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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

2 Answers 2