Q.If $A=\begin{bmatrix}\sin\alpha & \cos\alpha\\-\cos\alpha & \sin\alpha\end{bmatrix}$, then verify that $A'A=I$.

Orthogonal Matrix Verification – From Intuition to Precision

An orthogonal matrix is a square matrix that, multiplied by its own transpose, gives back the identity. Why care? Think of a rotation in 2D: rotating the coordinate axes preserves the lengths of vectors and the angles between them. A matrix that preserves both lengths and angles is orthogonal. "Verification" just means checking whether a given matrix has this property.

The Intuition: What Does "Orthogonal" Mean Here?

"Orthogonal" means "at right angles." For matrices it refers to the columns:

• Each column vector has length 1 (a unit vector).
• Any two different columns are perpendicular (their dot product is zero).

So the columns form an orthonormal set. The same holds for the rows.

The Precise Definition

A square $n \times n$ matrix $A$ is orthogonal if and only if:

where $A^T$ is the transpose and $I$ the $n \times n$ identity.

Verification: How to Check

Compute $A^T A$ and check whether it equals the identity.

Example: Check $A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$.

So this rotation matrix is orthogonal.

Why This Works: The Column Interpretation

Let the columns of $A$ be $c_1, \dots, c_n$. The $(i,j)$ entry of $A^T A$ is the dot product $c_i \cdot c_j$.

• When $i = j$: the entry is $\|c_i\|^2$. For it to equal 1, each column must have length 1.
• When $i \neq j$: the entry is $c_i \cdot c_j$. For it to equal 0, different columns must be orthogonal.

So $A^T A = I$ is exactly the condition that the columns are orthonormal.

Common Mistakes to Avoid …