A matrix is an arrangement of objects, numbers or expressions of same kind in a rectangular form. A matrix has its elements arranged in rows and columns. A square matrix has same number of rows as that of columns. There are many concepts defined on the basis of square matrices. The orthogonal matrix is one of them. After successful completion of this chapter, you will have an understanding of the following notions : Proof of the equivalence of transpose and inverse of orthogonal matrix.(2) Determinant of orthogonal matrix.(3) |

An orthogonal matrix is defined as a square matrix whose product with its transpose matrix is equal to the identity matrix. Let a matrix $P$ be a square matrix with $n\ \times\ n$ dimension. Then, $P$ is said to be an orthogonal matrix if

$P\ P^{T}$ = $I$

Recall that the transpose matrix is obtained by interchanging the rows and columns of the original matrix.

We may define orthogonal matrix in component form as under :

$a^{-1}_{ij}$ = $a_{ji}$

The basic properties of orthogonal matrices are discussed below.

**(1)**As we learn that for an orthogonal matrix, $P^{-1}$ = $P^{T}$. Therefore, If $P$ is orthogonal, then $P^{-1}$ and $P^{T}$ will also be orthogonal.

**(2)**If two matrices, say $P$ and $Q$ are orthogonal, then their product will also be orthogonal.

If $P\ P^{T}$ = $I$ and $Q\ Q^{T}$ = $I$, Then

$(PQ)^{T}\ (PQ)$ = $(Q^{T}P^{T})\ PQ$ = $Q^{T}\ (P^{T}P)\ Q$ = $Q^{T}Q$ = $I$

**(3)**An orthogonal matrix has a determinant of either $1$ or $-1$.

**(4)**An orthogonal matrix has any real eigenvalue as absolute value $1$.

**(5)**A rotation matrix $R_{\theta}$ for any $\theta$ is an orthogonal matrix.

Recall that $R_{\theta}$ is defined as:

$R_{\theta} = \begin{bmatrix} cos \theta & -sin \theta\\ sin \theta & cos \theta \end{bmatrix}$

From the definition of orthogonal matrix, it is evident that its transpose must be equal to its inverse, i.e.

If $P$ be an orthogonal matrix, then

$P\ P^{T}$ = $I$

Multiplying by $P^{-1}$, we get

$P^{-1}\ P\ P^{T}$ = $P^{-1}\ I$

$I\ P^{T}$ = $P^{-1}$

$P^{T}$ = $P^{-1}$ or $P^{-1}$ = $P^{T}$

In other words, the transpose and inverse of an orthogonal matrix are equivalent. If we find transpose of an orthogonal matrix, we end up getting its inverse.

For an orthogonal matrix, the value of its determinant is $+\ 1$ or $-1$. How? Confused? Let's prove this.

Let P be an orthogonal matrix, then we must have

$P\ P^{T}$ = $I$

$det(P\ P^{T})$ = $det\ (I)$

$det(P)\ det(P^{T})$ = $det\ (I)$

We know that determinant of identity matrix is $1$. Hence,

$det(P)\ det(P^{T})$ = $1$

Also, determinant of a matrix and that of its transpose are same, i.e. $det(P)$ = $det(P^{T})$

So, $det(P)\ det(P)$ = $1$

$[det(P)]^{2}$ = $1$

$det(P)$ = $\pm 1$

**Step 1 :**Find the transpose of given matrix. To do so, interchange the positions of row and column elements.

**Step 2 :**Calculate the product of given matrix and its transpose.

**Step 3 :**Check if it is equal to identity matrix. If yes, then given matrix would be an orthogonal matrix, otherwise not.

Let us discuss an example based on orthogonal matrices.

**Example:**

Check if $\begin{bmatrix} cos \theta & 0 & -sin \theta\\ 0 & 1 & 0\\ sin \theta & 0 & cos \theta \end{bmatrix}$ is an orthogonal matrix.

**Solution:**

Let P = $\begin{bmatrix} cos \theta & 0 & -sin \theta\\ 0 & 1 & 0\\ sin \theta & 0 & cos \theta \end{bmatrix}$

Find transpose matrix

$P^{T} = \begin{bmatrix} cos \theta & 0 & sin \theta\\ 0 & 1 & 0\\ -sin \theta & 0 & cos \theta \end{bmatrix}$

Find product of both

$P\ P^{T}$ = $\begin{bmatrix} cos \theta & 0 & -sin \theta\\ 0 & 1 & 0\\ sin \theta & 0 & cos \theta \end{bmatrix} \begin{bmatrix} cos \theta & 0 & sin \theta\\ 0 & 1 & 0\\ -sin \theta & 0 & cos \theta \end{bmatrix}$

= $\begin{bmatrix} cos^{2} \theta + 0 + sin^{2} \theta & 0 + 0 + 0 & cos \theta sin \theta + 0 - sin \theta cos \theta \\ 0 + 0 + 0 & 0 + 1 + 0 & 0 + 0 + 0\\ sin \theta cos \theta + 0 - sin \theta cos \theta & 0 + 0 + 0 & sin^{2} \theta + 0 + cos^{2} \theta \end{bmatrix}$

= $\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$ = $I$

Hence, given matrix is an orthogonal matrix.