Question

Consider the bases B = left(begin{array}{c}begin{bmatrix}2

Alternate coordinate systems
ANSWERED
asked 2020-11-14

Consider the bases \(B = \left(\begin{array}{c}\begin{bmatrix}2 \\ 3 \end{bmatrix}, \begin{bmatrix}3 \\ 5 \end{bmatrix}\end{array}\right) of R^2 \ and\ C = \left(\begin{array}{c}\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix}1\\0 \\ 1 \end{bmatrix}\end{array}, \begin{bmatrix}0 \\ 1\\1 \end{bmatrix}\right) of R^3\).
and the linear maps \(S \in L (R^2, R^3) \ and\ T \in L(R^3, R^2)\) given given (with respect to the standard bases) by \([S]_{E, E} = \begin{bmatrix}2 & -1 \\ 5 & -3\\ -3 & 2 \end{bmatrix} \ and\ [T]_{E, E} = \begin{bmatrix}1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}\) Find each of the following coordinate representations. \(\displaystyle{\left({b}\right)}{\left[{S}\right]}_{{{E},{C}}}\)
\(\displaystyle{\left({c}\right)}{\left[{S}\right]}_{{{B},{C}}}\)

Answers (1)

2020-11-15

\(B = \left(\begin{array}{c}\begin{bmatrix}2 \\ 3 \end{bmatrix}, \begin{bmatrix}3 \\ 5 \end{bmatrix}\end{array}\right) of R^2, C = \left(\begin{array}{c}\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix}1\\0 \\ 1 \end{bmatrix}\end{array}, \begin{bmatrix}0 \\ 1\\1 \end{bmatrix}\right) of R^3\).

\(S \in L (R^2, R^3) , T \in L(R^3, R^2)\)
\([S]_{E, E} = \begin{bmatrix}2 & -1 \\ 5 & -3\\ -3 & 2 \end{bmatrix} , [T]_{E, E} = \begin{bmatrix}1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}\)

(b) To find \(\displaystyle{\left[{S}\right]}_{{{E},{C}}}:\)
\(S(\begin{bmatrix}1 \\ 0 \end{bmatrix}) = \begin{bmatrix}2 \\ 5\\ -3 \end{bmatrix} = a\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix} + b\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} + c\begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix}a + b \\ a + c \\ b + c \end{bmatrix}\)
\(\displaystyle\therefore{a}+{b}={2},{a}+{c}={5},{b}+{c}=-{3}\)
\(\displaystyle{b}={2}-{a}\Rightarrow{b}+{c}={2}-{a}+{c}=-{3}\Rightarrow-{a}+{c}=-{5}\)
\(\begin{array}{cc} \therefore a + c = 5 \\ + - a + c = -5 \\ \hline 2c = 0 \end{array} \begin{array}{cc} \Rightarrow c = 0, b = -3, a = 5 \\ \Rightarrow a = 5, b = -3, c = 0 \end{array}\)
\(S(\begin{bmatrix}0 \\ 1 \end{bmatrix}) = \begin{bmatrix}-1 \\ -3 \\ 2 \end{bmatrix} = a \begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix} + b\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} + c \begin{bmatrix} 0 \\ 1 \\ 1\end{bmatrix} = \begin{bmatrix}a + b \\ a + c \\ b + c \end{bmatrix}\)
\(\displaystyle\therefore{a}+{b}=-{1},{a}+{c}=-{3},{b}+{c}={2}\)
\(\displaystyle\therefore{b}=-{1}-{a}\Rightarrow{b}+{c}=-{1}-{a}+{c}={2}\Rightarrow-{a}+{c}={3}\)
\(\begin{array}{cc} \therefore a + c = -3 \\ + - a + c = 3 \\ \hline 2c = 0 \end{array} \begin{array}{cc} \Rightarrow c = 0, a = -3, b = 2 \\ \Rightarrow a = -3, b = 2, c = 0 \end{array}\)
\(\therefore [S]_{E, C} = \begin{bmatrix}5 & -3 \\ -3 & 2 \\ 0 & 0 \end{bmatrix}\)

c) To find \(\displaystyle{\left[{S}\right]}_{{{B},{C}}}\)
\(S(\begin{bmatrix}2 \\ 3 \end{bmatrix}) = \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} and\ S(\begin{bmatrix}3 \\ 5 \end{bmatrix}) = \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} .. (by (a))\)
\(\therefore S(\begin{bmatrix}2 \\ 3 \end{bmatrix}) = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = 1\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix} + 0 \begin{bmatrix}1 \\0 \\ 1 \end{bmatrix} + 0 \begin{bmatrix}0\\ 1 \\ 1 \end{bmatrix}\)
\(S(\begin{bmatrix}3 \\ 5 \end{bmatrix}) = \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} = 0 \begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix} + 1 \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} + 0 \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}\)
\([S]_{B, C} = \begin{bmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}\)

0
 
Best answer

expert advice

Need a better answer?

Relevant Questions

...