Question 4.4.2: Let T: R² → R³ be the linear transformation defined by T (v)...
Let T: R² → R³ be the linear transformation defined by
T (v) = T\left(\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} \right) = \begin{bmatrix} x_{2} \\x_{1} + x_{2} \\ x_{1}-x_{2} \end{bmatrix}
and let
B= \left\{\begin{bmatrix}1 \\ 2 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \end{bmatrix} \right\} B^{′}=\left\{\begin{bmatrix}1 \\ 0\\0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1\\0 \end{bmatrix}, \begin{bmatrix}1 \\ 1\\1 \end{bmatrix} \right\}
be ordered bases for R² and R³,respectively.
a. Find the matrix \left[T\right] ^{B^{\prime } }_{B}.
b. Let v =\begin{bmatrix} -3 \\ -2 \end{bmatrix}. Find T (v) directly and then use the matrix found in part (a).
a. We first apply T to the basis vectors of B, which gives
T\left(\begin{bmatrix} 1 \\ 2 \end{bmatrix} \right)= \begin{bmatrix} 2 \\ 3\\-1 \end{bmatrix} and T\left(\begin{bmatrix} 3 \\ 1 \end{bmatrix} \right)= \begin{bmatrix} 1 \\ 4\\2 \end{bmatrix}
Next we find the coordinates of each of these vectors relative to the basis B^{′}.
That is, we find scalars such that
a_{1}\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} +a_{2} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} +a_{3} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}= \begin{bmatrix} 2 \\ 3\\ -1 \end{bmatrix}
and
b_{1}\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} +b_{2} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} +b_{3} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}= \begin{bmatrix} 1 \\ 4\\ 2 \end{bmatrix}
The solution to the first linear system is
a_{1} = −1 a_{2} = 4 a_{3} = −1
and the solution to the second system is
b_{1} = −3 b_{2} = 2 b_{3} = 2
Thus,
\left[T\right] ^{B^{\prime } }_{B} = \begin{bmatrix} -1&-3 \\ 4&2\\ -1&2 \end{bmatrix}
b. Using the definition of T directly, we have
T\left(\begin{bmatrix} -3 \\ -2 \end{bmatrix} \right) = \begin{bmatrix} -2\\ -3-2\\ −3 + 2 \end{bmatrix}= \begin{bmatrix} -2\\ -5\\ -1 \end{bmatrix}
Now, to use the matrix found in part (a), we need to find the coordinates of v relative to B. Observe that the solution to the equation
a_{1}\begin{bmatrix} 1\\ 2 \end{bmatrix} +a_{2} \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} -3 \\ -2 \end{bmatrix} is a_{1}= – \frac{3}{5} a_{2}=-\frac{4}{5}
Thus, the coordinate vector of \begin{bmatrix} -3 \\ -2 \end{bmatrix}. relative to B is
\begin{bmatrix} -3 \\-2 \end{bmatrix} _{B} = \begin{bmatrix} – \frac{3}{5} \\- \frac{4}{5} \end{bmatrix}
We can now evaluate T, using matrix multiplication, so that
\left[T(v)\right] _{B^{\prime } }=\begin{bmatrix} -1&-3 \\ 4&2 \\ -1&2 \end{bmatrix} \begin{bmatrix} – \frac{3}{5} \\- \frac{4}{5} \end{bmatrix} = \begin{bmatrix} 3 \\ -4 \\ -1 \end{bmatrix}
Hence,
T (v) = 3 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} -4\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} -\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -2 \\ -5 \\ -1 \end{bmatrix}
which agrees with the direct computation.