Home page for accesible maths 3 Inverting matrices 3 Inverting matrices 3.2 Inverting matrices

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

3.1. Identity matrices and inverses

In this subsection, we focus on important properties of square matrices. Recall that the products $A B$ and $B A$ of two matrices $A$ and $B$ both exist if and only if there exist $m,n\in{\mathbb{N}}$ with $A\in\operatorname{M}_{{n\times m}}({{\mathbb{R}}})$ and $B\in\operatorname{M}_{{m\times n}}({{\mathbb{R}}})$ . In particular, $A^{2}$ is defined if and only if $A$ is square.

Recall from Example 1.4 that for any positive integer $n$ , the identity matrix $I_{n}\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ is the square matrix whose coefficients take values $1$ on the diagonal and $0$ elsewhere. If the size $n$ is obvious, we write $I$ instead of $I_{n}$ .

Remark 3.1.1.

The Kronecker symbol, usually denoted $\delta_{ij}$ , for integers $i, j$ , is defined by

\delta_{ij}=\left\{\begin{array}[]{rl}1&\quad\hbox{if $i=j$,}\\ 0&\quad\hbox{otherwise.}\end{array}\right.

In particular, the identity matrix $I_{n}$ has $(i,j)$ coefficient $(I_{n})_{ij}=\delta_{ij}$ , for all $1\leq i,j\leq n$ . Also, the vectors $e_{j}\in{\mathbb{R}}^{n}$ have coefficients $(e_{j})_{i}=\delta_{ij}$ , for all $1\leq i,j\leq n$ .

The main property of the identity matrix, as its name suggests, is that it is a multiplicative identity. That is,

\quad\hbox{for any matrix $A\in\operatorname{M}_{{n\times m}}({{\mathbb{R}}})$% , we have}\quad AI_{m}=A=I_{n}A.

In any algebraic structure containing an identity element for the multiplication, we can speak of the existence or not of a multiplicative inverse of an element.

Definition 3.1.2.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ . We say that $A$ is invertible (or non-singular) if there exists $B\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ such that

AB=BA=I_{n}.

In this case, $B$ is called the inverse of $A$ and we write $B={A}^{-1}$ .

Remark 3.1.3.

(i)

From the definition, we note that only square matrices may possibly be invertible.
(ii)

Suppose that a matrix $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ has a ‘left’ inverse, in the sense that there exists $B\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ with $BA=I_{n}$ . Then, $AB=I_{n}$ as well (we will omit the proof).
(iii)

If we know that there exist $A,B,C\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ with $BA=AC=I_{n}$ , then $B=C$ . Here is the proof:

$B=BI_{n}=B(AC)=(BA)C=I_{n}C=C.$
(iv)

We cannot use the fraction notation, i.e. $\frac{I_{n}}{A}$ instead of ${A}^{-1}$ , because matrix multiplication is not commutative.
(v)

If $A$ is invertible, then it is the inverse of its inverse. That is,

$A=({A}^{-1})^{-1}.$

Example 3.1.4.

Let

$A=\begin{pmatrix}1&-4\\ -1&3\end{pmatrix}\quad\hbox{and}\quad B=\begin{pmatrix}-3&-4\\ -1&-1\end{pmatrix}.$

Then, $AB=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}=BA$ so that $B={A}^{-1}$ and $A={B}^{-1}$ . So $A$ and $B$ are inverse of each other.

Even though we do not know how to find the inverse (if it exists) of a matrix yet, let us prove an essential property about products of invertible matrices.

Theorem 3.1.5.

Let $A,B\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ . If $A$ and $B$ are invertible, then $A B$ is invertible with

{(AB)}^{-1}={B}^{-1}{A}^{-1}.

Moreover, if $A B$ is invertible, then both $A$ and $B$ must also be invertible.

Proof.

Assume that $A,B\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ are invertible. Then, by associativity of multiplication,

(AB)({B}^{-1}{A}^{-1})=A(B{B}^{-1}){A}^{-1}=AI_{n}{A}^{-1}=I_{n}.

Thus, by Remark 3.1.3 on uniqueness of inverses, we conclude that ${B}^{-1}{A}^{-1}={(AB)}^{-1}$ as required.

To prove the second statement, assume $A B$ is invertible. So there exists a matrix $C$ such that $(AB)C=I_{n}$ . By associativity of matrix multiplication, this is the same as $A(BC)=I_{n}$ . Therefore, $A$ is invertible. A similar argument shows $B$ is also invertible. ∎

Theorem 3.1.6.

Suppose that $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ is invertible. Then so is $A^{t}$ , and ${(A^{t})}^{-1}=({A}^{-1})^{t}$ .

Proof.

We need to check that $A^{t}({A}^{-1})^{t}=I_{n}=({A}^{-1})^{t}A^{t}$ . We prove the first equality and leave the other as an exercise. By Theorem 1.5.4(ii), we have

A^{t}({A}^{-1})^{t}=\big{(}{A}^{-1}A\big{)}^{t}=I_{n}^{t}=I_{n}.

∎

Suppose that $A$ is a square matrix of size $n$ . How do we go about deciding whether $A$ is invertible or not? And if it is invertible, how do we find its inverse? As a matter of introduction, let us answer these questions in the case of a $2\times 2$ matrix.

Theorem 3.1.7.

Let

A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in\operatorname{M}_{{2}}({{\mathbb{R}}})\;.

(i)

If $ad-bc=0$ then $A$ is not invertible,
(ii)

If $ad-bc\neq 0$ then $A$ is invertible with

${A}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}\;.$

Proof.

Let $B=\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}$ . Then

AB=\begin{pmatrix}ad-bc&0\\ 0&ad-bc\end{pmatrix}=(ad-bc)I_{2}=BA.

Suppose that $ad=bc$ . Then, $AB=0=BA$ . If $A=0$ then $A$ is clearly not invertible, so suppose that $A$ , hence also $B$ , is non-zero. If $A$ has an inverse, that is if ${A}^{-1}$ exists, then

B=BI_{2}=B(A{A}^{-1})=(BA){A}^{-1}=0{A}^{-1}=0,

a contradiction! So $A$ cannot be invertible.

Now suppose that $ad\neq bc$ and set $\displaystyle{C=\frac{1}{ad-bc}B}$ . Then matrix multiplication yields

CA=\frac{1}{ad-bc}BA=\frac{1}{ad-bc}\begin{pmatrix}ad-bc&0\\ 0&ad-bc\end{pmatrix}=I_{2}.

Similarly $AC=I_{2}$ so $A$ is invertible with ${A}^{-1}=C$ . ∎