Prove Least Squares Equation Has Solution

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The discussion centers on proving that the least squares equation A^T A x* = A^T b always has a solution. It distinguishes between two cases: when A^T A is invertible and when it is not. In the invertible case, the solution is straightforward using the inverse. For the non-invertible case, the conversation suggests using the pseudo-inverse and minimum norm criteria, but emphasizes that a unique solution is not required. A key point raised is the need to demonstrate that the columns of A^T b lie within the column space of A^T A.
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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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