2.6. The Identity and Inverses

Definition - Main Diagonal

The main diagonal of a matrix is the line of entries that goes from the top left corner to the bottom right corner of the matrix. The main diagonal consists of the elements that have the same row and column index. That is, the elements that are of the form \(a_{ii}\), where \(i\) is any positive integer [Kuttler and Farah, 2020, Nicholson, 2018].

For example, if

\[\begin{split} A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \end{split}\]

then the main diagonal of \(A\) is

\[\begin{split} \begin{bmatrix} \color{red}{a_{11}} & a_{12} & a_{13} \\ a_{21} & \color{red}{a_{22}} & a_{23} \\ a_{31} & a_{32} & \color{red}{a_{33}} \end{bmatrix} \end{split}\]

Definition - Symmetric and Skew Symmetric Matrices

An \(n\times n\) matrix \(A\) is

• symmetric if \(A = A^T\), which means that the entry in the \(i\)th row and \(j\)th column of \(A\) is equal to the entry in the \(j\)th row and \(i\)th column of \(A\) for all \(i\) and \(j\). Symmetric matrices have real eigenvalues, orthogonal eigenvectors, and can be diagonalized by an orthogonal matrix [Kuttler and Farah, 2020, Nicholson, 2018].

• skew-symmetric if \(A = -A^T\), which means that the entry in the \(i\)th row and \(j\)th column of \(A\) is the negative of the entry in the \(j\)th row and \(i\)th column of \(A\) for all \(i\) and \(j\). Skew-symmetric matrices have zero diagonal entries, purely imaginary eigenvalues, and can be diagonalized by a unitary matrix [Kuttler and Farah, 2020, Nicholson, 2018].

Remark

Note that a symmetric matrix has to be a square matrix. If a symmetric \(A\) is \(m\times n\), then \(A^T\) is \(n\times m\), so \(A=A^T\) forces \(n=m\). For example, the matrix

\[\begin{split} A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} \end{split}\]

is symmetric because it is equal to its transpose:

\[\begin{split} A^T = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} \end{split}\]

However, the matrix

\[\begin{split} B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \end{split}\]

is not symmetric because it is not a square matrix and it is not equal to its transpose:

\[\begin{split} B^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix} \end{split}\]

Example: Examples of symmetric and skew symmetric matrices.

\[\begin{align*} &\begin{bmatrix}1 & 2\\ 2 & 1\end{bmatrix} &\text{(Symmetric)},\\ &\begin{bmatrix}1 & 2 & 3\\ 2 & 1 & 3\\ 3 & 3 & 1\end{bmatrix} &\text{(Symmetric)},\\ &\begin{bmatrix} 0 & 2 & 3 & 13\\ -2 & 0 & 10 & 8\\ -3 & -10 & 0 & 12\\ -13 & -8 & -12 & 0 \end{bmatrix}&\text{(Skew-symmetric)},\\ &\begin{bmatrix}0 & 24 & 1 & 8 & 15\\ -24 & 0 & 7 & 14 & 16\\ -1 & -7 & 0 & 20 & 22\\ -8 & -14 & -20 & 0 & 3\\ -15 & -16 & -22 & -3 & 0\end{bmatrix} &\text{(Skew-symmetric)}, \end{align*}\]

Remark

The entries on the main diagonal of a skew-symmetric matrix are zero. This can be shown by using the definition of a skew-symmetric matrix, which is \(A = -A^T\). If we denote the entry in the \(i\)th row and \(i\)th column of \(A\) by \(a_{ii}\), then we have

\[ a_{ii} = -a_{ii}^T \]

But the transpose of a scalar is the same scalar, so we have

\[ a_{ii} = -a_{ii} \]

This implies that \(a_{ii} = 0\) for all \(i\). Therefore, the entries on the main diagonal of a skew-symmetric matrix are zero.

Definition - The Identity matrix

The identity matrix is (always) a square matrix with 1s on its main diagonal and 0s elsewhere, and it is shown with \(I_n\) for \(n\geq 1\). The identity matrix acts as the multiplicative identity in matrix multiplication, which means that for any matrix \(A\) of size \(n\times m\), we have

\[ I_n A = A I_m = A \]

The identity matrix is also the inverse of itself, which means that

\[ I_n I_n = I_n \]

The identity matrix can be seen as a special case of a diagonal matrix, which has non-zero entries only on the main diagonal.

For example, the identity matrix of size \(3\times 3\) is

\[\begin{split} I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{split}\]

Example: Some identity matrices of various sizes

\[\begin{align*} I_1=\begin{bmatrix} 1 \end{bmatrix},~I_2=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}~I_3=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix},\ldots \end{align*}\]

Lemma: Multiplication by the Identity Matrix

The identity matrix is a special matrix that plays a similar role to the number 1 in arithmetic. It is (always) a square matrix with 1s on its main diagonal and 0s elsewhere, and it is shown with \(I_n\) for \(n\geq 1\). Assume that \(A\) is an \(m\times n\) matrix and \(I_n\) is the \(n\times n\) identity matrix. Then

\[\begin{align*} AI_n = A. \end{align*}\]

If \(I_m\) is the \(m\times m\) identity matrix, then

\[\begin{align*} I_m A = A. \end{align*}\]

In other words, it is the only matrix such that when multiplied by itself the result is itself. This property makes the identity matrix useful for finding the inverse of a matrix, which is another matrix that when multiplied by the original matrix gives the identity matrix [Kuttler and Farah, 2020, Nicholson, 2018].

For example, if \(A\) is an invertible matrix, then there exists a matrix \(A^{-1}\) such that

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

The inverse of a matrix can be used to solve systems of linear equations, find determinants, and perform other operations in matrix algebra.

Note

We generally use \(I\) instead of \(I_m\) and \(I_n\).

Definition - The Inverse of a Matrix

A square matrix \(A\) is said that have an inverse \(A^{-1}\) (read it \(A\) inverse) if and only if {cite:p}`nicholson2018linear,kuttler2020first

\[\begin{align*} AA^{-1} = A^{-1} A = I_n \end{align*}\]

If matrix \(A\) has an inverse, then the matrix \(A\) is called invertible. Not every square matrix is invertible, however. A matrix is invertible if and only if its determinant is not zero. The determinant is a scalar value that measures the area or volume spanned by the columns or rows of a matrix.

For example, the matrix

\[\begin{split} A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \end{split}\]

is invertible because its determinant is \(1\times 4 - 2\times 3 = -2 \neq 0\). Its inverse is

\[\begin{split} A^{-1} = \frac{1}{-2}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \end{split}\]

However, the matrix

\[\begin{split} B = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \end{split}\]

is not invertible because its determinant is \(1\times 4 - 2\times 2 = 0\). This means that the columns or rows of \(B\) are linearly dependent, which means that they are not linearly independent and do not span the whole space.

Theorem: Uniqueness of Inverse

If matrix \(A\) has an inverse, then its inverse is unique. That is, if \(B\) and \(C\) are both inverses of \(A\), then \(B =C\).

Example: Show that \(A=\begin{bmatrix} 1 & 2\\ 0 & 1 \end{bmatrix}\) is the inverse of \(B=\begin{bmatrix} 1 & -2\\ 0 & 1 \end{bmatrix}\).

Solution: We need to show that \(AB = I\) and \( BA = I\). Therefore,

\[\begin{align*} AB &=\begin{bmatrix}1 & 2\\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & -2\\ 0 & 1\end{bmatrix}= \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} & \surd \\ BA &=\begin{bmatrix} 1 & -2\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1 \end{bmatrix}= \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} & \surd \end{align*}\]

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Note

1. Not every (square) matrix is invertible. For example, \(\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\) (See page 72 of the textbook for the reason).

2. Zero matrix has no inverse.