Orthogonal matrices — properties, det = ±1: Visual walkthrough

Step 1 — What a matrix does to a picture

Call the East arrow $e_{1}$ and the North arrow $e_{2}$ . These are the standard basis vectors: $e_{1} = (10), e_{2} = (01) .$

The top number is "how far East", the bottom number "how far North". So $e_{1}$ is one step East, zero North.

A $2 \times 2$ matrix $Q$ has two columns. Column 1 is where $e_{1}$ lands; column 2 is where $e_{2}$ lands. Nothing more mysterious than that.

Figure — Orthogonal matrices — properties, det = ±1

Step 2 — The transpose and the identity, drawn first

Before we can even write the orthogonality rule we need two pieces of notation. We will not use them until you have seen their pictures.

That last sentence is the whole reason transpose is useful to us: when we multiply $A^{T}$ by $A$ , we are literally sliding each column of $A$ (now lying flat as a row of $A^{T}$ ) across the columns of $A$ — and "row-dotted-into-column" is exactly a dot product. So $A^{T} A$ is the table of all column-vs-column dot products. Hold that thought for Step 3.

Step 3 — The rule that makes $Q$ orthogonal

Name the columns $q_{1}$ (where $e_{1}$ lands) and $q_{2}$ (where $e_{2}$ lands). Using the dot product from Step 2, the rule in symbols is: $q_{1}^{T} q_{1} = 1, q_{2}^{T} q_{2} = 1, q_{1}^{T} q_{2} = 0.$

$q_{1}^{T} q_{1}$ is the dot product of $q_{1}$ with itself (its flat row-copy times its column). Multiply matching entries, add: it equals the squared length of $q_{1}$ . Setting it to $1$ means length $1$ .
$q_{1}^{T} q_{2}$ is the dot product of the two different columns. It equals $∥ q_{1} ∥ ∥ q_{2} ∥ cos θ$ , so it is $0$ exactly when the angle $θ$ between them is $9 0^{\circ}$ .

Step 4 — Orthonormal columns force the tile to be a unit square

Watch the tile: it never stretches (sides stay length $1$ ) and never leans (corner stays $9 0^{\circ}$ ). It can only turn or turn-and-flip.

Step 5 — WHY $det = a d - b c$ is the signed area

Look at the figure. Enclose the parallelogram in the $(a + c) \times (b + d)$ rectangle. Its area is $(a + c) (b + d)$ . Now subtract the two little rectangles and the four triangles that stick out around the parallelogram. When the dust settles, all the cross-terms cancel and you are left with: $area = (a + c) (b + d) - 2 b c - ab - c d = a d - b c .$

$a d$ is the "co-rotating" contribution ( $q_{1}$ East-part times $q_{2}$ North-part).
$- b c$ subtracts the "counter" contribution ( $q_{1}$ North-part times $q_{2}$ East-part).
Their difference $a d - b c$ is positive when $q_{2}$ sits counter-clockwise from $q_{1}$ , negative when clockwise — exactly the signed area.

Step 6 — Two determinant facts, seen not just stated

Fact 1 — areas multiply: $det (A B) = det A \cdot det B$ . Apply machine $B$ : it scales every area by $det B$ . Then apply machine $A$ : it scales that already-scaled area again by $det A$ . Feeding the unit square through both multiplies the two factors — the tile's final area is $det A \cdot det B$ . (See the left panel: a $\times 2$ stretch followed by a $\times 3$ stretch gives a $\times 6$ tile.)

Fact 2 — transpose keeps area: $det (A^{T}) = det A$ . Transposing (Step 2's diagonal flip) swaps the off-diagonal entries $b \leftrightarrow c$ . But the determinant $a d - b c$ only depends on the product $b c$ , and $b c = c b$ — swapping $b$ and $c$ leaves the product untouched. So the signed area is identical. (See the right panel: the flipped tile has the same area.)

Step 7 — The algebra: squeeze the sign out of $Q^{T} Q = I$

Take the determinant of both sides of the orthogonality rule: $det (Q^{T} Q) = det I .$

Left side, by Fact 1 (areas multiply): $det (Q^{T}) det (Q)$ .
By Fact 2 (transpose keeps area): $det (Q^{T}) = det Q$ , so the left side is $(det Q)^{2}$ .
Right side: the identity leaves the tile untouched, area $1$ , so $det I = 1$ (Step 2).

Putting it together: $(det Q)^{2} = 1 ⟹ det Q = \pm 1.$

A real number whose square is $1$ can only be $+ 1$ or $- 1$ . Done.

Step 8 — Case $+ 1$ : the rotation (orientation kept)

The clean example is the rotation matrix $Q_{θ} = (cos θ sin θ - sin θ cos θ), det Q_{θ} = cos^{2} θ + sin^{2} θ = 1.$

Column 1 is $e_{1}$ turned by angle $θ$ ; column 2 is $e_{2}$ turned by the same $θ$ — the whole square spins as one piece.

Step 9 — Case $- 1$ : every reflection, not just one

The simplest example is the flip across the $x$ -axis: $R_{0} = (10 0 - 1), det R_{0} = (1) (- 1) - 0 = - 1.$

Column 1 stays $e_{1}$ (points on the mirror line don't move); column 2 is $e_{2}$ sent to its mirror image $- e_{2}$ .

The general reflection. Tilt the mirror line to make angle $ϕ$ with the East axis. Reflecting across that line is the orthogonal matrix $R_{ϕ} = (cos 2 ϕ sin 2 ϕ sin 2 ϕ - cos 2 ϕ), det R_{ϕ} = - cos^{2} 2 ϕ - sin^{2} 2 ϕ = - 1.$

Whatever the mirror angle $ϕ$ , the determinant is always $- 1$ : all reflections flip orientation, none stretch. The left panel of the figure shows three different mirror lines, each with its tile flipped.

Reflection = rotation × flip. Any $det = - 1$ orthogonal matrix can be written as one fixed flip followed by a rotation: $R_{ϕ} = Q_{2 ϕ} R_{0}$ . So the " $- 1$ world" is just the " $+ 1$ world" (rotations) with one mirror stapled on — the right panel shows this composition. That is the entire other half of $O (2)$ .

Step 10 — Why the sign can never be $0$ (or anything but $\pm 1$ )

Area $0$ would mean the two columns lie on the same line (one is a multiple of the other). But orthonormal columns are perpendicular, so they can never be parallel — no collapse. Hence $det \neq = 0$ , and indeed $Q$ is always invertible.
Any area $\neq = 1$ would mean a column is longer or shorter than $1$ , contradicting "unit length". So $∣ det ∣ = 1$ exactly.
Combined with Step 7's $(det Q)^{2} = 1$ , the only survivors are $+ 1$ (Step 8) and $- 1$ (Step 9). No third option exists.

Recall Why

$(det Q)^{2} = 1$ leaves only two values Over the real numbers, $x^{2} = 1$ has solutions? ::: exactly $x = + 1$ and $x = - 1$ — no other real number squares to $1$ . Could $det Q$ be an imaginary number like $i$ ? ::: No — $Q$ is a real matrix, so $det Q$ is a real number; $i$ is excluded.

The one-picture summary

The whole argument on one canvas: orthonormal columns ⟶ the image of the unit square is a unit square (area magnitude $1$ ) ⟶ taking $det$ of $Q^{T} Q = I$ gives $(det Q)^{2} = 1$ ⟶ spin keeps the sign $(+ 1)$ , flip reverses it $(- 1)$ .

Recall Feynman retelling — the walkthrough in plain words

Picture a sheet of graph paper with two arrows drawn on it: one pointing East, one pointing North, each one square long. A matrix picks these two arrows up and drops them somewhere new — and the whole grid follows.

The little " $T$ " (transpose) is just the sheet flipped across its diagonal; laying the columns flat as rows and sliding them over the columns spells out every dot product at once. The "do-nothing" matrix $I$ has ones down the diagonal and leaves the square exactly where it was — that's why its area factor is $1$ .

An orthogonal matrix has a promise: the two dropped arrows are still each exactly one square long, and they still meet at a perfect right angle. So the little square they build is still a perfect unit square — the machine can only spin the square or flip it over, never squash or stretch it.

The determinant is just "the signed area of that square", and cutting the parallelogram out of its bounding rectangle shows that area is $a d - b c$ . Area is always $1$ because it's a unit square, so the number can only be $+ 1$ or $- 1$ . If the square merely spun, we kept the same handedness: $+ 1$ . If we had to flip it like a pancake to match — across any mirror line — handedness reversed: $- 1$ .

The algebra is the same idea wearing a suit: take $det$ of both sides of $Q^{T} Q = I$ , use "areas multiply" and "flipping across the diagonal keeps area", and you land on $(det Q)^{2} = 1$ , so $det Q = \pm 1$ . Same truth, two languages — the picture and the equation agree.

See also: Rotations and reflections in $\mathbb{R}^2$ and $\mathbb{R}^3$, Determinants — properties, Orthonormal bases & Gram–Schmidt, Eigenvalues and eigenvectors.