Proof that $\exists x \in C$ that $A - x\cdot I_n$ is invertible We know that $A \in \mathbb C^{n,n}$
Proof that $\exists x \in C$ that $A - x\cdot I_n$ is invertible. How many such x exists? 
I am thinking about this problem and generally I have only few little concepts...
We should consider two cases: 
1. $A$ is reversible. 
Then we can put $ x = 0 $ and $$ A - x\cdot I_n = A $$ so it works.

2. $A$ is not invertible - and there I have problem. If A is invertible It means that $\det_n A \neq 0$. Ok, but does it give me something? We can take such $x$ to make at least one element to 0 on diagonal - But if we have $ 0 $ on diagonal, It does not guarantee that $A - x\cdot I_n = A$ will be invertible... thanks for your time
 A: The matrix $A-x\operatorname{Id}$ is invertible if and only if $\det(A-x\operatorname{Id})\neq0$. But $\det(A-x\operatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.
A: Suppose $A-xI_n$ is not invertible. Then there exists $v_x\ne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.
Suppose $x\ne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
$$
av_x+bv_y=0
$$
Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
$$
axv_x-ayv_x=0
$$
and since $x\ne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.
This can be generalized: suppose we have $x_1,x_2,\dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
$$
\{v_{x_1},v_{x_2},\dots,v_{x_m}\}
$$
is linearly independent. Since the vectors belong to $\mathbb{C}^n$, we see that $m\le n$.
Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.
