Okay this may be a bit long. Only the last question is an issue for me but I can't verify whether the earlier questions (which affect the last) are done correctly.
Given
\begin{eqnarray*}
T(x,y,z) = (x+2y-z,x+z,4x-4y+5z)
\end{eqnarray*}
for $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$.
1) Find the matrix of $T$ with respect to the standard basis $\{(1,0,0),(0,1,0),(0,0,1)\}$.
My answer:
\begin{eqnarray*}
(x,2y,-z) &=& x(1,0,0)+ 2y(0,1,0) -z(0,0,1)\\
(x,0,z) &=& x(1,0,0)+ 0(0,1,0) +z(0,0,1)\\
(4x,-4y,5z) &=& 4x(1,0,0)-4y(0,1,0) +5z(0,0,1)\\
\end{eqnarray*}
$T = \left(
\begin{matrix}
1&2&-1\\
1&0&1\\
4&-4&5
\end{matrix}
\right).$
2) Find the matrix $A$ of $T$ with respect to the ordered basis $e = \{(-1,1,2),(-2,1,4),(-1,1,4)\}$.
My answer:
\begin{eqnarray*}
(x,2y,-z) &=& a_1(-x,y,2z)+ b_1(-2x,y,4z) +c_1 (-x,y,4z)\\
(x,0,z) &=& a_2(-x,y,2z)+ b_2(-2x,y,4z) + c_2(-x,y,4z)\\
(4x,-4y,5z) &=& a_3(-x,y,2z)+ b_3(-2x,y,4z) +c_3(-x,y,4z).
\end{eqnarray*}
$A =
\left(
\begin{matrix}
a_1&b_1&c_1\\
a_2&b_2&c_2\\
a_3&b_3&c_3
\end{matrix}
\right)=
\left(
\begin{matrix}
4.5&-3&0.5\\
-0.5&-1&1.5\\
-10.5&0&6.5
\end{matrix}
\right).$
3) Find the coordinate vector $[v]_e$ when $v = (1,0,2)$.
My answer:
\begin{eqnarray*}
(1,0,2) &=& a(-1,1,2) + b(-2,1,4)+ c(-1,1,4)
\end{eqnarray*}
$[v]_e =
\left[
\begin{matrix}
a\\b\\c
\end{matrix}
\right]
=
\left[
\begin{matrix}
-1\\-1\\2
\end{matrix}
\right].$
4) Find $T(v)$ and $[T(v)]_e$.
\begin{eqnarray*}
T(v) &=&
\left(
\begin{matrix}
1&2&-1\\
1&0&1\\
4&-4&5
\end{matrix}
\right)
\left(
\begin{matrix}
1\\0\\2
\end{matrix}
\right)\\
&=&
\left(
\begin{matrix}
-1\\3\\4
\end{matrix}
\right).\\
\text{[}T(v)\text{]}_e
&=&
\left(
\begin{matrix}
1&2&-1\\
1&0&1\\
4&-4&5
\end{matrix}
\right)
\left(
\begin{matrix}
-1\\-1\\2
\end{matrix}
\right)\\
&=&
\left(
\begin{matrix}
-5\\1\\10
\end{matrix}
\right).
\end{eqnarray*}
5. Verify that $[T(v)]_e = A[v]_e:$
\begin{eqnarray*}
A[v]_e &=&
\left(
\begin{matrix}
4.5&-3&0.5\\
-0.5&-1&1.5\\
-10.5&0&6.5
\end{matrix}
\right)
\left(
\begin{matrix}
-1\\-1\\2
\end{matrix}
\right)\\
&=&
\left(
\begin{matrix}
-0.5\\4.5\\23.5
\end{matrix}
\right)\\
&\neq& [T(v)]_e
\end{eqnarray*}
I think I misunderstood the difference between $T[v]_e$ and $[T(v)]_e$ (I assumed they're the same) but I don't know what is the difference.
Best Answer
$$T(v)=Tv=[Tv]_e^T\left(\begin{array}{c} -1 & 1& 2\\ -2 & 1& 4\\ -1 & 1& 4\\ \end{array}\right)$$ $$Tv=A[v]_e^T\left(\begin{array}{c} -1 & 1& 2\\ -2 & 1& 4\\ -1 & 1& 4\\ \end{array}\right)$$ I suggest you read the chapter of linear thranformation about any linear algebra textbook.