Invertible Matrix Theorem

## Invertible Matrix Theorem

Let $A$ be a square $n \times n$ matrix. Then the following statements are equivalent. That is, for a given $A$, the statements are either all true or all false.

- $A$ is an invertible matrix.
- $A$ is row equivalent to the $n \times n$ identity matrix.
- $A$ has $n$ pivot position.
- The equation $A \vec{x} = \vec{0}$ has only the trivial solution.
- The columns of $A$ form a linearly independent set.
- The linear transformation $\vec{x}$ to $A \vec{x}$ is one-to-one.
- The equation $A \vec{x} = \vec{b}$ has at least one solution for each $\vec{b}$ in $\mathbb{R}^{n}$.
- The columns of $A$ span $\mathbb{R}^{n}$.
- The linear transformation $\vec{x}$ to $A \vec{x}$ maps $\mathbb{R}^{n}$ onto $\mathbb{R}^{n}$.
- There is an $n \times n$ matrix $C$ such that $CA = I$.
- There is an $n \times n$ matrix $D$ such that $AD = I$.
- $A^{T}$ is an invertible matrix.
- The columns of $A$ form a basis of $\mathbb{R}^{n}$.
- $\mathrm{Col} \ A = \mathbb{R}^{n}$.
- $\mathrm{dim} \ \mathrm{Col} \ A = n$.
- $\mathrm{rank} \ A = n$.
- $\mathrm{Nul} \ A = \left \{ \vec{0} \right \}$.
- $\mathrm{dim} \ \mathrm{Nul} \ A = 0$.
- The number $0$ is not an eigenvalue of $A$.
- The determinant of $A$ is
*not*zero; $\mathrm{det} \ A \neq 0$.

page revision: 2, last edited: 29 Apr 2015 00:20