S\quad f(x)=0 \}
$$
for some closed subset $S$ in $X$.



> 
> Is the same true for the two-sided ideals in the $C^\*$-algebra $C(X,M\_n)$ of continuous maps $f:X\to M\_n$ into the algebra $M\_n$ of all $n\times n$ complex matrices?
> 
> 
> 



Closed two-sided ideals in $C(X,M_n)$
I'm assuming your compact space $X$ is Hausdorff. Then the answer is yes.


Suppose $I$ is a closed two-sided ideal in $C(X, M\_n)$.
For each $x \in X$, $I(x) = \{f(x): f \in I\}$ is a two-sided ideal in $M\_n$. But 
$M\_n$ is a simple ring: it has no two-sided ideals except itself and $\{0\}$.  

Let $S = \{x \in X: \; I(x) = \{0\}\}$. Then $I \subseteq I(S) = \{f \in C(X, M\_n):\; f|\_S = 0\}$.


Now suppose $f \in I(S)$. I need to show $f \in I$. For each $x \notin S$, $I(x) = M\_n$ so there is some $g\_x \in I$ with $g\_x(x) = f(x)$. Given $\epsilon > 0$, we can find a finite partition of unity $\rho\_1, \ldots, \rho\_N$ of $X$ and $g\_1, \ldots, g\_N \in I$ such that $\|f(x) - \sum\_j \rho\_j(x) g\_j(x)\| < \epsilon$ for all $x \in X$. Since $I$ is closed, $f \in I$. 


