12.6: The Identity and Inverses

There is a special matrix, denoted $I$, which is called to as the identity matrix. The identity matrix is always a square matrix, and it has the property that there are ones down the main diagonal and zeroes elsewhere. Here are some identity matrices of various sizes.

$$\left[ 1\right] ,\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] ,\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] ,\left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right]\nonumber$$

The first is the $1\times 1$ identity matrix, the second is the $2\times 2$ identity matrix, and so on. By extension, you can likely see what the $n\times n$ identity matrix would be. When it is necessary to distinguish which size of identity matrix is being discussed, we will use the notation $I_n$ for the $n \times n$ identity matrix.

The identity matrix is so important that there is a special symbol to denote the $ij^{th}$ entry of the identity matrix. This symbol is given by $I_{ij}=\delta _{ij}$ where $\delta _{ij}$ is the Kronecker symbol defined by $$\delta _{ij}=\left\{ \begin{array}{c} 1 \text{ if }i=j \\ 0\text{ if }i\neq j \end{array} \right.\nonumber$$

$I_n$ is called the identity matrix because it is a multiplicative identity in the following sense.

Lemma $\PageIndex{1}$: Multiplication by the Identity Matrix

Suppose $A$ is an $m\times n$ matrix and $I_{n}$ is the $n\times n$ identity matrix. Then $AI_{n}=A.$ If $I_{m}$ is the $m\times m$ identity matrix, it also follows that $I_{m}A=A.$

Proof

The $(i,j)$-entry of $AI_n$ is given by: $$\sum_{k}a_{ik}\delta _{kj}=a_{ij}\nonumber$$ and so $AI_{n}=A.$ The other case is left as an exercise for you.

We now define the matrix operation which in some ways plays the role of division.

Definition $\PageIndex{1}$: The Inverse of a Matrix

A square $n\times n$ matrix $A$ is said to have an inverse $A^{-1}$ if and only if

$$AA^{-1}=A^{-1}A=I_n\nonumber$$

In this case, the matrix $A$ is called invertible.

Such a matrix $A^{-1}$ will have the same size as the matrix $A$. It is very important to observe that the inverse of a matrix, if it exists, is unique. Another way to think of this is that if it acts like the inverse, then it $\textbf{is}$ the inverse.

Theorem $\PageIndex{1}$: Uniqueness of Inverse

Suppose $A$ is an $n \times\ n$ matrix such that an inverse $A^{-1}$ exists. Then there is only one such inverse matrix. That is, given any matrix $B$ such that $AB=BA=I$, $B=A^{-1}$.

Proof

In this proof, it is assumed that $I$ is the $n \times n$ identity matrix. Let $A, B$ be $n \times n$ matrices such that $A^{-1}$ exists and $AB=BA=I$. We want to show that $A^{-1} = B$. Now using properties we have seen, we get:

$$A^{-1}=A^{-1}I=A^{-1}\left( AB\right) =\left( A^{-1}A\right) B=IB=B\nonumber$$

Hence, $A^{-1} = B$ which tells us that the inverse is unique.

The next example demonstrates how to check the inverse of a matrix.

Example $\PageIndex{1}$: Verifying the Inverse of a Matrix

Let $A=\left[ \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right] .$ Show $\left[ \begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array} \right]$ is the inverse of $A.$

Solution

To check this, multiply $$\left[ \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right] \left[ \begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array} \right] = \ \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right] = I\nonumber$$ and $$\left[ \begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array} \right] \left[ \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right] = \ \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right] = I\nonumber$$ showing that this matrix is indeed the inverse of $A.$

Unlike ordinary multiplication of numbers, it can happen that $A\neq 0$ but $A$ may fail to have an inverse. This is illustrated in the following example.

Example $\PageIndex{2}$: A Nonzero Matrix With No Inverse

Let $A=\left[ \begin{array}{rr} 1 & 1 \\ 1 & 1 \end{array} \right] .$ Show that $A$ does not have an inverse.

Solution

One might think $A$ would have an inverse because it does not equal zero. However, note that $$\left[ \begin{array}{rr} 1 & 1 \\ 1 & 1 \end{array} \right] \left[ \begin{array}{r} -1 \\ 1 \end{array} \right] =\left[ \begin{array}{r} 0 \\ 0 \end{array} \right]\nonumber$$ If $A^{-1}$ existed, we would have the following $$\begin{aligned} \left[ \begin{array}{r} 0 \\ 0 \end{array} \right] &= A^{-1}\left( \left[ \begin{array}{r} 0 \\ 0 \end{array} \right] \right) \\ &= A^{-1}\left( A\left[ \begin{array}{r} -1 \\ 1 \end{array} \right] \right) \\ &=\left( A^{-1}A\right) \left[ \begin{array}{r} -1 \\ 1 \end{array} \right] \\ &=I\left[ \begin{array}{r} -1 \\ 1 \end{array} \right] \\ &=\left[ \begin{array}{r} -1 \\ 1 \end{array} \right]\end{aligned}$$ This says that $$\left[ \begin{array}{r} 0 \\ 0 \end{array} \right] = \left[ \begin{array}{r} -1 \\ 1 \end{array} \right]\nonumber$$ which is impossible! Therefore, $A$ does not have an inverse.

In the next section, we will explore how to find the inverse of a matrix, if it exists.