Theorem (Invertible Matrix Theorem).
Let \(A\) be an \(n\times n\) matrix and \(T\) be the linear transformation \(\overrightarrow{x} \mapsto A\overrightarrow{x}\). Then the following are equivalent.
\(A\) is invertible.
\(A\overrightarrow{x}=\overrightarrow{b}\) has a unique solution for each \(\overrightarrow{b}\in \mathbb R^n\).
The RREF of \(A\) is \(I_n\).
\(T\) (i.e., \(\overrightarrow{x} \mapsto A\overrightarrow{x}\)) is an isomorphism.
\(T\) (i.e., \(\overrightarrow{x} \mapsto A\overrightarrow{x}\)) is one-to-one.
Each row and column of \(A\) has a pivot position.
(a), (b), and (c) are equivalent by the second theorem
in the section Inverse of a Matrix.
Also (d)-(l) are equivalent by the last theorem
in the section Linear Transformtions.
Since \(A\) is a square matrix, (c) and (l) are equivalent.
Example.
What can we say about \(\operatorname{CS}\left(A\right),\operatorname{NS}(A),\operatorname{rank}(A),
\operatorname{nullity}(A)\), and pivot positions of a \(3\times 3\) invertible matrix? What about
\(\overrightarrow{x} \mapsto A\overrightarrow{x}\)?
Solution. By the IMT, \(\operatorname{CS}\left(A\right)=\mathbb R^3,\operatorname{NS}(A)=\{\overrightarrow{0_3}\},\operatorname{rank}(A)=3,
\operatorname{nullity}(A)=0\), \(A\) has 3 pivot positions, and \(\overrightarrow{x} \mapsto A\overrightarrow{x}\) is
an isomorphism, i.e., a one-to-one linear transformation from \(\mathbb R^3\) onto \(\mathbb R^3\). Also
\(A\overrightarrow{x}=\overrightarrow{b}\) has a unique solution \(A^{-1}\overrightarrow{b}\) for each
\(\overrightarrow{b}\in \mathbb R^3\).
Remark.
In general the conditions in the IMT are not equivalent for a non-square matrix.
Example.
The linear transformation \(T:\mathbb R^3 \to \mathbb R^2\) defined by \(T(x_1,x_2,x_3)=(x_1,x_2)\) has
\(2\times 3\) standard matrix \(A=\left[\begin{array}{rrr} 1&0&0\\ 0&1&0 \end{array} \right]\). Note that \(T\) is
onto but not one-to-one. Equivalently the columns of \(A\) span \(\mathbb R^2\) but they are not linearly independent.
The linear transformation \(T:\mathbb R^2 \to \mathbb R^3\) defined by \(T(x_1,x_2)=(x_1,x_2,0)\) has \(3\times 2\)
standard matrix \(A=\left[\begin{array}{rr} 1&0\\ 0&1\\ 0&0 \end{array} \right]\). Note that \(T\) is one-to-one but not onto.
Equivalently the columns of \(A\) are linearly independent but they do not span \(\mathbb R^3\).
Definition.
A linear transformation \(T:\mathbb R^n \to \mathbb R^n\) is invertible if there is another linear transformation
\(S:\mathbb R^n \to \mathbb R^n\) such that
\[T(S(\overrightarrow{x}))=S(T(\overrightarrow{x}))=\overrightarrow{x} \text{ for all } \overrightarrow{x}\in \mathbb R^n.\]
This \(S\) is called the inverse of \(T\), denoted by \(T^{-1}\), for which \(T\circ T ^{-1}=T ^{-1} \circ T=I\),
the identity function on \(\mathbb R^n\).
Remark.
It is well-known that a function is invertible if it is one-to-one and onto. So a linear transformation \(T:\mathbb R^n \to \mathbb R^n\) is an isomorphism if and only if it invertible.
Example.
The linear transformation \(T:\mathbb R^2 \to \mathbb R^2\) defined by \(T(x_1,x_2)=(x_1+2x_2,3x_1+5x_2)\) is
one-to-one and onto consequently invertible. How to find \(T^{-1}:\mathbb R^2 \to \mathbb R^2\)?
Theorem.
Let \(T:\mathbb R^n \to \mathbb R^n\) be a linear transformation with the standard matrix \(A\). Then \(T\) is
invertible if and only if \(A\) is invertible. Also \(T^{-1}:\mathbb R^n \to \mathbb R^n\) is given by
\[T^{-1}(\overrightarrow{x})=A^{-1}\overrightarrow{x}.\]
\(T\) is invertible (i.e., an isomorphism) if and only if \(A\) is invertible by the IMT. Let \(S:\mathbb R^n \to \mathbb R^n\)
be a linear transformation defined by \(S(\overrightarrow{x})=A^{-1}\overrightarrow{x}\). Then for all
\(\overrightarrow{x}\in \mathbb R^n\),
\[\begin{align*}
T(S(\overrightarrow{x})) &=T(A^{-1}\overrightarrow{x})=A(A^{-1}\overrightarrow{x})=I_n \overrightarrow{x}=\overrightarrow{x} \text{ and }\\
S(T(\overrightarrow{x})) &=S(A\overrightarrow{x})=A^{-1}(A\overrightarrow{x})=I_n \overrightarrow{x}=\overrightarrow{x}.
\end{align*}\]
Thus \(S=T^{-1}\).
Example.
The isomorphism \(T:\mathbb R^2 \to \mathbb R^2\) defined by \(T(x_1,x_2)=(x_1+2x_2,3x_1+5x_2)\) has the standard matrix
\(A=[T(\overrightarrow{e_1})\: T(\overrightarrow{e_2}) ]
=\left[\begin{array}{rr} 1&2\\ 3&5\end{array} \right]\).
Since \(A^{-1}=\left[\begin{array}{rr} -5&2\\ 3&-1\end{array} \right]\),
\(T^{-1}:\mathbb R^2 \to \mathbb R^2\) is given by
\(T^{-1}(\overrightarrow{x})=A^{-1}\overrightarrow{x}\), i.e., \(T^{-1}(x_1,x_2)=(-5x_1+2x_2,3x_1-x_2)\).
Verify that for all \([x_1,x_2]^T\in \mathbb R^2\),
\[\begin{align*}
T(T^{-1}(x_1,x_2))& =T(-5x_1+2x_2,3x_1-x_2)=(x_1,x_2) \text{ and }\\
T^{-1}(T(x_1,x_2)) &=T^{-1}(x_1+2x_2,3x_1+5x_2)=(x_1,x_2).
\end{align*}\]