Computing via linear dependence

in $ℝ^{4}$

$ℝ^{4}$

φ ((\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix})) = (\begin{matrix} 16 x_{1} + 5 x_{2} - 26 x_{3} - 12 x_{4} \\ - 10 x_{1} - 3 x_{2} + 16 x_{3} + 8 x_{4} \\ 3 x_{1} - x_{2} - 7 x_{3} - 4 x_{4} \\ 7 x_{1} + 3 x_{2} - 10 x_{3} - 5 x_{4} \end{matrix})

$ℝ^{4}$

φ ((\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix})) = (\begin{matrix} 16 x_{1} + 5 x_{2} - 26 x_{3} - 12 x_{4} \\ - 10 x_{1} - 3 x_{2} + 16 x_{3} + 8 x_{4} \\ 3 x_{1} - x_{2} - 7 x_{3} - 4 x_{4} \\ 7 x_{1} + 3 x_{2} - 10 x_{3} - 5 x_{4} \end{matrix})

φ ((\begin{matrix} 1 \\ 1 \\ 1 \\ -1 \end{matrix})) = (\begin{matrix} 7 \\ -5 \\ -1 \\ 5 \end{matrix}) φ ((\begin{matrix} 5 \\ -3 \\ 1 \\ 2 \end{matrix})) = (\begin{matrix} 15 \\ -9 \\ 3 \\ 6 \end{matrix}) = 3 (\begin{matrix} 5 \\ -3 \\ 1 \\ 2 \end{matrix})

$ℝ^{4}$

φ ((\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix})) = (\begin{matrix} 16 x_{1} + 5 x_{2} - 26 x_{3} - 12 x_{4} \\ - 10 x_{1} - 3 x_{2} + 16 x_{3} + 8 x_{4} \\ 3 x_{1} - x_{2} - 7 x_{3} - 4 x_{4} \\ 7 x_{1} + 3 x_{2} - 10 x_{3} - 5 x_{4} \end{matrix})

\overset{\overset{{\vec{𝒖}}_{0}}{⏞}}{(\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix})} \overset{\overset{{\vec{𝒖}}_{1} = φ ({\vec{𝒖}}_{0})}{⏞}}{(\begin{matrix} 4 \\ -2 \\ -1 \\ 2 \end{matrix})} \overset{\overset{{\vec{𝒖}}_{2} = φ ({\vec{𝒖}}_{1})}{⏞}}{(\begin{matrix} 56 \\ -34 \\ 13 \\ 22 \end{matrix})} \overset{\overset{{\vec{𝒖}}_{3} = φ ({\vec{𝒖}}_{2})}{⏞}}{(\begin{matrix} 124 \\ -74 \\ 23 \\ 50 \end{matrix})} \overset{\overset{{\vec{𝒖}}_{4} = φ ({\vec{𝒖}}_{3})}{⏞}}{(\begin{matrix} 416 \\ -250 \\ 85 \\ 166 \end{matrix})}

$ℝ^{4}$

(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}) = 0 (\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix}) - 3 (\begin{matrix} 4 \\ -2 \\ -1 \\ 2 \end{matrix}) - 2 (\begin{matrix} 56 \\ -34 \\ 13 \\ 22 \end{matrix}) + (\begin{matrix} 124 \\ -74 \\ 23 \\ 50 \end{matrix})

construct

\begin{matrix} p (x) & = & 0 - 3 x - 2 x^{2} + x^{3} \\ = & (x - 0) (x - 3) (x + 1) \end{matrix}

$ℝ^{4}$

\vec{𝒐} = (φ - 0 𝒊𝒅) \overset{\overset{{\vec{𝒛}}_{2}}{⏞}}{(φ - 3 𝒊𝒅) \underset{\underset{{\vec{𝒛}}_{1}}{⏟}}{(φ + 1 𝒊𝒅) \vec{𝒖}}}

compute

\overset{\overset{{\vec{𝒛}}_{0}}{⏞}}{(\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix})} \overset{\overset{{\vec{𝒛}}_{1} = φ ({\vec{𝒛}}_{0}) + 𝒊𝒅 ({\vec{𝒛}}_{0})}{⏞}}{(\begin{matrix} 5 \\ -2 \\ -1 \\ 3 \end{matrix}) = (\begin{matrix} 4 \\ -2 \\ -1 \\ 2 \end{matrix}) + (\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix})} \overset{\overset{{\vec{𝒛}}_{2} = φ ({\vec{𝒛}}_{1}) - 3 𝒊𝒅 ({\vec{𝒛}}_{1})}{⏞}}{(\begin{matrix} 45 \\ -30 \\ 15 \\ 15 \end{matrix}) = (\begin{matrix} 60 \\ -36 \\ 12 \\ 24 \end{matrix}) - 3 (\begin{matrix} 5 \\ -2 \\ -1 \\ 3 \end{matrix})}

$ℝ^{4}$

\vec{𝒐} = (φ + 1 𝒊𝒅) \overset{\overset{{\vec{𝒛}}_{2}}{⏞}}{(φ - 3 𝒊𝒅) \underset{\underset{{\vec{𝒛}}_{1}}{⏟}}{(φ - 0 𝒊𝒅) \vec{𝒖}}}

compute

\overset{\overset{{\vec{𝒛}}_{0}}{⏞}}{(\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix})} \overset{\overset{{\vec{𝒛}}_{1} = φ ({\vec{𝒛}}_{0}) - 0 𝒊𝒅 ({\vec{𝒛}}_{0})}{⏞}}{(\begin{matrix} 4 \\ -2 \\ -1 \\ 2 \end{matrix}) = (\begin{matrix} 4 \\ -2 \\ -1 \\ 2 \end{matrix}) - 0 (\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix})} \overset{\overset{{\vec{𝒛}}_{2} = φ ({\vec{𝒛}}_{1}) - 3 𝒊𝒅 ({\vec{𝒛}}_{1})}{⏞}}{(\begin{matrix} 44 \\ -28 \\ 16 \\ 16 \end{matrix}) = (\begin{matrix} 56 \\ -34 \\ 13 \\ 22 \end{matrix}) - 3 (\begin{matrix} 4 \\ -2 \\ -1 \\ 2 \end{matrix})}

Computing via linear dependence

in $ℝ^{3}$

$ℝ^{3}$

L = (\begin{matrix} 7 & -6 & -4 \\ 3 & -2 & -2 \\ 3 & -3 & -1 \end{matrix})

\overset{\overset{{\vec{𝒖}}_{0}}{⏞}}{(\begin{matrix} 1 \\ 2 \\ 0 \end{matrix})} \overset{\overset{{\vec{𝒖}}_{1} = L {\vec{𝒖}}_{0}}{⏞}}{(\begin{matrix} -5 \\ -1 \\ -3 \end{matrix})} \overset{\overset{{\vec{𝒖}}_{2} = L {\vec{𝒖}}_{1}}{⏞}}{(\begin{matrix} -17 \\ -7 \\ -9 \end{matrix})} \overset{\overset{{\vec{𝒖}}_{3} = L {\vec{𝒖}}_{2}}{⏞}}{(\begin{matrix} -41 \\ -19 \\ -21 \end{matrix})}

$ℝ^{3}$

(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) = 1 (\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}) + \frac{-3}{2} (\begin{matrix} -5 \\ -1 \\ -3 \end{matrix}) + \frac{1}{2} (\begin{matrix} -17 \\ -7 \\ -9 \end{matrix})

construct

\begin{matrix} p (x) & = & 2 - 3 x + x^{3} \\ = & (x - 2) (x - 1) \end{matrix}

$ℝ^{3}$

\vec{𝒐} = \overset{\overset{{\vec{𝒛}}_{2}}{⏞}}{(φ - 2 𝒊𝒅) \underset{\underset{{\vec{𝒛}}_{1}}{⏟}}{(φ - 1 𝒊𝒅) \vec{𝒖}}}

compute

\overset{\overset{{\vec{𝒛}}_{0}}{⏞}}{(\begin{matrix} 1 \\ 2 \\ 0 \end{matrix})} \overset{\overset{{\vec{𝒛}}_{1} = φ ({\vec{𝒛}}_{0}) + 𝒊𝒅 ({\vec{𝒛}}_{0})}{⏞}}{(\begin{matrix} -6 \\ -3 \\ -3 \end{matrix}) = (\begin{matrix} -5 \\ -1 \\ -3 \end{matrix}) - 1 (\begin{matrix} 1 \\ 2 \\ 0 \end{matrix})} \overset{\overset{{\vec{𝒛}}_{2} = φ ({\vec{𝒛}}_{1}) - 2 𝒊𝒅 ({\vec{𝒛}}_{1})}{⏞}}{(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) = (\begin{matrix} -12 \\ -6 \\ -6 \end{matrix}) - 2 (\begin{matrix} -6 \\ -3 \\ -3 \end{matrix})}