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This is for a graduate level course on computational linear systems. I want to do it myself, without help, but it's been ten years since I took linear algebra and I'm not sure I understand what the first part is asking for. The problem states:

Find a solution $x = x^* = (x_1^*, x_2^*, x_3^*, x_4^*)^T$ to the LLS problem:

$\left\| \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 0 & -e & 0 \\ e & 0 & 0 & 0 \\ 0 & 0 & 0 & e \\ 0 & e & 0 & 0 \\ -e & 0 & 0 & 0 \\ 0 & -e & 0 & 0 \\ 0 & 0 & 0 & e \\ 0 & 0 & e & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \right\|_2 \to \min $

as a function of $e$.

  1. Find a function $f(x) = f(x_1, x_2, x_3, x_4)$ such that $x$ minimizes $f(x)$ is solution $x^*$ to problem

In case it isn't clear from the formatting, we have an (4x9) matrix with a variable e (not the log base) multiplied by a vertical vector of x1,x2,x3,x4, take the difference of that and another vertical vector with one 1 and 8 zeros and minimize. At one point in the assignment, he calls them Normal Equations. I've looked those up in Wolfram Alpha, but I don't think it helps.

I'm stumped by the question labeled 1. What is the teacher asking for? This is just the beginning part. I think once I understand the question he's asking here I can move on to solving Cholesky's algorithm, etc.

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    Ok, one last clarification: is$x$in$f(x)$the vector of [x1,x2,x3,x4]?2011-02-18

1 Answers 1

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Okay, let's call the big matrix $A$ and the vector with 9 elements $b$.

Then, of course, you want a solution $x$ such that $Ax = b$, because then $Ax - b = 0$ and $\|Ax-b\| = 0$. My guess is that this is not possible for all $e$ (it is possible for $e = 0$, easy to check).

Now, what you then have is a linear least squares problem. That is, you can "solve" for the $x$ such that $Ax$ is as close as it gets to $b$.

In question 1, the teacher probably wants you to write down a function $f$, such that if $f(x)$ reaches it minimum at $x$, then $\|Ax - b\|$ is minimal too.

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    It is certainly possible there's a typo, but that wouldn't help my understanding.2011-02-18