Answer
The image of $(1,1)$ is $(0,2,1)$ and the preimage of $w$ is $(-6,4)$
Work Step by Step
We are given
$T\left[ {\begin{array}{*{20}{c}} {{v_1}}\\ {{v_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{\sqrt 2 }}{2}{v_1} - \frac{{\sqrt 2 }}{2}{v_2}}\\ {{v_1} + {v_2}}\\ {2{v_1} - {v_2}} \end{array}} \right]$
a) The image of $v=(1,1)$ is
$\begin{array}{l}
T\left[ {\begin{array}{*{20}{c}}
{{v_1}}\\
{{v_2}}
\end{array}} \right]\\
= T\left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right]\\
= \left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 2 }}{2}(1) - \frac{{\sqrt 2 }}{2}(1)}\\
{1 + 1}\\
{2(1) - 1}
\end{array}} \right]\\
= \left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}}\\
{1 + 1}\\
{2 - 1}
\end{array}} \right]\\
= \left[ {\begin{array}{*{20}{c}}
0\\
2\\
1
\end{array}} \right]
\end{array}$
b) Let the preimage of $w=(-5\sqrt{2},-2,-16)$ be $(a,b)$.
$\begin{array}{l}
\Rightarrow T\left[ {\begin{array}{*{20}{c}}
a\\
b
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ - 5\sqrt 2 }\\
{ - 2}\\
{ - 16}
\end{array}} \right]\\
\Rightarrow \left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 2 }}{2}a - \frac{{\sqrt 2 }}{2}b}\\
{a + b}\\
{2a - b}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ - 5\sqrt 2 }\\
{ - 2}\\
{ - 16}
\end{array}} \right]
\end{array}$
Comparing each component of both vectors, we have
$\begin{array}{l}
\frac{{\sqrt 2 }}{2}a - \frac{{\sqrt 2 }}{2}b = - 5\sqrt 2, \quad (1) \\
a + b = - 2, \quad (2)\\
2a - b = - 16, \quad (3)
\end{array}$
Adding equation $(2)$ and equation $(3)$, we get,
$\begin{array}{l}
(a + b = - 2) + (2a - b = - 16)\\
\Rightarrow 3a = - 18\\
\Rightarrow a = - 6
\end{array}$
Putting this in equation $(2)$ we get, $b=-2-a=-2-(-6)=4$
Therefore, the preimage of $w$ is $(-6,4)$
