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For

A=$\begin{pmatrix}0&1&1&1\\ \:1&-1&0&1\\ \:1&1&1&1\\ \:0&-1&0&1\end{pmatrix}$ and b=$\begin{pmatrix}5\\ \:3\\ \:-1\\ \:1\end{pmatrix}$

How would one show that a least-squares solution to Ax=b is not unique and solve the normal equations to find all of the least-squares solutions?

What I have done:

So I found $A^TA=\begin{pmatrix}3&0&3&0\\ 0&3&1&2\\ 3&1&4&0\\ 0&2&0&2\end{pmatrix}$ and $A^Tb=\begin{pmatrix}3\\ 3\\ 8\\ -2\end{pmatrix}$

I then solved by row reduction and got that $\begin{pmatrix}3&0&3&0&3\\ \:0&3&1&2&3\\ \:3&1&4&0&8\\ \:0&2&0&2&-2\end{pmatrix}$ reduced to $\begin{pmatrix}3&0&3&0&3\\ 0&3&1&2&3\\ 0&0&\frac{2}{3}&-\frac{2}{3}&4\\ 0&0&0&0&0\end{pmatrix}$

I am not sure where to go from here...

user520403
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  • Do these help https://math.stackexchange.com/questions/901484/unique-least-square-solutions/2170702#2170702 and https://math.stackexchange.com/questions/1718183/is-a-least-squares-solution-to-ax-b-necessarily-unique?noredirect=1&lq=1? – Moo Mar 28 '18 at 23:00
  • Yes those are super helpful for proving the solution is not unique thanks! I am just not sure how to solve to find all the least squares solutions. – user520403 Mar 28 '18 at 23:11

1 Answers1

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Complete your solution: continue reducing the matrix until it’s in row-reduced echelon form. Once you have that, you can then simply read the general solution to the equation $A^TA\hat x=A^Tb$ from it. So, picking up where you left off, the rref is $$\left[\begin{array}{cccc|c}1&0&0&1&-5 \\ 0&1&0&1&-1 \\ 0&0&1&-1&6 \\ 0&0&0&0&0\end{array}\right],$$ therefore the general solution is $\hat x=(-5,-1,6,0)^T+\lambda(1,1,-1,-1)^T.$

amd
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