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An invertible Matrix $A \in \Bbb R ^{2 \times2}$ and $b \in \Bbb R ^{2}$ are to $x \in \Bbb R^2$ the definite solution $Ax=b$. $x_N$ is the solution to the normalization equation $$A^TAx_N = A^Tb $$ and $x_Q \in \Bbb R^2$ the solution to $$QRx_Q = b $$ whereas $QR = A$ is a $QR$ decomposition of $A$. By exact calculation $x_N = x_Q = x $. However, how do I find a Matrix $A \in \Bbb R ^{2 \times 2}$ and $b \in \Bbb R^2$, with $\delta_N := \frac{\Vert x_N - x \Vert}{\Vert x \Vert} \ge 10^-1$ and $\delta_Q := \frac{\Vert x_Q - x \Vert}{\Vert x \Vert} \le 10^{-6}$? How do I even start?

Maxim
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  • The point is that the numerical solution to $x_Q$ is directly solving a problem with a lower condition number than the numerical solution to $x_N$. This is because typically the condition number of $A^T A$ is basically the square of that of $A$. So you want to find an invertible matrix $A$ such that $A^T A$ has a large condition number, then choose pretty much any $b$ that can't be exactly represented in floating point arithmetic. – Ian Dec 17 '17 at 00:03
  • I'm a bit perplexed why they are having you do it this way in the case of an invertible 2x2 matrix, though. Usually you would do this in the context of mxn with m much larger than n and rank n (the usual overdetermined problem). – Ian Dec 17 '17 at 00:04

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