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## Invertible Matrix Theorem

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.

(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.

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 $T^{-1}(\overrightarrow{x})=A^{-1}\overrightarrow{x}.$

$$T$$ 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*}

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