Everyone who has taken a linear algebra or physics course at university remembers this strange dualism. We were taught that vectors have TWO types of products. The first, scalar, takes two vectors and outputs a number. Geometrically, it's about projections and angles. The second, vector, also takes two vectors and... suddenly spits out a third vector, perpendicular to the first two. And this trick only works in 3D and 7D.

It always seemed like some kind of mathematical 'crutch'.

Why is it so complicated? Why two different products for different tasks? Why does one depend on the cosine and the other on the sine?

What if I told you that they really are 'crutches'? That there is a single, universal, and elegant geometric product, which includes both of these cases (and much more), and which is based on a single, crystal-clear idea. An idea that changes the way we look at the very essence of mathematics.

This article is an invitation to the world of Geometric Algebra. We are going to reinvent multiplication.

A note on notation.

Since there are several types of multiplication in this article, we will distinguish them as follows

or no sign — the new geometric product

— scalar product

∧ — exterior product

Part 1. The Philosophy — A Transformation for Every Object

After all, there's something to this student joke, popular at MIPT: "a vector up every lecturer's..., and for Dmitry Vladimirovich — an orthonormal basis" .

This refers to Beklemishev, thanks to whom the teaching of linear algebra became so closely linked with analytic geometry, not only at MIPT but throughout the entire USSR. When he was just a 25-year-old graduate student at the MSU Faculty of Mechanics and Mathematics, he had the bright idea to teach complex abstractions using a visual science, and he began to implement it at MIPT. He was met with incredible success — by the age of 40, the completely new course created by Beklemishev had spread in various forms throughout the USSR, and he himself taught vector algebra on Soviet television.

It's quite possible that new courses based on geometric algebra could gain similar popularity in our time. After all, here we are reaching a completely new level — we will learn to operate not just with vectors, but with entire bases at once.

Let's start with a simple but profound thought. What if every mathematical object is not just a static entity, but an operator that performs a transformation?

The number 5: it's not just 'five apples'. It's the operator 'increase by a factor of 5'.
The complex number i: it's not just √-1. It's the operator 'rotate by 90 degrees'. Multiply by i again — you'll rotate by 180. Hence i² = -1.

This philosophy transforms mathematics from a descriptive language into a language of action. Objects are verbs, not nouns.

Okay, but what then is a vector? What fundamental geometric transformation should it represent through multiplication?

It can't be a translation — addition is responsible for that (v + a). It can't be simple scaling either. What is the most basic, most elementary geometric action one can perform in space?

Reflection.

The key idea: let's assume that a vector is not an arrow. A vector is a mirror.

A unit vector n defines the orientation of a mirror placed at the origin and perpendicular to this vector. The whole essence of a vector is to reflect other objects.

Вектор n действует как зеркало, отражающее вектор v — Vector n acts as a mirror, reflecting vector v

But why is n — specifically the perpendicular to the mirror?

The vectors v and n lie in the same plane and define a single plane (except when they are parallel). So, the vector n can define any position of the mirror in this plane. It's much more convenient to make the vector n the mirror itself.

Вектор n сам стал зеркалом — Vector n itself has become the mirror

The entire power of geometric algebra will grow from this simple assumption.

Part 2. Inventing Multiplication from Scratch

So, a vector is a mirror. How do we describe the operation of reflection mathematically? We want to find a 'multiplication' () that allows us to reflect one vector v with respect to another vector n.

But instead of fitting a formula to the answer, let's act like true researchers. Let's demand just one, but very important, property from our multiplication: invertibility.

In ordinary algebra, the equation 5x = 10 is solved by division. Division is multiplication by the inverse element (x = 10 * 5⁻¹). We want to solve geometric equations the same way! If ax = b, we want to find x as a⁻¹b.

For this to be possible, for any non-zero vector a, there must exist an inverse a⁻¹. It's logical to assume that a⁻¹ is somehow related to a. The simplest option is that a⁻¹ is proportional to a.

This means that the product of a vector with itself, a*a, must be a simple scalar. After all, if a*a = k (where k is a number), then we can easily find a⁻¹ = a/k. What is the most important scalar associated with a vector? Its squared length, of course!

Postulate 1 (and the only one):

The product of a vector with itself is a scalar equal to its squared length.

That's it. We don't need anything else. We will derive everything else.

Let's see what this seemingly modest postulate leads to. Take two vectors a and b and expand the square of their sum, just like in school:

Using our postulate, let's replace a*a with |a|² and b*b with |b|²:

On the other hand, (a+b) is also a vector. And the square of any vector, according to our postulate, is its squared length:

And we know the squared length of the sum of vectors from school by the law of cosines:

But is exactly the scalar product !

We get: .

Now let's equate both expressions for :

We cancel out the extra terms and get:

The symmetric part of our new multiplication (ab+ba) turned out to be the good old scalar product! But what is ab itself?

Let's consider the most interesting case: a and b are perpendicular.

Let our basis vectors be e₁ and e₂. Their scalar product is zero.

Let's substitute this into our formula

And this means...

We didn't make this up. We didn't pull it out of thin air. We derived the anticommutativity for orthogonal vectors from the simplest requirement of multiplication invertibility.

Part 3. Exploring the Plane — The Birth of a New Number

Let's stay in the plane. We have a basis and three rules of the game:

Let's take two arbitrary vectors and and just multiply them head-on, like polynomials:

Let's apply our rules: , and :

Look what happened! Our geometric product ab has split into two parts on its own:

Scalar part: (a₁b₁ + a₂b₂). This is exactly the scalar product a·b!
Non-scalar part: (a₁b₂ - a₂b₁)(e₁e₂).
The coefficient (a₁b₂ - a₂b₁) is a value proportional to the area of the parallelogram formed by the vectors a and b (and the sine of the angle between them).

And what is this mysterious beast e₁e₂?

It's not a scalar.
It's not a vector.
Let's calculate its square:
Using anticommutativity:

An object that squares to -1.

We've seen this somewhere before... Yes, it's an analog of the imaginary unit i!

This object I = e₁e₂ is called a bivector. It's not an arrow. It is the oriented plane e₁e₂ itself.

It says: 'I am the plane in which rotation goes from e₁ to e₂'.

Our geometric product ab is a multivector.

(scalar part) + (bivector part)

Here is the exterior product.

It represents an oriented area.

Orientation here means that if you swap the two vectors, it gets multiplied by .

Part 4. Discovering the Reflection Formula. A Path of Trial and Error.

So, we agreed that the vector a is a mirror, and reflection is its main purpose. Let's now find the mathematical language for this.

Our goal is to find a formula that takes an arbitrary vector v and reflects it in the mirror defined by the unit vector a.

For simplicity, let's take our plane with the basis e₁, e₂.

Let the mirror be the e₁ axis itself. Which vector v will we reflect?

Let's take the most general one:

What do we expect to get after reflecting in the 'mirror' e₁? The component along e₁ should remain, and the component along e₂ (perpendicular to the mirror) should be inverted. Expected result:

Now let's try to find this formula using our new geometric product.

Attempt #1: Left Multiplication

The simplest thing that comes to mind is to just multiply v by e₁ on the left:

(As a reminder, e₁e₁ = 1 and e₁e₂ = -e₂e₁)

What is this? A scalar v₁ minus a bivector -v₂e₁e₂. It's not even a vector. An interesting result, but definitely not our reflection. Miss.

Attempt #2: Right Multiplication

Okay, let's try multiplying on the right:

Again, a scalar plus a bivector. Another miss.

Attempt #3: The 'Sandwich'

It seems simple multiplication is not enough. We are getting objects of a higher 'grade' (bivectors). Maybe we need to get rid of them somehow? What if we 'wrap' our vector v from both sides? Let's try the construction e₁ve₁:

Bingo!

We got exactly the result we expected.

The Reflection Formula:

the reflection of vector v in the mirror defined by the unit vector a, is described by the sandwich product:

Does this work for any vector, not just the basis vector e₁?

Yes! We can prove that if we decompose v into components and relative to any unit vector a, the operation ava will always preserve and invert , which is the definition of reflection.

Proof: Why ava is the Universal Reflection Formula

Okay, our 'sandwich' trick e₁ve₁ worked perfectly for reflection across the e₁ axis. But was it just a lucky coincidence? Let's prove that the formula v' = ava works for reflection with respect to any unit vector a.

This proof will not just show that the formula is correct. It will reveal how deeply our new multiplication 'understands' geometry.

Step 1: Decomposing the Vector into Components

Let's decompose our vector v into two components relative to the mirror-vector a:

v_∥ (v-parallel).
The component v that is parallel to a. This is simply the projection of v onto a.
v_⊥ (v-perpendicular).
The component v that is perpendicular to a. This is what remains of v if you subtract the parallel part from it.

Thus:

Definition of Reflection:

we want our transformation to leave unchanged (since it already lies on the 'line' of the mirror) and invert (since it is perpendicular to the mirror).

Our goal is to prove that

Step 2: Let's see how ava acts on each component separately

Since our multiplication is distributive, we can write:

Now let's analyze each term.

Analysis of (the parallel part):

By definition, is parallel to a. This means that is simply a multiplied by some scalar.
.
Key property: scalars commute with everything. They can be factored out and moved anywhere.
Let's calculate: .
What is aaa? It's (aa)a. And since a is a unit vector, then aa = a² = 1.
Therefore, aaa = (1)a = a.
Returning to our expression:

Result A:

The operation ava preserves the parallel component. The first part is proven!

Analysis of (the perpendicular part):

By definition, and a are orthogonal.
And what do we know about the geometric product of two orthogonal vectors? They anticommute! This is a fundamental property we derived earlier.

Now let's apply this to our expression We can group it as .
Let's replace with its anticommuted equivalent
Factor out the minus sign and regroup: .
And again, since a is a unit vector, aa = a² = 1.
We get: .

Result B:

The operation ava inverts the perpendicular component. The second part is proven!

Step 3: Putting It All Together

Now let's return to our sum:

We have just shown that

Substituting these results:

This is precisely the mathematical definition of the reflection of vector v across the line defined by vector a.

Proof complete.

Part 5. Two Reflections = A Rotation. The Birth of the Rotor.

We started by saying that a vector is a mirror, and reflection is a fundamental operation. What happens if you perform two reflections in a row? Any schoolkid who has played with two mirrors, to cheat on a test with them, knows: it will be a rotation!

(first reflection)

Let's group the parentheses differently, using the associativity of our multiplication:

Let's look at this new object

This is the result of the geometric product of two vectors. It's no longer just a vector or just a bivector, but their sum — a multivector. Let's call it a rotor.

And what is ab in this formula? Let's multiply it by ba:

This means ab is exactly the inverse element of ba! That is, ab = (ba)⁻¹ = R⁻¹.

Our formula for two reflections (that is, for a rotation!) turns into something incredibly elegant.

The Rotation Formula:

We didn't use any sines, cosines, or matrices. We just said, 'a vector is a mirror,' found the formula for one reflection, and applied it twice. Rotations came to us practically for free, as a byproduct.

Part 6. What's the Deception in the Vector Cross Product?

So far, we've been saying 'vector a is a mirror'. And the formula v' = ava worked. But, as we found out, it describes a reflection across the line defined by vector a.

But what about our everyday 'mirror', which is a plane?

For that, as we recall, we need the formula with a minus sign: v' = -ava. Here, the vector a defines the normal to the mirror-plane.

So which formula is 'correct'? Both! They describe two different types of reflections.

It turns out that any complex motion (rotation, translation) in n-dimensional space can be decomposed into a sequence of simple reflections. This is a fundamental result known as Householder transformations.

Reflections are the atoms from which all motions are composed.

And here a surprise awaits us, related to the dimension of space.

In 2D space: a 'mirror-plane' is just a line. Its normal is a vector. The line and its normal vector uniquely define each other.
In 3D space: a 'mirror-plane' is a plane. Its normal is a vector. The plane and its normal vector also uniquely define each other. What else defines a plane? A bivector! In 3D space, every bivector (plane) can be associated with a unique perpendicular vector (pseudovector). This is the very trick on which the vector cross product is based. Because of this 'accidental' coincidence in 3D, you can substitute bivectors for vectors and vice versa.
In 4D and higher: a 'mirror-plane' is a hyperplane (for example, a 3D volume in 4D space). Its normal is a vector. But a bivector still only defines a 2D plane and rotations within it!

Conclusion.

In spaces of dimension higher than 3, bivectors (generators of rotations) and hyperplanes (mirrors for reflections) are completely different objects. A bivector defines a rotation in a specific 2D plane, while a reflection occurs in the entire (n-1)-dimensional space at once.

The deceptive simplicity of our 3D world, where a plane can be described by both a bivector and a normal vector, long concealed this fundamental difference and forced us to use the 'crutch' of the vector cross product.

Geometric algebra puts everything in its proper place.

Part 7. Pauli Matrices from Thin Air

Those familiar with quantum mechanics know the Pauli matrices σ₁, σ₂, σ₃. This is a set of 2x2 matrices that form the basis for describing spin. They have amusing properties: σ₁² = σ₂² = σ₃² = I (the identity matrix) and they anticommute (σ₁σ₂ = iσ₃ = -σ₂σ₁).

Where do they come from? In geometric algebra, they appear on their own.

Let's just identify our basis vectors e₁, e₂, e₃ with the Pauli matrices!

Let's check our rules:

? Yes, .
for ? Yes, they anticommute.

What then are the bivectors?

And the trivector (pseudoscalar)?

.

The trivector of the space is the imaginary unit i multiplied by the identity matrix!

It turns out the Pauli matrices are not some abstract mathematical trick. They are the matrix representation of the algebra of our three-dimensional Euclidean space. All those i in their products are simply a consequence of the fact that the product of two vectors generates a bivector, which in is dual to a vector. Geometric algebra reveals the geometric meaning behind the spin matrices.

Part 8. Let's Derive the Elements of the Pauli Matrices Ourselves.

Our task: to find three 2x2 matrices that behave exactly like our orthonormal vectors e₁, e₂ and e₃. That is, they must satisfy the 'rules of the game' of geometric algebra:

(where I is the 2x2 identity matrix).
for (anticommutativity).

We will be working with column vectors of the form [α, β], where α and β are complex numbers. This is the standard basis for describing spin 'up' and spin 'down' .

Step 1. σ₃ — The 'Measurement' Operator along the Z-axis

In quantum mechanics, (or ) is the operator that measures the spin projection onto the Z-axis. Its eigenstates are spin 'up' and spin 'down'.

When σ₃ acts on the 'up' state ↑ ([1, 0]), it should yield .
When σ₃ acts on the 'down' state ↓ ([0, 1]), it should yield

Let's write this in matrix form.

$\sigma_3 = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

$\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \left(\begin{array}{c} 1 \\0 \end{array}\right) = \left(\begin{array}{c} a \cdot 1 + b \cdot 0 \\ c \cdot 1 + d \cdot 0 \end{array}\right) = \left(\begin{array}{c} a \\ c \end{array}\right)$

We want to get . This means

$\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \left(\begin{array}{c} 0 \\ 1 \end{array}\right) = \left(\begin{array}{c} a \cdot 0 + b \cdot 1 \\ c \cdot 0 + d \cdot 1 \end{array}\right) = \left(\begin{array}{c} b \\ d \end{array}\right)$

We want to get . This means .

Assembling the matrix:

$\sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

Let's check:

$σ₃² = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} * \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}= I.$

The property holds. The first matrix was found purely from its physical meaning.

Step 2: σ₁ — The 'Flip' Operator

What should the operator e₁ (or σ₁) do? In our geometric world, it is perpendicular to e₃. What is the simplest action it can perform on the 'up/down' states? Flip them!

σ₁ should turn 'up' into 'down': σ₁ * [1, 0] = [0, 1].
σ₁ should turn 'down' into 'up': σ₁ * [0, 1] = [1, 0].

Again, we look for the matrix

$\sigma_1 = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ :

$\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \left(\begin{array}{c} 1 \\0 \end{array}\right) = \left(\begin{array}{c} a \cdot 1 + b \cdot 0 \\ c \cdot 1 + d \cdot 0 \end{array}\right) = \left(\begin{array}{c} a \\ c \end{array}\right)$

$\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \left(\begin{array}{c} 0 \\ 1 \end{array}\right) = \left(\begin{array}{c} a \cdot 0 + b \cdot 1 \\ c \cdot 0 + d \cdot 1 \end{array}\right) = \left(\begin{array}{c} b \\ d \end{array}\right)$

We want to get . This means
We want to get . This means

Assembling the matrix:

$\sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Let's check :

$σ₁² = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} * \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}= I.$

Done.

You can check the anticommutativity with on your own.

Step 3: σ₂ — The Magic of Geometric Algebra

We could try to guess the geometric meaning of , but we have a better way. In Geometric Algebra, objects generate each other!

We know that in 3D, the bivector is dual to the vector (up to a sign and ).

Let's define through other vectors and the pseudoscalar

For example, .

In matrix representation, . Then

Let's just multiply our 'vectors' σ₁ and σ₃ and see what we get.

Their product should be somehow related to

$σ₁σ₃ = \begin{pmatrix} 0 & -1 \\ 1& 0 \end{pmatrix}.$

This is not , it's the bivector .

Let's use the property .

From this, .

$σ₂ = \begin{pmatrix} 0 & -i*(-1) \\ -i*1& 0 \end{pmatrix} = \begin{pmatrix} 0 & -i*(-1) \\ -i*1& 0 \end{pmatrix} = \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}.$

Wait a minute, let's change the order. .

From this, .

$σ₂ = i * \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}$

What if we define through the product of and ? That is,

?

In matrix form:

$\sigma_1 \sigma_3 = i \sigma_2?$

Then

$\sigma_2 = -i \sigma_1 \sigma_3 = -i \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}$

Wait a minute. The standard Pauli matrix σ₂ has the form $\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}.$

Where did we go wrong?

Nowhere! The choice of signs (σ₁σ₂ = iσ₃ or σ₁σ₂ = -iσ₃) is a matter of chirality convention (right-handed or left-handed coordinate system). Let's just define e₂ through the product that gives the standard result.

Let's recall that the bivector e₃e₁ corresponds to iσ₂. Let's demand just that:

$σ₂ = -i * \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$

$\sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$

There it is, the third matrix, obtained as a product of the interaction of the other two.

$\begin{pmatrix} 0 & -i \\ i& 0 \end{pmatrix}$

Let's check :

$\begin{pmatrix} 0 & -i \\ i& 0 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i& 0 \end{pmatrix} = \begin{pmatrix} -i*i & 0 \\ 0& -i*i \end{pmatrix} = \begin{pmatrix} -(-1) & 0 \\ 0& -(-1) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0& 1 \end{pmatrix} = I.$

Done. Anticommutativity with the others is also easily checked.

Conclusion

We didn't 'find' the Pauli matrices. We constructed them.

we built from the requirement of being the 'up/down' measurement operator.
we built from the requirement of being the 'flip' operator between 'up' and 'down'.
we obtained as the geometric product of σ₃ and σ₁ (with normalization by i).

The Pauli matrices are not magic. They are the inevitable consequence of trying to represent three orthogonal directions and their products as 2x2 matrices. Geometric algebra reveals their internal logic and shows that they are just one of the possible 'embodiments' of the eternal geometry of space.

Part 9. Let's Summarize.

Let's summarize what we have discovered:

There is only one multiplication. The geometric product ab. It contains all the information about the relative orientation of vectors.
It splits into parts. . Its symmetric part — is the scalar product (projection). Its antisymmetric part— is the bivector (oriented area).
It generates new objects. By multiplying vectors, we get bivectors, which in 2D behave like and define the plane of rotation.
It encodes transformations. The product of two vectors creates a rotor — a rotation operator. The formula works always and everywhere, from 2D graphics to the special theory of relativity.

We started with a simple idea and, by following logic, reassembled all of vector algebra, making it simpler, more powerful, and more intuitive. This is geometric algebra. And we have only just opened the door to this amazing world.

This isn't some esoteric mathematics.

GA is actively used in computer graphics, robotics, electrodynamics, and quantum mechanics. Perhaps if universities started with it, we would have many more engineers and physicists who truly feel geometry.

And you can also rewrite all of physics with it!

Let's forget everything about scalar and vector products. There's a much better way