LINEAR-ALGEBRA · Unit 16 · Video 1 · Interactive Practice

When Transpose Equals Inverse

IKey Formulas

Formula	Name	Description
$\mathbf{q}_i^T \mathbf{q}_j = 0 \;(i \neq j)$	Orthogonality	Columns are perpendicular
$\mathbf{q}_i^T \mathbf{q}_i = 1$	Normalization	Each column has unit length
$Q^T Q = I \;\Rightarrow\; Q^T = Q^{-1}$	Orthogonal matrix (square $Q$ )	Transpose is the inverse
$Q = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$	2×2 rotation matrix	Rotates vectors by angle $\theta$

IIVisualization 1 — Orthonormal Columns of a Rotation Matrix

The rotation columns $\mathbf{q}_1 = (\cos\theta,\; \sin\theta)$ and $\mathbf{q}_2 = (-\sin\theta,\; \cos\theta)$ stay orthonormal for every angle $\theta$ .

Angle θ 43°

IIIVisualization 2 — Orthogonal Matrices Preserve Lengths

Multiplying by an orthogonal $Q$ rotates $\mathbf{v}$ without changing its length: $\|Q\mathbf{v}\| = \|\mathbf{v}\|$ .

Rotation θ 60°

💡 The length cannot change because $\|Q\mathbf{v}\|^2 = (Q\mathbf{v})^\top(Q\mathbf{v}) = \mathbf{v}^\top Q^\top Q\,\mathbf{v} = \mathbf{v}^\top\mathbf{v} = \|\mathbf{v}\|^2$ — the identity $Q^\top Q = I$ does all the work.

IVVisualization 3 — The Transpose Undoes the Rotation

$Q$ rotates by $+\theta$ and $Q^\top$ rotates by $-\theta$ , so $Q^\top(Q\mathbf{v}) = \mathbf{v}$ .

Angle θ 72° Scrub

💡 Rotating $+\theta$ then $-\theta$ is doing nothing, so $Q^\top Q = I$ : the transpose is the inverse — no elimination, cofactors, or determinant needed.

VQuiz Questions

Question 1

What conditions must a set of vectors satisfy to be called orthonormal?

✅ Correct! Orthonormal = orthogonal (perpendicular, dot product 0) + normalized (unit length).

❌ Not quite. Remember: 'orthonormal' combines two separate requirements — perpendicularity and unit length.

Show solution

Solution:

Orthonormal = Orthogonal + Normalized:

Orthogonal means all pairwise dot products are zero: $\mathbf{q}_i^T \mathbf{q}_j = 0$ for $i \neq j$
Normalized means each vector has unit length: $\mathbf{q}_i^T \mathbf{q}_i = 1$

Both conditions are required. Orthogonal alone (just perpendicular) isn't sufficient — the vectors must also have length 1. Note: while orthonormal vectors are always linearly independent, that's a consequence, not the definition.

Question 2

A matrix $Q$ has size $3 \times 2$ (3 rows, 2 columns) with orthonormal columns. Which of the following is true?

✅ Correct! The columns being orthonormal gives QᵀQ = I₂, but 'orthogonal matrix' requires Q to be square.

❌ Not quite. Think about the dimensions: Q is 3×2, so what size is QᵀQ? And can a non-square matrix be called 'orthogonal'?

Show solution

Solution:

For a $3 \times 2$ matrix $Q$ with orthonormal columns:

$Q^T$ is $2 \times 3$ , so $Q^TQ$ is $2 \times 2$ . Since the columns are orthonormal, $Q^TQ = I_2$ ✓
$QQ^T$ is $3 \times 3$ but is not $I_3$ — it's a projection matrix ✗
The term "orthogonal matrix" applies only to square matrices where $Q^T = Q^{-1}$

From the video: the $3 \times 2$ example $Q = \frac{1}{3}\begin{bmatrix}1 & -2 \\ 2 & -1 \\ 2 & 2\end{bmatrix}$ satisfies $Q^TQ = I_2$ but $QQ^T \neq I_3$ . Without squareness, $Q^T$ is only a left inverse.

Question 3

The $2 \times 2$ Hadamard matrix is $H = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}$ .

What is the dot product of the two columns of $H$ ?

✅ Correct! The +1 and -1 entries cause the terms to cancel: 1/2 - 1/2 = 0. The columns are orthogonal.

❌ Not quite. Don't forget the 1/√2 scaling factor on each column when computing the dot product.

Show solution

Solution:

Column 1 of $H$ : $\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}$ , Column 2 of $H$ : $\frac{1}{\sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix}$

Dot product:

$\mathbf{q}_1^T \mathbf{q}_2 = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(1)(1) + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(1)(-1) = \frac{1}{2} - \frac{1}{2} = 0$

The dot product is 0, confirming the columns are orthogonal. Combined with the fact that each column has unit length ( $\frac{1}{2} + \frac{1}{2} = 1$ ), the columns are orthonormal, so $H$ is an orthogonal matrix.

Question 4

True or False: Every orthogonal matrix has determinant equal to 1.

✅ Correct! The determinant of an orthogonal matrix is ±1, not always +1. Reflection matrices are orthogonal but have det = −1.

❌ Not quite. Consider the reflection matrix [[1,0],[0,-1]]. It's orthogonal — what is its determinant?

Show solution

Solution: False

If $Q$ is orthogonal, then $Q^TQ = I$ , so:

$\det(Q^TQ) = \det(I) = 1$ $\det(Q^T) \cdot \det(Q) = 1$ $[\det(Q)]^2 = 1$ $\det(Q) = \pm 1$

The determinant can be +1 (pure rotations) or -1 (reflections).

For example, the reflection matrix $Q = \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$ satisfies $Q^TQ = I$ (it's orthogonal), but $\det(Q) = -1$ .

Orthogonal matrices with $\det = +1$ are called special orthogonal (pure rotations). Those with $\det = -1$ include a reflection component.

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