Matrix Invertibility

Identity Matrix

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 $F$:

$$I_n\stackrel{\text{def}}{=}\left[\begin{array}{ccc}{1}_F& \cdots & 0_F\\ ⋮& \ddots & ⋮\\ 0_F& \cdots & 1_F\end{array}\right]$$

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

Matrix Invertibility

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 $A{A}^{-1}$ and $A^{-1}A$ are equal to the identity matrix $I_n$.

$$A{A}^{-1}=A^{-1}A=I_n$$

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.

• Thetranspose 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

Finding Inverses

Algorithm: Matrix Inversion

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

1. Notate an $n\times 2n$-matrix $(A\mid I_n)$ by sticking the identity matrix $I_n$ to the right of $A$.

2. 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 $A=\left[\begin{array}{ccc}6& 8& 3\\ 4& 7& 3\\ 1& 2& 1\end{array}\right]$. Notate

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

Perform Gauss-Jordan elimination:

$$\left[\begin{array}{cccccc}6& 8& 3& 1& 0& 0\\ 4& 7& 3& 0& 1& 0\\ 1& 2& 1& 0& 0& 1\end{array}\right]\approx \left[\begin{array}{cccccc}1& 2& 1& 0& 0& 1\\ 6& 8& 3& 1& 0& 0\\ 4& 7& 3& 0& 1& 0\end{array}\right]\approx \left[\begin{array}{cccccc}1& 2& 1& 0& 0& 1\\ 0& -4& -3& 1& 0& -6\\ 0& -1& -1& 0& 1& -4\end{array}\right]$$ $$\left[\begin{array}{cccccc}1& 2& 1& 0& 0& 1\\ 0& -4& -3& 1& 0& -6\\ 0& -1& -1& 0& 1& -4\end{array}\right]\approx \left[\begin{array}{cccccc}1& 0& -1& 0& 2& -7\\ 0& 1& 1& 0& -1& 4\\ 0& 0& 1& 1& -4& 10\end{array}\right]\approx \left[\begin{array}{cccccc}1& 0& 0& 1& -2& 3\\ 0& 1& 0& -1& 3& -6\\ 0& 0& 1& 1& -4& 10\end{array}\right]$$ $$A^{-1}=\left[\begin{array}{ccc}1& -2& 3\\ -1& 3& -6\\ 1& -4& 10\end{array}\right]$$

Theorem: Inverting $2\times 2$-Matrices

A $2\times 2$-matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is invertible if and only if

$$ad-bc\ne 0$$

If $A$ is invertible matrix, then

$$A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$$

PROOF

TODO