While I am reading lecture notes on Poles and Zeros of MIMO systems, I find the following example, which is not clear for me.
$$ H(s) =\pmatrix{1 & \dfrac{1}{s-3} \\ 0 & 1 } $$
The explanation given as:
It is clear that $ H(s) $ has a pole at $ s = 3 $ but it may not be obvious that it also has a zero at $ s = 3 $. Observe that for $ s $ approaching $ 3 $, the second column of $ H(s) $ approaches alignment with the first column, so the rank of $H(s)$ approaches $ 1 $.
I am thinking that what ever the value $ \frac{1}{s-3} $ takes the matrix will have rank $ 2 $, because of two linearly independent columns. Kindly help me out.
What the explanation is trying to say is that as $s \to 3$, $H\left( s\right)$ becomes nearly singular.
For $s = 3+\epsilon$, the singular values of $H\left( s\right)$ are given by $$\sigma _1 = \frac{1}{\epsilon} + \epsilon + O\left(\epsilon ^2\right)$$ $$\sigma _2 = \epsilon + O\left(\epsilon ^2\right)$$ hence the matrix clearly becomes degenerate, effectively "losing" one dimension in its range as $\sigma_2 \to 0$.
In simpler terms, as $s \to 3$, the first column of the matrix becomes an increasingly good approximation of a scalar multiple of the second column in the matrix.