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| + | [[Category:ECE]] | ||
| + | [[Category:QE]] | ||
| + | [[Category:CNSIP]] | ||
| + | [[Category:problem solving]] | ||
| + | [[Category:automatic control]] | ||
| + | [[Category:optimization]] | ||
=QE2012_AC-3_ECE580-3= | =QE2012_AC-3_ECE580-3= | ||
| − | + | :[[QE2012_AC-3_ECE580-1|Part 1]],[[QE2012_AC-3_ECE580-2|2]],[[QE2012_AC-3_ECE580-3|3]],[[QE2012_AC-3_ECE580-4|4]],[[QE2012_AC-3_ECE580-5|5]] | |
| Line 139: | Line 145: | ||
<br> | <br> | ||
| + | |||
| + | ---- | ||
| + | ---- | ||
| + | <font face="serif"></font><math>\color{blue}\text{Related Problem: }</math> | ||
| + | |||
| + | <math> | ||
| + | \text{Given the matrix A and vector b:} | ||
| + | A=\begin{bmatrix} | ||
| + | 2 & 1 \\ | ||
| + | 3 & 1 \\ | ||
| + | 4 & 1 | ||
| + | \end{bmatrix} | ||
| + | b=\begin{bmatrix} | ||
| + | 3\\ | ||
| + | 4 \\ | ||
| + | 15 | ||
| + | \end{bmatrix} | ||
| + | </math> | ||
| + | |||
| + | |||
| + | '''Find the vector <math>x^{(\ast)}</math> that minimizes <math>\| Ax -b\|^{2}_2</math> ''' | ||
| + | |||
| + | |||
| + | '''Solution:''' | ||
| + | |||
| + | <math> | ||
| + | x^{(\ast)}=A^{\dagger}=(A^T A)^{-1}A^T b= \begin{bmatrix} | ||
| + | -0.5 & 0 & 0.5 \\ | ||
| + | 11/6 & 1/3 & 7/6 \\ | ||
| + | \end{bmatrix} | ||
| + | \begin{bmatrix} | ||
| + | 3\\ | ||
| + | 4 \\ | ||
| + | 15 | ||
| + | \end{bmatrix}=\begin{bmatrix} | ||
| + | 6 \\ | ||
| + | -\frac{32}{3} \\ | ||
| + | 0 | ||
| + | \end{bmatrix} | ||
| + | </math> | ||
Latest revision as of 09:12, 13 September 2013
QE2012_AC-3_ECE580-3
Solutions:
$ A = BC = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 &-1 \\ 0 & 1 & 0 \end{bmatrix} $
$ B^{\dagger} = (B^T B)^{-1}B^T = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}^{-1} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} \end{bmatrix} $
$ C^{\dagger} = C^T(CC^T)^{-1} =\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \\ -\frac{1}{2} & 0 \end{bmatrix} $
$ A^{\dagger} = C^{\dagger}B^{\dagger} =\begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \\ -\frac{1}{2} & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & 0 & 0 \end{bmatrix} $
$ x^{\ast} = A^{\dagger} b = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & 0 & 0 \end{bmatrix}\begin{bmatrix} 0 \\ \frac{1}{2} \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{2} \\ 0 \end{bmatrix} $
Solution 2:
$ x^{(\ast)}=A^{\dagger}b $
Since $ A = BC = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 &-1 \\ 0 & 1 & 0 \end{bmatrix} $
$ B^{\dagger} = (B^T B)^{-1}B^T = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}^{-1} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} \end{bmatrix} $
$ C^{\dagger} = C^T(CC^T)^{-1} =\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \\ -\frac{1}{2} & 0 \end{bmatrix} $
$ A^{\dagger} = C^{\dagger}B^{\dagger} =\begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \\ -\frac{1}{2} & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & 0 & 0 \end{bmatrix} $
$ x^{\ast} = A^{\dagger} b = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & 0 & 0 \end{bmatrix}\begin{bmatrix} 0 \\ \frac{1}{2} \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{2} \\ 0 \end{bmatrix} $
$ \color{blue} \text{ The pseudo inverse of a matrix has the property } (BC)^{\dagger}=C^{\dagger}B^{\dagger} $
$ \color{blue}\text{Related Problem: } $
$ \text{Given the matrix A and vector b:} A=\begin{bmatrix} 2 & 1 \\ 3 & 1 \\ 4 & 1 \end{bmatrix} b=\begin{bmatrix} 3\\ 4 \\ 15 \end{bmatrix} $
Find the vector $ x^{(\ast)} $ that minimizes $ \| Ax -b\|^{2}_2 $
Solution:
$ x^{(\ast)}=A^{\dagger}=(A^T A)^{-1}A^T b= \begin{bmatrix} -0.5 & 0 & 0.5 \\ 11/6 & 1/3 & 7/6 \\ \end{bmatrix} \begin{bmatrix} 3\\ 4 \\ 15 \end{bmatrix}=\begin{bmatrix} 6 \\ -\frac{32}{3} \\ 0 \end{bmatrix} $
