Prove Least Squares Equation Has Solution

dirk_mec1
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Homework Statement



In the least squares method the vector x* that is the best approximation to b statisfies the Least squares equation:

A^T A x^*= A^T b

Prove that there's always a solution to this equation.

Homework Equations


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The Attempt at a Solution


I distinct 2 situations A^T A is invertible and it isn't invertible. If it's invertible then there's no problem x^*= (A^T A)^{-1} A^T b

But how I prove that it works in the non-invertible case?
 
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dirk_mec1 said:

Homework Statement



In the least squares method the vector x* that is the best approximation to b statisfies the Least squares equation:

A^T A x^*= A^T b

Prove that there's always a solution to this equation.

Homework Equations

The Attempt at a Solution


I distinct 2 situations A^T A is invertible and it isn't invertible. If it's invertible then there's no problem x^*= (A^T A)^{-1} A^T b

But how I prove that it works in the non-invertible case?

If it is not invertible you need another criteria to get a unique solution. For instance you could require that x has minimum norm. In which case use could use the pseudo inverse which is based on singular value decomposition. However, you do not required a unique solution in the above question. So perhaps you could try showing that the columns of A^T form the same column space as the columns of A^TA.
 
John Creighto said:
If it is not invertible you need another criteria to get a unique solution. For instance you could require that x has minimum norm. In which case use could use the pseudo inverse which is based on singular value decomposition. However, you do not required a unique solution in the above question. So perhaps you could try showing that the columns of A^T form the same column space as the columns of A^TA.

Don't you mean the columns of A^T b are in the span of the columns of A^T A? If so I don't understand how to prove such a thing.
 
I've thought about it and I seriously don't know how to prove that the A^Tb is in the column space of A^TA. Can someone help me?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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