Let \(T: \IR^n \rightarrow \IR^m\) be a linear map with standard matrix \(A\text{.}\) Sort the following items into three groups of statements: a group that means \(T\) is injective, a group that means \(T\) is surjective, and a group that means \(T\) is bijective.
\(A\vec x=\vec b\) has a solution for all \(\vec b\in\IR^m\)
\(A\vec x=\vec b\) has a unique solution for all \(\vec b\in\IR^m\)
\(A\vec x=\vec 0\) has a unique solution.
The columns of \(A\) span \(\IR^m\)
The columns of \(A\) are linearly independent
The columns of \(A\) are a basis of \(\IR^m\)
Every column of \(\RREF(A)\) has a pivot
Every row of \(\RREF(A)\) has a pivot
\(m=n\) and \(\RREF(A)=I\)
Activity4.2.5.
Let \(T: \IR^3 \rightarrow \IR^3\) be the linear transformation given by the standard matrix \(A=\left[\begin{array}{ccc} 2 & -1 & 0 \\ 2 & 1 & 4 \\ 1 & 1 & 3 \end{array}\right]\text{.}\)
Write an augmented matrix representing the system of equations given by \(T(\vec x)=\vec{0}\text{,}\) that is, \(A\vec x=\left[\begin{array}{c}0 \\ 0 \\ 0 \end{array}\right]\text{.}\) Then solve \(T(\vec x)=\vec{0}\) to find the kernel of \(T\text{.}\)
Definition4.2.6.
Let \(T: \IR^n \rightarrow \IR^n\) be a linear map with standard matrix \(A\text{.}\)
If \(T\) is a bijection and \(\vec b\) is any \(\IR^n\) vector, then \(T(\vec x)=A\vec x=\vec b\) has a unique solution.
So we may define an inverse map \(T^{-1} : \IR^n \rightarrow \IR^n\) by setting \(T^{-1}(\vec b)\) to be this unique solution.
Let \(A^{-1}\) be the standard matrix for \(T^{-1}\text{.}\) We call \(A^{-1}\) the inverse matrix of \(A\text{,}\) so we also say that \(A\) is invertible.
Activity4.2.7.
Let \(T: \IR^3 \rightarrow \IR^3\) be the linear transformation given by the standard matrix \(A=\left[\begin{array}{ccc} 2 & -1 & -6 \\ 2 & 1 & 3 \\ 1 & 1 & 4 \end{array}\right]\text{.}\)
(a)
Write an augmented matrix representing the system of equations given by \(T(\vec x)=\vec{e}_1\text{,}\) that is, \(A\vec x=\left[\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right]\text{.}\) Then solve \(T(\vec x)=\vec{e}_1\) to find \(T^{-1}(\vec{e}_1)\text{.}\)
(b)
Solve \(T(\vec x)=\vec{e}_2\) to find \(T^{-1}(\vec{e}_2)\text{.}\)
(c)
Solve \(T(\vec x)=\vec{e}_3\) to find \(T^{-1}(\vec{e}_3)\text{.}\)
(d)
Write \(A^{-1}\text{,}\) the standard matrix for \(T^{-1}\text{.}\)
Observation4.2.8.
We could have solved these three systems simultaneously by row reducing the matrix \([A\,|\,I]\) at once.
Find the inverse \(A^{-1}\) of the matrix \(A=\left[\begin{array}{cc} 1 & 3 \\ 0 & -2 \end{array}\right]\) by row-reducing \([A\,|\,I]\text{.}\)
Activity4.2.10.
Is the matrix \(\left[\begin{array}{ccc} 2 & 3 & 1 \\ -1 & -4 & 2 \\ 0 & -5 & 5 \end{array}\right]\) invertible? Give a reason for your answer.
Observation4.2.11.
An \(n\times n\) matrix \(A\) is invertible if and only if \(\RREF(A) = I_n\text{.}\)
Activity4.2.12.
Let \(T:\IR^2\to\IR^2\) be the bijective linear map defined by \(T\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)=\left[\begin{array}{c} 2x -3y \\ -3x + 5y\end{array}\right]\text{,}\) with the inverse map \(T^{-1}\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)=\left[\begin{array}{c} 5x+ 3y \\ 3x + 2y\end{array}\right]\text{.}\)
\(T^{-1}\circ T=T\circ T^{-1}\) is the identity map for any bijective linear transformation \(T\text{.}\) Therefore \(A^{-1}A=AA^{-1}\) equals the identity matrix \(I\) for any invertible matrix \(A\text{.}\)
Assume \(A\) is an \(n \times n\) matrix. Prove the following are equivalent. Some of these results you have proven previously.
\(A\) is non-singular.
\(A\) row reduces to the identity matrix.
For any choice of \(\vec{b} \in \mathbb{R}^n\text{,}\) the system of equations represented by the augmented matrix \([A|\vec{b}]\) has a unique solution.
The columns of \(A\) are a linearly independent set.
The columns of \(A\) form a basis for \(\mathbb{R}^n\text{.}\)
The rank of \(A\) is \(n\text{.}\)
The nullity of \(A\) is 0.
\(A\) is invertible.
The linear transformation \(T\) with standard matrix \(A\) is injective and surjective. Such a map is called an isomorphism.
Exploration4.2.15.
Assume \(T\) is a square matrix, and \(T^4\) is the zero matrix. Prove that \((I - T)^{-1} = I + T + T^2 + T^3.\) You will need to first prove a lemma that matrix multiplication distributes over matrix addition.
Generalize your result to the case where \(T^n\) is the zero matrix.