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.

1. \(A\) is invertible.


2. \(A\overrightarrow{x}=\overrightarrow{b}\) has a unique solution for each \(\overrightarrow{b}\in \mathbb R^n\).


3. The RREF of \(A\) is \(I_n\).


4. \(T\) (i.e., \(\overrightarrow{x} \mapsto A\overrightarrow{x}\)) is an isomorphism.


5. \(T\) (i.e., \(\overrightarrow{x} \mapsto A\overrightarrow{x}\)) is one-to-one.


6. \(\ker T=\operatorname{NS}(A)=\{\overrightarrow{0_n}\}\).


7. \(\operatorname{nullity}(T)=\operatorname{nullity}(A)=0\).


8. The columns of \(A\) are linearly independent.


9. \(T\) (i.e., \(\overrightarrow{x} \mapsto A\overrightarrow{x}\)) is onto.


10. \(\operatorname{im} T=\operatorname{CS}\left(A\right)=\mathbb R^n\).


11. \(\operatorname{rank}(T)=\operatorname{rank}(A)=n\).


12. Each row and column of \(A\) has a pivot position.

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.

1. 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.


2. 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 \[T^{-1}(\overrightarrow{x})=A^{-1}\overrightarrow{x}.\]

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 \([x_1,x_2]^T\in \mathbb R^2\), \[\begin{align*} T(T^{-1}(x_1,x_2))& =T(-5x_1+2x_2,3x_1-x_2)=(x_1,x_2) \text{ and }\\ T^{-1}(T(x_1,x_2)) &=T^{-1}(x_1+2x_2,3x_1+5x_2)=(x_1,x_2). \end{align*}\]

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