# Consider the linear transformation U: R^3 rightarrow R^3

Consider the linear transformation $U:{R}^{3}\to {R}^{3}$ defined by $U\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}z-y\\ z+y\\ 3z-x-y\end{array}\right)$ and the bases $ϵ=\left\{\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\},\gamma =\left\{\left(\begin{array}{c}1-i\\ 1+i\\ 1\end{array}\right),\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\}$, Compute the four coordinate matrices ${\left[U\right]}_{ϵ}^{\gamma },{\left[U\right]}_{\gamma }^{\gamma },$

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

irwchh

$U\left(\begin{array}{c}1-i\\ 1+i\\ 1\end{array}\right)=\left(\begin{array}{c}-i\\ 2+i\\ 1\end{array}\right)=1\cdot {v}_{1}+1\cdot {v}_{2}+0\cdot {v}_{3}$
$U\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)=0\cdot {v}_{1}+1\cdot {v}_{2}+0\cdot {v}_{3}$
$U\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 2\\ 2\end{array}\right)=0\cdot {v}_{1}+0\cdot {v}_{2}+2\cdot {v}_{3}$ So, $\left[U{\right]}_{\gamma }^{\gamma }=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 0& 2\end{array}\right]$
$U\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right).Since\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\text{⧸}\in \left\{\gamma \right\}$ So, $\left[U{\right]}_{ϵ}^{\gamma }$ does not exist $U\left(\begin{array}{c}1-i\\ 1+i\\ 1\end{array}\right)={i}_{1}\cdot {ϵ}_{1}+\left(2+i\right)\cdot {ϵ}_{2}+1\cdot {ϵ}_{2}$
$U\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)=-1\cdot {ϵ}_{1}+1\cdot {ϵ}_{2}+0\cdot {ϵ}_{3}$
$U\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)=0\cdot {ϵ}_{1}+2\cdot {ϵ}_{2}+1\cdot {ϵ}_{3}$ So, $\left[U{\right]}_{ϵ}^{\gamma }=\left[\begin{array}{ccc}-i& -1& 0\\ 2+i& 1& 2\\ 1& 0& 1\end{array}\right]$