Invertible Matrix Theorem

Theorem

$$\text{Given an n x n matrix } A = \begin{pmatrix}\overrightarrow{a_1} & ... & \overrightarrow{a_n} \end{pmatrix} \text{, the following are equivalent:} \\ \text{1. A is invertible: } A * A^{-1} = A^{-1} * A = I_n \\ \text{2. A has n pivots} \\ \text{3. } T(\overrightarrow{x}) = A * \overrightarrow{x} \text{ is onto} \\ \text{4. } Span\{\overrightarrow{a_1}, ..., \overrightarrow{a_n} \} = \mathbb{R}^n\\ \text{5. } A\overrightarrow{x} = \overrightarrow{b} \text{ is consistent for every }\overrightarrow{b} \\ \text{6. } A\overrightarrow{x} = \overrightarrow{b} \implies \overrightarrow{x} = \overrightarrow{0} \\ \text{7. } \{\overrightarrow{a_1}, ..., \overrightarrow{a_n} \} \text{ is linearly independent} \\ \text{8. } T(\overrightarrow{x}) = A * \overrightarrow{x} \text{ is one-to-one} \\ \text{9. } BA = I_n \text{ for some B} \\ \text{10. } AB = I_n \text{ for some B} \\ \text{11. } A\overrightarrow{x} = \overrightarrow{b} \text{ has exactly one solution for every } \overrightarrow{b}$$

Note that 3-5 is the same as the Row Theorem and 4-6 is the same as the Column Theorem.

Examples

$$\text{Prove that 1 implies 11} \\ A\overrightarrow{x} = \overrightarrow{b} \\ A^{-1}\overrightarrow{b} = A^{-1}A\overrightarrow{x} = I_n * \overrightarrow{x} = \overrightarrow{x}$$ $$\text{Q. If A is 3 x 3 and its columns span } \mathbb{R}^3 \text{, are the columns linearly independent?} \\ \text{A. Yes because of the Invertible Matrix Theorem (4 and 7)} \\ \text{Q. If A is 4 x 4 and } A\overrightarrow{x} = \overrightarrow{0} \text{ has a nontrivial solution, does } A\overrightarrow{x} = \overrightarrow{0} \text{ have a unique solution for every } \overrightarrow{b} \text{?}\\ \text{A. No, as it fails condition 6 so it fails all conditions (specifically 11)}$$

Additional properties of inverses

1. If A is invertible, so is A^-1, and (A^-1)^-1 = A

2. If A and B are invertible, so is AB and (AB)^-1 = B^-1A^-1
   • This is because the identity matrix is only gotten through ABB^-1A^-1 or B^-1A^-1AB
3. If A ias invertible, so is A^T and (A^T)^-1 = (A^-1)^T

2 x 2 Example

$$A = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \text{, and if A is invertible, then } A^{-1} = \frac{1}{ad - bc} * \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \\ \text{A is invertible } \iff ad - bc = det(A) \neq 0$$

The determinant of a 2 x 2 matrix is simple, but larger matrices are extremely difficult to calculate the determinant for.

$$\text{Consider } A = \begin{pmatrix} 1 & 1 \\ 3 & 4 \\ \end{pmatrix} \\ A^{-1} = \frac{1}{4 - 3} * \begin{pmatrix} 4 & -1 \\ -3 & 1 \\ \end{pmatrix} = \begin{pmatrix} 4 & -1 \\ -3 & 1 \\ \end{pmatrix}$$

Reliance on Square Matrices

$$\text{Consider } A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix} \text{, } B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} \\ AB = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \text{, but } BA = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{pmatrix}$$

Subspaces

$$\text{A subspace of } \mathbb{R}^n \text{ is a subset } H \subset \mathbb{R}^n \text{ such that} \\ \overrightarrow{0} \in H \\ \overrightarrow{u}, \overrightarrow{v} \in H \implies \overrightarrow{u} + \overrightarrow{v} \in H \\ \overrightarrow{u} \in H, r \in \mathbb{R} \implies r * \overrightarrow{u} \in H$$

Spans are an example of a subspaces

$$A = \begin{pmatrix} \overrightarrow{a_1} & ... & \overrightarrow{a_n}\end{pmatrix} \text{, } Col(A) = Span\{\overrightarrow{a_1}, ..., \overrightarrow{a_n}\} \text{ is a subspace}$$

Null Subspace

$$\text{m x n matrix A, } Null(A) = \{\overrightarrow{x} \in \mathbb{R}^n | A\overrightarrow{x} = \overrightarrow{0}\} \\ A\overrightarrow{0} = \overrightarrow{0} \\ A\overrightarrow{u} = \overrightarrow{0} \text{ and } A\overrightarrow{v} = \overrightarrow{0} \text{, } A(\overrightarrow{u} + \overrightarrow{v}) = A\overrightarrow{u} + A\overrightarrow{v} = \overrightarrow{0} + \overrightarrow{0} = \overrightarrow{0} \\ A\overrightarrow{u} = \overrightarrow{0}, r \in \mathbb{R}, A(r\overrightarrow{u}) = r A(\overrightarrow{u}) = r\overrightarrow{0} = \overrightarrow{0}$$

Examples

$$A = \begin{pmatrix} 1 & 0 & 2 \\ 3 & 5 & -1 \end{pmatrix}\\ \text{What is Nul(A) and Col(A) as spans?} \\ \text{By the row theorem, Col(A) is } \mathbb{R}^2 \\ \overrightarrow{x}_0 = \begin{pmatrix} -2 \\ \frac{7}{5} \\ 1 \end{pmatrix}, Nul(A) = Span\{\begin{pmatrix} -2 \\ \frac{7}{5} \\ 1 \end{pmatrix}\}$$