Assume that $A$ is a unital $C^{\*}$ algebra. Is there a subvector space $Y\subset A$ of finite codimension which does not contain any invertible element?


Let $n(A)$ be the infimum of such codimensions. For example $n(A)=1$ if $A$ is commutative. Or $n(M\_{n}(\mathbb{C}))=n$.


For a commutative $A$, is it true to say $n(A\otimes M\_{n}(\mathbb{C}))=n\times 1=n$? More generaly, can one express $n(A\otimes B)$ in term of $n(A)$ and $n(B)$? 


**Edit:** According to the comments on this post, we ask:"What is an example of a simple $C^{\*}$ algebra $A$ for which $n(A)$ is finite? Of course for such algebra, $n(A)\neq 1$.


**Note:** This question is somehow a reverse question to the following famous problem:


What is the maximim possible dimension for those subvector space of $M\_{n}(\mathbb{R})$ which consist only invertible matrices(except zero matrix). For what valuse of $n$ the sharp dimension would occure?



Finite codimensional subvector space of $C^{*}$ algebras which contains no invertible elements
Let $A$ be a unital C\*-algebra.
As was already noted in the comments, we have $n(A)=1$ if and only if $A$ has a character.


Let $M\_n=M\_n(\mathbb{C})$.
It is easy to see that $n(M\_n)\leq n$. Conversely, a linear subspace of $M\_n$ of codimension less than $n$ contains an invertible matrix, as shown by Dieudonné, [1], see also [2]. Thus $n(M\_n)=n$.


If $\pi\colon A\to B$ is a surjective ${}^\*$-homomorphism onto another C\*-algebra $B$, then $n(A)\leq n(B)$. Indeed, let $B\_0$ be a subspace of $B$ of codimension $n(B)$ such that $B^{-1}\cap B\_0=\emptyset$. Set $A\_0:=\pi^{-1}(B\_0)$. Then $A\_0$ has codimension $n(B)$ in $A$ and $A^{-1}\cap A\_0=\emptyset$. Therefore $n(A)\leq n(B)$.


Next, let $X$ be a compact, Hausdorff space and consider $A=C(X,M\_n)=C(X)\otimes M\_n$. We show $n(A)=n$.


First, for every $x\in X$, there is a surjective ${}^\*$-homomorphism $\pi\_x\colon A\to M\_n$ given by evaluating at $x$. It follows $n(A)\leq n$. Let us show that the converse inequality also holds. So assume there exists a subspace $A\_0$ of noninvertible functions of codimension $\leq n-1$. Then there are at most $n-1$ points in $X$ such that $\pi\_x(A\_0)\neq M\_n$, say $x\_1,\ldots,x\_k$. For each $j\in 1,\ldots,k$, since $A\_0$ has codimension $\leq n-1$, we have that $\pi\_{x\_j}(A\_0)$ has codimension $\leq n-1$, and so $\pi\_{x\_j}(A\_0)$ contains an invertible matrix. Since the set $\{x\_1,\ldots,x\_k\}$ is discrete in $X$, we can easily construct an invertible element in $A\_0$, a contradiction.


One could now ask if $n(A)<\infty$ if and only if $A$ has a (nonzero) finite-dimensional representation (equivalently, a nonzero, finite-dimensional quotient C\*-algebra).


[1] Dieudonné: Sur une généralisation du groupe orthogonal à quatre variables, Arch. Math. 1, 282-287 (1949)


[2] Fillmore, Laurie, Radjavi: On matrix spaces with zero determinant, Linear multilinear algebra 18, 255-266 (1985)


