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$Ax=b$ has two possible difficulties.

(1)Dependent rows
$Ax=b$ may have no solution when $b$ is outside the column space.
Instead of $Ax=b$, we solve $A^{T}A\hat{x}=A^{T}b$.

(2)Dependent columns
In this case, the solution of $A^{T}A\hat{x}=A^{T}b$ may not be unique.
The optimal solution of $Ax=b$ is the minimum length of $A^{T}A\hat{x}=A^{T}b$.

I can't understand the second one.
If $A$ has dependent columns, why the solution of $A^{T}A\hat{x}=A^{T}b$ may not be unique?

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    Why don't you make up an example where the system is inconsistent and the columns are dependent, and see what happens?2012-12-08

2 Answers 2

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Here, this would mean that $A^T A$ is singular. If you consider its generalized inverse, which is not unique, then there is a solution for each one of these inverses.

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I do not think it is true that A^{T}Ax=A^{T}b will always have a solution. Note that A^{T}A is a square matrix and it might not always be invertible. If A^{T}A is not invertible it follows, from the invertible matrix theorem, that the transformation it represents is neither onto nor one-to-one.