9.13
Determine if the function \(T: M_{m \times n}(\mathbb{R}) \rightarrow M_{n \times m}(\mathbb{R})\) defined by \(T(A) = A^T\) is a linear transformation.
Solution:
(1) Let \(A, B \in M_{m \times n}(\mathbb{R})\), then:
(2) Let \(A \in M_{m \times n}(\mathbb{R})\) and \(\alpha \in \mathbb{R}\), then:
Therefore, \(T\) is a linear transformation.
\(P_m(\mathbb{R}) = \{a_0 + a_1x + \dots + a_m x^m : a_i \in \mathbb{R} \text{ for all } i \}\)
Consider the functions \(T: \mathbb{R}^2 \rightarrow P_1(\mathbb{R})\) and \(S: P_1(\mathbb{R}) \rightarrow P_2(\mathbb{R})\) defined by:
Show that \(T, S\) are linear transformations and find \(S \circ T: \mathbb{R}^2 \rightarrow P_2(\mathbb{R})\)
\(T\left[ \begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} \right] = T\begin{pmatrix} a + c \\ b + d \end{pmatrix} = a + c + (b + d)x = a + bx + c + dx = T\begin{pmatrix} a \\ b \end{pmatrix} + T\begin{pmatrix} c \\ d \end{pmatrix}\)
\(T\left( \alpha \begin{pmatrix} a \\ b \end{pmatrix} \right) = T\begin{pmatrix} \alpha a \\ \alpha b \end{pmatrix} = \alpha a + \alpha bx = \alpha T\begin{pmatrix} a \\ b \end{pmatrix}\)
\(S\left\lbrack\left(a+bx\right)+\left(c+dx\right)\right\rbrack=\left\lbrack\left(a+bx\right)+\left(c+dx)\right\rbrack x=\left(a+bx+c+\mathrm{d}x\right)x=ax+bx^2+cx+dx^2=S(a+bx)+S(c+dx)\right\rbrack\)
\(S(\alpha(a+bx))=(\alpha a+\alpha bx)x=\alpha ax+\alpha bx^2=\alpha\left(ax+bx^2\right)=\alpha\left(S\left(a+bx\right)\right)\)
\(S\left(a+b\left(T\begin{pmatrix}a\\ b\end{pmatrix}\right)\right)=x\left(a+b\left(\right)T\begin{pmatrix}a\\ b\end{pmatrix}\right)=ax+bx(a+bx)=ax+abx^2+b^2x^2=b^2x^2+a(b+1)x\)
If \(T: \mathbb{R}^2 \rightarrow \mathbb{R}\) is a linear transformation with
\(T\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = 5\) and \(T\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 2\), find \(T\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}\).
Solution:
We need to find \(\alpha, \beta \in \mathbb{R}\) such that \(\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} = \alpha \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + \beta \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\)
Expanding, this gives:
\(\alpha \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + \beta \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3\alpha + \beta \\ -\alpha \\ 2\alpha + \beta \end{pmatrix}\)
Equating components:
\(3\alpha + \beta = -1\)
\(-\alpha = 1\)
\(2\alpha + \beta = 0\)
Solving the system, we get:
\(\alpha = -1, \quad \beta = 2\).
Thus,
\(\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} = -\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\).
Since \(T\) is a linear transformation, we have:
\(T\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} = T\left( -\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \right) = -T\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + 2 T\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\).
Substituting the known values for \(T\):
\(= -5 + 2(2) = -5 + 4 = -1\).
Thus,
\(T\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} = -1\).
Determine if the function \(T: M_{2 \times 2}(\mathbb{R}) \rightarrow \mathbb{R}\) defined by \(T(A) = \det(A)\) is a linear transformation.
Solution:
\(T\) is not a linear transformation since if we consider
\(A=\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix},B=\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}\), then
\(A+B=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\)
with \(T(A + B) = \det(A + B) = 1 \neq 0 = \det(A) + \det(B)\)
\(\|v\| = \sqrt{\langle v, v \rangle}\)
Pythagorean Theorem:
If \(u, v\) are orthogonal vectors in \(\mathbb{K}^n\) then
Proof:
If \(u\) and \(v\) are orthogonal vectors, then \(\langle u, v \rangle = 0\). Now,
Let \(v\) be a nonzero vector in \(\mathbb{K}^n\). Prove that the vector
is a unit vector and has the same direction as \(v\).
Find a unit vector in the direction of the vector