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In two dimensions we have a system of two linear equations like :

2 x1 + 3 x2 = 5

3 x1 - 4 x2 = -1

Then we can solve for an exact solution, if it exists, by usual methods or by matrix method as :

In reality we have a number of points in two dimensions and we have to fit a curve among them, which best fits the given data. No solution exists so that A X = b is satisfied exactly. This is the same for 4 or more dimensions and when the matrix is rectangular and has m x n dimensions with a rank n. We cannot find A inverse in that case.

In such cases, we find the best fit, by taking projections from each data point on to the** best fit and minimize the sum of squares of the projections **from points on to the solution we give.

**This is the least squares problem**. We find the nearest solution to the data given.

In n-dimensional space: A X = b A is rectangular = m x n size and rank n.

Column vector space of A = W (Let)

The projection of vector b on to W = P = A X gives the solution for X

The*vector Orthogonal to the projection P*

= n-dimensional distance (*projection*) from data to the n-dimensional plane W

=*b - A X* .

Then from orthogonality of vectors :

Thus the unique solution of an n-dimensional least squares problem is given by the above matrix equation. Here it can be proved that (A^TA) is non-singular and has an inverse.

2 x1 + 3 x2 = 5

3 x1 - 4 x2 = -1

Then we can solve for an exact solution, if it exists, by usual methods or by matrix method as :

In reality we have a number of points in two dimensions and we have to fit a curve among them, which best fits the given data. No solution exists so that A X = b is satisfied exactly. This is the same for 4 or more dimensions and when the matrix is rectangular and has m x n dimensions with a rank n. We cannot find A inverse in that case.

In such cases, we find the best fit, by taking projections from each data point on to the

In n-dimensional space: A X = b A is rectangular = m x n size and rank n.

Column vector space of A = W (Let)

The projection of vector b on to W = P = A X gives the solution for X

The

= n-dimensional distance (

=

Then from orthogonality of vectors :

Thus the unique solution of an n-dimensional least squares problem is given by the above matrix equation. Here it can be proved that (A^TA) is non-singular and has an inverse.