## M5 Special matrices

It is helpful to understand the definition of a number of different types of “special” matrices.

### Transpose of a matrix

The transpose of a matrix $$\mathbf{A}$$ is denoted $$\mathbf{A}^{T}$$and is found by interchanging the rows and the columns.

The first row becomes the first column, the second row becomes the second column etc.

If $$\mathbf{A}$$ is an $$m\times n$$ matrix, then $$\mathbf{A}^{T}$$ is an $$n\times m$$ matrix.

#### Examples

1. If \begin{align*} \mathbf{A} & =\left[\begin{array}{cc} 1 & 2\\ 3 & 4\\ 5 & 6 \end{array}\right] \end{align*} then by interchanging the rows and columns we get \begin{align*} \mathbf{A}^{T} & =\left[\begin{array}{ccc} 1 & 3 & 5\\ 2 & 4 & 6 \end{array}\right]. \end{align*}

In this case $$\mathbf{A}$$ is a $$3\times2$$ matrix and its transpose $$\mathbf{A}^{T}$$ is a $$2\times3$$ matrix.

1. If \begin{align*} \mathbf{B} & =\left[\begin{array}{cc} 1 & 2\\ 7 & 5 \end{array}\right] \end{align*} then \begin{align*} \mathbf{B}^{T} & =\left[\begin{array}{cc} 1 & 7\\ 2 & 5 \end{array}\right]. \end{align*}

2. If \begin{align*} \boldsymbol{\mathbf{C}} & =\left[\begin{array}{cc} 5 & 2\\ 2 & 4 \end{array}\right] \end{align*} then \begin{align*} \mathbf{C}^{T} & =\left[\begin{array}{cc} 5 & 2\\ 2 & 4 \end{array}\right]. \end{align*}

Note that in this case $$\mathbf{C}=\mathbf{C}^{T}$$ and $$\mathbf{C}$$ is called a symmetric matrix.

### Symmetric Matrix

A symmetric matrix is a square matrix which is equal to its transpose. It is also symmetric about its leading diagonal (top left to bottom right).

#### Examples

1. The matrix

\begin{align*} \mathbf{D} & =\left[\begin{array}{cc} 1 & 3\\ 3 & 4 \end{array}\right] \end{align*} is symmetric because \begin{align*} \mathbf{D}^{T} & =\left[\begin{array}{cc} 1 & 3\\ 3 & 4 \end{array}\right]\\ & =\mathbf{D}. \end{align*} 2. The matrix

\begin{align*} \mathbf{E} & =\left[\begin{array}{ccc} 3 & 4 & 5\\ 4 & -2 & -3\\ 5 & -3 & 1 \end{array}\right] \end{align*} is symmetric because \begin{align*} \mathbf{E}^{T} & =\left[\begin{array}{ccc} 3 & 4 & 5\\ 4 & -2 & -3\\ 5 & -3 & 1 \end{array}\right]\\ & =\mathbf{E}. \end{align*} 3. The matrix \begin{align*} \mathbf{F} & =\left[\begin{array}{ccc} 1 & -2 & 4\\ -2 & 3 & -4\\ 4 & -4 & 2 \end{array}\right] \end{align*} is symmetric because \begin{align*} \mathbf{F}^{T} & =\left[\begin{array}{ccc} 1 & -2 & 4\\ -2 & 3 & -4\\ 4 & -4 & 2 \end{array}\right]\\ & =\mathbf{F}. \end{align*}

### Orthogonal Matrix

A square matrix is orthogonal if $$\mathbf{A}^{T}\mathbf{A}=\mathbf{A}\mathbf{A}^{T}=\mathbf{I}$$ where $$\mathbf{I}$$ is the unit matrix (also called the identity matrix).

Because $$\mathbf{A}^{-1}\mathbf{A}=\boldsymbol{\mathbf{A}}\mathbf{A}^{-1}=\mathbf{I}$$ it follows that for an orthogonal matrix $$\mathbf{A}^{T}=\mathbf{A}^{-1}$$ . This can be useful for finding the inverse of an orthogonal matrix as it is usually easier to find the transpose than the inverse of a matrix.

Note that the determinant of an orthhogonal matrix is either equal to $$+1$$ (a rotation matrix) or $$-1$$ (a reflection matrix).

The rows of an orthogonal matrix are mutually orthogonal (perpendicular) unit vectors. The columns of an orthogonal matrix are also mutually orthogonal unit vectors.

#### Examples

The rotation matrix $$\mathbf{A}=\left[\begin{array}{cc} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{array}\right]$$ is an orthogonal matrix that rotates points, lines and regions through an angle of $$\theta^{\circ}$$ anticlockwise. 1 Note that in the following sections, we use the trigonometric identity \begin{align*} \cos^{2}\theta+\sin^{2}\theta & =1. \end{align*}

The determinant, $$\det\left|\mathbf{A}\right|=\left|\begin{array}{cc} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{array}\right|=\cos^{2}\theta+\sin^{2}\theta=1.$$

$$\mathbf{A}^{T}=\left[\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array}\right]$$

\begin{align*} \mathbf{A}^{T}\mathbf{A} & =\left[\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array}\right]\left[\begin{array}{cc} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{array}\right]\\ & =\left[\begin{array}{cc} \cos^{2}\theta+\sin^{2}\theta & \cos\theta\sin\theta-\sin\theta\cos\theta\\ \sin\theta\cos\theta-\cos\theta\sin\theta & \sin^{2}\theta+\cos^{2}\theta \end{array}\right]\\ & =\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right] \end{align*} and \begin{align*} \mathbf{A}\mathbf{A}^{T} & =\left[\begin{array}{cc} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{array}\right]\left[\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array}\right]\\ & =\left[\begin{array}{cc} \cos^{2}\theta+\sin^{2}\theta & -\cos\theta\sin\theta+\sin\theta\cos\theta\\ -\sin\theta\cos\theta+\cos\theta\sin\theta & \sin^{2}\theta+\cos^{2}\theta \end{array}\right]\\ & =\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]. \end{align*}

That is to say $$\mathbf{A}^{T}\mathbf{A}=\mathbf{A}\mathbf{A}^{T}=\mathbf{I}$$. Hence

\begin{align*} \mathbf{A}^{-1} & =\frac{1}{\det\mathbf{A}}\left[\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array}\right]\\ & =\frac{1}{1}\left[\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array}\right]\\ & =\mathbf{A}^{T}. \end{align*}

The reflection matrix $$\mathbf{B}=\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]$$ is an orthogonal matrix that reflects points, lines and regions in the $$x$$-axis.

$$\left|\mathbf{B}\right|=-1$$ and $$\mathbf{B}^{T}=\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]$$

$$\mathbf{B}^{T}\mathbf{B}=\mathbf{B}\mathbf{B}^{T}=\mathbf{I}$$ and $$\mathbf{B}^{T}=\mathbf{B}^{-1}$$

Note that in this case $$\mathbf{B}$$ is both orthogonal and symmetric.

### Exercise 1

Given \begin{align*} \mathbf{A} & =\left[\begin{array}{cc} 3 & -2\end{array}\right]\quad\mathbf{B}=\left[\begin{array}{cc} 1 & -3\\ -3 & 0 \end{array}\right]\quad\mathbf{C}=\left[\begin{array}{ccc} 2 & 0 & 5\\ 0 & -2 & 2\\ 5 & 2 & 1 \end{array}\right]\quad\mathbf{D=\left[\mathrm{\begin{array}{ccc} 2 & -1 & 5\\ 5 & -1 & 2 \end{array}}\right]\quad E=\left[\mathrm{\begin{array}{ccc} 0 & 1 & -1\\ 1 & 0 & 1\\ 1 & -1 & 0 \end{array}}\right]}, \end{align*}

1. Write down the transpose of each of the matrices.

2. Which of the given matrices are symmetric?

#### Answers

\begin{align*} \text{a. }\mathbf{A}^{T} & =\left[\begin{array}{c} 3\\ -2 \end{array}\right]\quad\mathbf{B}^{T}=\left[\begin{array}{cc} 1 & -3\\ -3 & 0 \end{array}\right]\quad\mathbf{C}^{T}=\left[\begin{array}{ccc} 2 & 0 & 5\\ 0 & -2 & 2\\ 5 & 2 & 1 \end{array}\right]\quad\mathbf{D}^{T}=\left[\begin{array}{cc} 2 & 5\\ -1 & -1\\ 5 & 2 \end{array}\right]\quad\mathbf{E}=\left[\begin{array}{ccc} 0 & 1 & 1\\ 1 & 0 & -1\\ -1 & 1 & 0 \end{array}\right] \end{align*}

$$\text{\text{b.} }\mathbf{B}$$ and $$\mathbf{C}$$ are symmetric.

### Exercise 2

Given \begin{align*} \mathbf{L} & =\left[\begin{array}{cc} 1 & -1\\ -1 & 0 \end{array}\right]\ \ \mathbf{M}=\left[\begin{array}{ccc} \frac{3}{5} & -\frac{4}{5} & 0\\ \frac{4}{5} & \frac{3}{5} & 0\\ 0 & 0 & 1 \end{array}\right]\ \ \mathbf{N}=\dfrac{1}{3}\left[\begin{array}{ccc} 2 & -2 & 1\\ 1 & 2 & 2\\ 2 & 1 & -2 \end{array}\right]\ \ \mathbf{P}=\left[\begin{array}{ccc} 0 & 1 & -1\\ 1 & 0 & 1\\ 1 & -1 & 0 \end{array}\right]\ \ \mathbf{Q}=\left[\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & -1 \end{array}\right], \end{align*}

1. Which of the given matrices are orthogonal? If the matrix is orthogonal, write down its inverse.

2. Which of the given matices is a rotation matrix?

#### Answers

1. $$\mathbf{M}$$, $$\mathbf{N}$$ and $$\mathbf{Q}$$ are orthogonal

\begin{align*} \mathbf{\qquad M}^{-1} & =\left[\begin{array}{ccc} \frac{3}{5} & \frac{4}{5} & 0\\ -\frac{4}{5} & \frac{3}{5} & 0\\ 0 & 0 & 1 \end{array}\right]\ \ \mathbf{N}^{-1}=\dfrac{1}{3}\left[\begin{array}{ccc} 2 & 1 & 2\\ -2 & 2 & 1\\ 1 & 2 & -2 \end{array}\right]\ \ \mathbf{Q}^{-1}=\left[\begin{array}{ccc} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & -1 \end{array}\right] \end{align*}

1. $$\mathbf{M}$$ is a rotation matrix. $$\left|\mathbf{M}\right|=1$$.

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