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Revision as of 05:28, 26 January 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} $
