A quantitative measure of rank of a matrix The rank of a matrix is only defined as integers. Is there some other criteria that is more quantitative.
E.g.
$$A =
\begin{bmatrix}
1 & 1\\
1 & 0\\
\end{bmatrix}
$$
$$B=
\begin{bmatrix}
1 & 1\\
1 & 0.999\\
\end{bmatrix}
$$
$A$ is "better ranked" than $B$
EDIT:
Context:
I am judging the observability $O$ of a system. I have two matrices
$$ dim(A) = n\times n$$
$$ dim(C) = m\times n$$
$$O = \begin{bmatrix} C \\ C A \\ C A^2 \\ \vdots \\ C A^{n-1} \end{bmatrix}$$
The system is observable if $O$ is full ranked. What I want is to be able to differentiate between observabilities of systems
 A: There is something like the condition of a matrix. This is a measurement of how strong errors in data inflict the error of your result. This is for $n\times n$ Matrixes definied which are invertible, when you have Matrixnorm. 
$$\operatorname{cond}(A)=\|A\|\cdot \|A^{-1}\|$$ 
where $\|\cdot\|$ is a matrix norm.
This is relevant for numeric as calculators often have an error of $10^{-16}$ (machine precision). So you would like to know what happens, how strong is the difference between your solution and the real solution.
As mentioned for nonsquare Matrix you can use $A^T A$ which is invertible when $A$ has maximum rank. 
A: I would try to build a criterion based on the singular values. See this Wikipedia article for starters. The singular values of a matrix $A$ are the square roots of the eigenvalues of $AA^T$. With your matrix $A$ we get
$$
AA^T=\left(\begin{array}{cc}2&1\\1&1\end{array}\right).
$$
The eigenvalues of this matrix are 
$$\lambda_1^2=\frac{3+\sqrt{5}}2\qquad\text{and}\qquad \lambda_2^2=\frac{3-\sqrt{5}}2.$$
While $\lambda_2$ is clearly smaller than $\lambda_1$, it is not dangerously close to zero.
With your other matrix on the other hand we get
$$
BB^T=\left(\begin{array}{cc}2&1.999\\1.999&1.998001\end{array}\right).
$$
The smaller eigenvalue of this matrix is very close to zero (at least in comparison to the bigger one) - sorry I don't have a CAS on this laptop yet, so can't give you the approximate values :-)
Singular values work better than eigenvalues, because the characteristic polynomial of a matrix fails to distinguish between
$$
\left(\begin{array}{cc}1&100\\0&1\end{array}\right)\qquad\text{and}\qquad
\left(\begin{array}{cc}1&0\\0&1\end{array}\right).
$$
I don't have a definite criterion in mind. May be the Wikiarticle helps or somebody else can give a well studied solution.
