Matrix Invertibility#

Definition: Identity Matrix

The \(n\times n\)-identity matrix \(I_n \in F^{n \times n}\) over some field \(F\) is the square matrix which has the multiplicative identity of \(F\) as its entries on the diagonal and whose other entries are the additive identity of \(F\):

\[ I_n \overset{\text{def}}{=} \begin{bmatrix} 1_F & \cdots & 0_F \\ \vdots & \ddots & \vdots \\ 0_F & \cdots & 1_F\end{bmatrix} \]

Theorem: Multiplication with the Identity Matrix

The product of any \(m\times n\)-matrix \(A\) with the identity matrix \(I_m\) on the left or the identity matrix \(I_n\) is \(A\) itself.

\[ I_m \cdot A = A = A \cdot I_n \]

Proof

TODO

Definition: Invertible Matrix

A square matrix \(A \in F^{n \times n}\) is invertible if there exists a square matrix \(A^{-1} \in F^{n \times n}\) such that the matrix products \(AA^{-1}\) and \(A^{-1}A\) are equal to the identity matrix \(I_n\).

\[ AA^{-1} = A^{-1} A = I_n \]

The matrices \(A\) and \(A^{-1}\) are called inverses of each other.

Theorem: The Invertible Matrix Theorem

The following statements are equivalent for every square matrix \(A \in F^{n \times n}\):

\(A\) is invertible.
The transpose of \(A\) is invertible.
The determinant of \(A\) is not zero, i.e. \(\det(A) \ne 0_F\).
The reduced row echelon form of \(A\) is the identity matrix \(I_n\).
The system of linear equations \(A \vec{x} = \vec{b}\) has a single solution for each \(\vec{b} \in F^n\).
The column space of \(A\) is the vector space \(F^n\).
The row space of \(A\) is the vector space \(F^{1 \times n}\).
The rank of \(A\) is \(n\).

Proof

TODO

Theorem: Antidistributivity of Matrix Inversion

Matrix inversion is antidistributive over matrix products - if \(A, B \in F^{n \times}\) and their matrix product \(AB\) are invertible, then:

\[ (AB)^{-1} = B^{-1} A^{-1} \]

Proof

TODO

Algorithm: Matrix Inversion

To find the inverse of an invertible matrix \(A \in F^{n \times n}\):

Notate an \(n\times 2n\)-matrix \((A\mid I_n)\) by sticking the identity matrix \(I_n\) to the right of \(A\).
Perform Gauss-Jordan elimination on \((A \mid I_n)\). If \(A\) is indeed invertible, the final result will be \((I_n \mid A^{-1})\).

Example

Let's find the inverse of \(A = \begin{bmatrix}6 & 8 & 3 \\ 4 & 7 & 3 \\ 1 & 2 & 1\end{bmatrix}\). Notate

\[ [A\mid I_3] = \left[\begin{array}{ccc|ccc}6 & 8 & 3 & 1 & 0 & 0\\ 4 & 7 & 3 & 0 & 1 & 0 \\ 1 & 2 & 1 & 0 & 0 & 1\end{array}\right] \]

Perform Gauss-Jordan elimination:

\[ \left[\begin{array}{ccc|ccc}6 & 8 & 3 & 1 & 0 & 0\\ 4 & 7 & 3 & 0 & 1 & 0 \\ 1 & 2 & 1 & 0 & 0 & 1\end{array}\right] \approx \left[\begin{array}{ccc|ccc}1 & 2 & 1 & 0 & 0 & 1 \\ 6 & 8 & 3 & 1 & 0 & 0\\ 4 & 7 & 3 & 0 & 1 & 0 \end{array}\right] \approx \left[\begin{array}{ccc|ccc}1 & 2 & 1 & 0 & 0 & 1 \\ 0 & -4 & -3 & 1 & 0 & -6\\ 0 & -1 & -1 & 0 & 1 & -4 \end{array}\right] \]

\[ \left[\begin{array}{ccc|ccc}1 & 2 & 1 & 0 & 0 & 1 \\ 0 & -4 & -3 & 1 & 0 & -6\\ 0 & -1 & -1 & 0 & 1 & -4 \end{array}\right] \approx \left[\begin{array}{ccc|ccc}1 & 0 & -1 & 0 & 2 & -7 \\ 0 & 1 & 1 & 0 & -1 & 4\\ 0 & 0 & 1 & 1 & -4 & 10 \end{array}\right] \approx \left[\begin{array}{ccc|ccc}1 & 0 & 0 & 1 & -2 & 3 \\ 0 & 1 & 0 & -1 & 3 & -6\\ 0 & 0 & 1 & 1 & -4 & 10 \end{array}\right] \]

\[ A^{-1} = \begin{bmatrix}1 & -2 & 3 \\ -1 & 3 & -6 \\ 1 & -4 & 10\end{bmatrix} \]

Theorem: Inverting \(2\times2\)-Matrices

A \(2\times 2\)-matrix \(A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}\) is invertible if and only if

\[ ad - bc \ne 0 \]

If \(A\) is invertible matrix, then

\[ A^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix} \]

Proof

TODO