231 lines
4.1 KiB
Markdown
231 lines
4.1 KiB
Markdown
# Example 12
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## Kallman Full Decomposition
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$$
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\begin{align*}
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A &= \begin{bmatrix}
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-4 & -3 & 0 & -2 \\
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6 & 5 & 0 & 2 \\
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4 & 1 & 1 & -6 \\
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-1 & -1 & 0 & -3 \\
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\end{bmatrix}\\
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B &= \begin{bmatrix}
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-1 \\
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1 \\
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2 \\
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0 \\
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\end{bmatrix}\\
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C &= \begin{bmatrix}
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-3 & -2 & 0 & 1 \\
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\end{bmatrix}\\
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D &= \begin{bmatrix}
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0
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\end{bmatrix}\\
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K_r &= \begin{bmatrix}
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-1 & 1 & -1 & 1 \\
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1 & -1 & 1 & -1 \\
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2 & -1 & 2 & -1\\
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0 & 0 & 0 & 0
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\end{bmatrix}\\
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K_{no} &= \begin{bmatrix}
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-3 &-2 &0 &1\\
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-1 &-2 &0 &-1\\
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-7 &-6 &0 &1\\
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-9 &-10& 0 &-1\\
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\end{bmatrix}\\
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\end{align*}\\
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\text{To find all the bases, for $P_{K_r}$ you should find the}\\
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\text{independent columns and find the system}\\
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\text{For $P_{K_{no}}$ you should find the system by looking at}\\
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\text{rows, solve it and then find some bases}\\
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\begin{align*}
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P_{K_r} &= \begin{bmatrix}
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0&1\\
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0&-1\\
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1&0\\
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0&0
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\end{bmatrix}\\
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X_r &= \begin{cases}
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x_1 =-x_2 \\
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x_4 = 0
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\end{cases}\\
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X_{no} &= \begin{cases}
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-3x_1 - 2x_2 + x_4 = 0 \\
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-x_1 -2x_2 -x_4 = 0
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\end{cases} \rightarrow \\
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&\rightarrow \begin{cases}
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x_1 = x_4 \\
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x_1 = -x_2
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\end{cases} \\
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P_{K_{no}} &= \begin{bmatrix}
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1 & 0\\
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-1 & 0\\
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0 & 1 \\
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1 & 0
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\end{bmatrix}
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\end{align*} \\
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\text{Now, let's get all $X_r$ and $X_{no}$ elements}\\
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\begin{align*}
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X_r &= \left\{
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\alpha \begin{bmatrix}
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0\\
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0\\
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1\\
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0
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\end{bmatrix}
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+
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\beta \begin{bmatrix}
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1\\
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-1\\
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0\\
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0
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\end{bmatrix}
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\right\}\\
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X_{no} &= \left\{
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\alpha \begin{bmatrix}
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1 \\
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-1\\
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0 \\
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1
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\end{bmatrix}
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+
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\beta \begin{bmatrix}
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0\\
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0\\
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1 \\
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0
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\end{bmatrix}
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\right\}
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\end{align*}\\
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$$
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Now, let's get $X_1$, $X_2$, $X_3$, $X_4$
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$$
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\begin{align*}
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X_1 &= X_r \cap X_{no} = \begin{cases}
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x_1 =-x_2 \\
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x_4 = 0 \\
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x_1 = x_4
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\end{cases}\\
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X_1 &= \begin{bmatrix}
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0 \\ 0 \\ x_3 \\ 0
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\end{bmatrix}\\
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0 &= X_1^T X_1^\perp \rightarrow \\
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&\rightarrow \begin{bmatrix}
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0 & 0 & x_3 & 0
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\end{bmatrix}
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\begin{bmatrix}
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x_a \\ x_b \\ x_c \\ x_d
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\end{bmatrix} = 0 \rightarrow \\
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&\rightarrow\begin{cases}
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x_c = 0
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\end{cases} \\
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X_1^\perp &= \begin{bmatrix}
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x_a \\ x_b \\ 0 \\ x_d
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\end{bmatrix}\\
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X_2 &= X_r \cap X_1^\perp = \begin{cases}
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x_1 = -x_2 \\
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x_3 = 0 \\
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x_4 = 0 \\
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\end{cases}\\
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X_2 &= \begin{bmatrix}
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x_1 \\ -x_1 \\ 0 \\ 0
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\end{bmatrix}\\
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X_3 &= X_{no} \cap X_1^\perp = \begin{cases}
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x_1 = -x_2 \\
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x_1 = x_4 \\
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x_3 = 0 \\
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\end{cases}\\
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X_3 &= \begin{bmatrix}
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x_1 \\ -x_1 \\ 0 \\ x_1
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\end{bmatrix}\\
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0 &= X_2^T X_2^\perp =
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\begin{bmatrix}
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1 & -1 & 0 & 0
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\end{bmatrix}
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\begin{bmatrix}
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x_a \\ x_b \\ x_c \\ x_d
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\end{bmatrix}\rightarrow \\
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&\rightarrow\begin{cases}
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x_a - x_b = 0
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\end{cases} \\
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\\
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0 &= X_3^T X_3^\perp =
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\begin{bmatrix}
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1 & -1 & 0 & 1
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\end{bmatrix}
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\begin{bmatrix}
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x_a \\ x_b \\ x_c \\ x_d
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\end{bmatrix}\rightarrow \\
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&\rightarrow\begin{cases}
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x_a - x_b + x_d = 0
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\end{cases} \\
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\\
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X_4 &= X_1^\perp \cap X_2^\perp \cap X_3^\perp = \\
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&= \begin{cases}
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x_3 = 0 \\
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x_1 = x_2 \\
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x_4 = 0\\
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\end{cases} = \\
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&= \begin{bmatrix}
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x_1 \\ x_2 \\ 0 \\ 0
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\end{bmatrix}
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\end{align*}
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$$
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Now we find the change state matrix:
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$$
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\begin{align*}
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Q^{-1} &= \begin{bmatrix}
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X_1 & X_2 & X_3 & X_4
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\end{bmatrix} = \\
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&= \begin{bmatrix}
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0 & 1 & 1 & 1\\
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0 & -1 & -1 & 1\\
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1 & 0&0&0\\
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0&0&1&0
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\end{bmatrix}
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\end{align*}
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$$
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Compute:
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$$
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\begin{align*}
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\hat{A} &= QAQ^{-1}\\
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\hat{B} &= QB \\
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\hat{C} &= CQ^{-1}
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\end{align*}
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$$
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