An orthogonal matrix is a square matrix with the following properties: - *Orthogonal basis vectors:* The column vectors (or row vectors) of the matrix are orthogonal to each other. This means that the [[Dot Product]] of any two distinct [[Basis Vectors]] is zero. $ a_i \cdot a_j =0 \quad \text{for}\: i \neq j$ - *Unit-length basis vectors:* Each basis vector has a magnitude (length) of $1$. This means that the dot product of a basis vector with itself is $1$. $ a_i \cdot a_j=1 \quad \forall i$ Combining these two conditions, an orthogonal matrix $A$ satisfies: $ A: \begin{cases} a_i\cdot a_j=0 \quad i \not =j \\ a_i \cdot a_i=1 \quad i=j \end{cases} $ **Key Property:** Given the above conditions, an orthogonal matrix has the property that multiplying the matrix by its [[Transposed Matrix|transpose]] results in the identity matrix $\mathbf I$. $ A^TA= \mathbf I $ **Proof:** Let $A = [a_1 \, \dots \, a_n]$ be a matrix whose columns are orthonormal vectors. $ \underbrace{\begin{bmatrix} \begin{pmatrix}\dots & a_1 & \dots \end{pmatrix} \\[4pt] \begin{pmatrix}\dots & a_2 & \dots \end{pmatrix}\\[4pt] \vdots\\[4pt] \begin{pmatrix}\dots & a_n & \dots \end{pmatrix} \end{bmatrix}}_{A^T} * \underbrace{\begin{bmatrix} \begin{pmatrix}\vdots \\a_1\\ \vdots \end{pmatrix} \begin{pmatrix}\vdots \\a_2\\ \vdots \end{pmatrix} \dots \begin{pmatrix}\vdots \\a_n\\ \vdots \end{pmatrix}\\ \end{bmatrix}}_{A}= \underbrace{\begin{bmatrix} 1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0\\ \vdots & \vdots & 1 &0\\ 0 & 0 & 0 & 1 \end{bmatrix}}_{\mathbf I} $ 1. Each element of $A^T A$ is the dot product of the $i$-th row of $A^T$ with the $j$-th column of $A$. 2. If $i = j$, the dot product is $1$ because $a_i \cdot a_i = 1$. 3. If $i \neq j$, the dot product is $0$ because $a_i \cdot a_j = 0$. Since the product of a matrix and its inverse equals the identity matrix $(A^{-1} A = \mathbf{I})$, and for orthogonal matrices $(A^T A = \mathbf{I})$, we have the special property: $ A^{-1}=A^T$