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\)
Section 2.2 Matrix transformations
We consider functions whose inputs are vectors and whose outputs are vectors defined through matrix-vector multiplication. That is, if \(A\) is a matrix and \(\xvec\) is a vector, the function \(T(\xvec) = A\xvec\) forms the product \(A\xvec\) as its output. Such a function is called a matrix transformation.
Proposition 2.2.1 . Linearity of matrix vector multiplication.
If \(A\) is a matrix, \(\vvec\) and \(\wvec\) vectors of the appropriate dimensions, and \(c\) a scalar, then
\(A\zerovec = \zerovec\text{.}\)
\(A(c\vvec) = cA\vvec\text{.}\)
\(A(\vvec+\wvec) = A\vvec + A\wvec\text{.}\)
Activity 2.2.1 .
In this activity, we will look at an example of matrix transformation \(T(\xvec) = A\xvec\) that takes two dimensional vectors to two dimensional vectors. \(A\) will necessarily be a \(2\times 2\) matrix. Furthermore since every two dimensional vector can be written as a linear combination of \(\hat{i}=\twovec{1}{0}\) and \(\hat{j}=\twovec{0}{1}\text{,}\) we let \(A\) be the matrix whose columns are \(T(\hat{i})\) and \(T(\hat{j})\text{.}\)
\begin{equation*}
A=\left[\begin{array}{rr}
T(\hat{i}) \amp T(\hat{j})
\end{array}\right]
\end{equation*}
To begin, suppose that \(A\) is the matrix
\begin{equation*}
A = \left[\begin{array}{rr}
2 \amp 1 \\
1 \amp 2 \\
\end{array}\right].
\end{equation*}
with associated matrix transformation \(T(\xvec) =
A\xvec\text{.}\)
What is \(T\left(\twovec{1}{-2}\right)\text{?}\)
What is \(T\left(\twovec{1}{0}\right)\text{?}\)
What is \(T\left(\twovec{0}{1}\right)\text{?}\)
Is there a vector \(\xvec\) such that \(T(\xvec) = \twovec{3}{0}\text{?}\) Hint: Use the interactive diagram in Activity 1.2 to answer this question.
Write \(T\left(\twovec xy\right)\) as a two-dimensional vector.
Suppose that \(A\) is a \(2\times 2\) matrix and that \(T(\xvec)=A\xvec\text{.}\) If
\begin{equation*}
T\left(\twovec{1}{0}\right) = \twovec{2}{0},
T\left(\twovec{0}{1}\right) = \twovec{2}{2}\text{,}
\end{equation*}
what is the matrix \(A\text{?}\)
Solution .
If \(A = \left[\begin{array}{rr} 2 \amp 1 \\
1 \amp 2 \end{array}\right]\text{,}\) then
\(T\left(\twovec{1}{-2}\right) =
A\twovec{1}{-2} = \twovec{0}{-3}\text{.}\)
\(T\left(\twovec{1}{0}\right) =
A\twovec{1}{0} = \twovec{2}{1}\text{.}\)
\(T\left(\twovec{0}{1}\right) =
A\twovec{0}{1} = \twovec{1}{2}\text{.}\)
\(T\left(\twovec xy\right) =
\twovec{2x+y}{x+2y}\text{.}\)