L12
Application of Least Squares Solution
Example: Find the "least squares linear fit" for the points (-1,2), (0,4), (1,1), and (2,7).
Given points (x_i, y_i), the least squares line is given by the equation:[ y = c_0 + c_1 x ]
Objective: Minimize [ n = \sum_{i=1}^{n} (y_i - (c_0 + c_1 x_i))^2 ]
Set Up the System
Set up the equation system as follows:
Coefficients [ c_0, c_1 ] are parameters to find.
Formulate:[ \begin{bmatrix} 1 & x_1 \ 1 & x_2 \ 1 & x_3 \ 1 & x_4 \ \end{bmatrix} \begin{bmatrix} c_0 \ c_1 \end{bmatrix} = \begin{bmatrix} y_1 \ y_2 \ y_3 \ y_4 \end{bmatrix} ]
Rearranged into:[ Ax = b ]
Least Squares Solution
Formula:
The least squares solution is found using: [ c = (A^TA)^{-1}A^Tb ]
Parameters calculated for the given example:
When calculated: [ c = \begin{pmatrix} -0.9 \ 2.2 \end{pmatrix} ]
The least squares fit line is: [ y = -0.9x + 2.2 ]
Invertibility of ATA
Proposition:
If [ A ] is a matrix, then [ A^TA ] is invertible if and only if rank(A) = n, meaning the columns of A are independent.
Proof:
If [ x \in N(A) ], then [ A^TAx = 0 \Rightarrow x \in N(A^TA) ].
Conversely, if [ x \in N(A^TA) ], then it implies that [ Ax = 0 ].
Thus, [ N(ATA) = N(A) ] follows.
Example 2: Quadratic Fit
Find a "least squares" quadratic fit for points (-1,2), (0,4), (2,7).
Set the equations to minimize:
[ \begin{pmatrix} 1 & -1 & 1 \ 1 & 0 & 0 \ 1 & 2 & 4 \ \end{pmatrix} \begin{pmatrix} c_0 \ c_1 \ c_2 \end{pmatrix} = \begin{pmatrix} 2 \ 4 \ 7 \end{pmatrix} ]
Conclusion: The columns of A are independent, leading to calculations yielding:
Ca = [ (A^TA)^{-1}A^Tb = egin{pmatrix} -0.75 \ -0.15 \ 2.95 \end{pmatrix} ]
Least Squares Quadratic Fit: [ y = -0.75x^2 - 0.15x + 2.95 ]
Projection Matrices
Definition: A projection matrix [ P ] projects vector [ b \in \mathbb{R}^m ] onto a subspace of [ \mathbb{R}^n ].
For least squares solution:
[ P = A(A^TA)^{-1}A^T ] which minimizes [ ||b - p|| ] over all [ p \in C_A ].
Properties of Projection Matrices:
Symmetric: [ P^T = P ]
Idempotent: [ P^2 = P ]
Special Cases
If [ m = n ] with rank(A) = n, then [ P = I ], where [ b ] is in the range and [ Pb = b ].
Exact Solution:
When [ A ] is invertible, the exact solution is given by: [ E = (A^TA)^{-1}A^Tb = A^T(A^TA)^{-1}b = A^T ]