Original problem $Ax=b$ (let's say A is 400x5 matrix)
$x$ can be found through SVD. Now we can create a new problem that will provide us with the same solution for x':
I_5 x' = x
Where $I_5$ is a 5x5 identity matrix
Let's now extend the problem by adding more rows to $A$ and $b$. Calling the new variables A' and b'. Same logic can then be applied: A' x' = b'
However, at this point we can notice that since the new values were simply extensions of the previous problem, then we can write A' as
A' =[\begin{matrix} A \\ A_{added} \end{matrix}]
From here I'd like to substitude the original $A$ with an identity matrix, and the values in $b$ with the original $x$, creating
\begin{matrix} I_5 \\ A_{added} \end{matrix} x' = \begin{matrix} x \\ b_{added} \end{matrix}
Question: Why does this not provide me with the same x' as solving the original "extended" problem? How can I achieve this reduction/compression of the matrix?
I can only assume it's some sort of "weighting" issue, but I don't even know where to begin. Should the identity matrix be multiplied by a constant? Or not a constant?
Thanks