Dot product

For the abstract scalar product or inner product, see Inner product space.

In mathematics, the dot product, also known as the scalar product, is an operation which takes two vectors over the real numbers R and returns a real-valued scalar quantity. It is the standard inner product of the Euclidean space.

Definition and examples

The dot product of two vectors (from an orthonormal vector space) a = [a₁, a₂, … , a_n] and b = [b₁, b₂, … , b_n] is by definition:

$\mathbf {a} \cdot \mathbf {b} =\sum _{i=1}^{n}a_{i}b_{i}=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}$

where Σ denotes summation notation.

For example, the dot product of two three-dimensional vectors [1, 3, −5] and [4, −2, −1] is

${\begin{bmatrix}1&3&-5\end{bmatrix}}\cdot {\begin{bmatrix}4&-2&-1\end{bmatrix}}=(1)(4)+(3)(-2)+(-5)(-1)=3.$

Using matrix multiplication and treating the (column) vectors as n×1 matrices, the dot product can also be written as:

$\mathbf {a} \cdot \mathbf {b} =\mathbf {a} ^{T}\mathbf {b} \,$

where a^T denotes the transpose of the matrix a.

Using the example from above, this would result in a 1×3 matrix (i.e., vector) multiplied by a 3×1 vector (which, by virtue of the matrix multiplication, results in a 1×1 matrix, i.e., a scalar):

${\begin{bmatrix}1&3&-5\end{bmatrix}}{\begin{bmatrix}4\\-2\\-1\end{bmatrix}}={\begin{bmatrix}3\end{bmatrix}}.$

For a complex vector, the dot product is defined as

$\mathbf {a} \cdot \mathbf {b} =\sum {a_{i}b_{i}^{*}}$

Geometric interpretation

|A|•cos(θ) is the scalar projection of A onto B

In Euclidean geometry, the dot product, length, and angle are related: For a vector a, a•a is the square of its length, and, more generally, if b is another vector

$\mathbf {a} \cdot \mathbf {b} =|\mathbf {a} |\,|\mathbf {b} |\cos \theta \,$

where

|a| and |b| denote the length (magnitude) of a and b

θ is the angle between them.

Since |a|cos(θ) is the scalar projection of a onto b, the dot product can be understood geometrically as the product of the length of this projection and the length of b.

As the cosine of 90° is zero, the dot product of two perpendicular vectors is always zero. If a and b have length one (i.e. they are unit vectors), the dot product simply gives the cosine of the angle between them. Thus, given two vectors, the angle between them can be found by rearranging the above formula:

$\theta =\arccos \left({\frac {{\mathbf {a}}\cdot {\mathbf {b}}}{|{\mathbf {a}}||{\mathbf {b}}|}}\right).$

Sometimes these properties are also used for defining the dot product, especially in 2 and 3 dimensions; this definition is equivalent to the above one. For higher dimensions the formula can be used to define the concept of angle.

The geometric properties rely on the basis of vectors being perpendicular and having unit length. Either we start with such a basis, or we use an arbitrary basis and define length and angle (including perpendicularity) with the above.

As the geometric interpretation shows, the dot product is invariant under isometric changes of the basis: rotations, reflections, and combinations, keeping the origin fixed.

In other words, and more generally for any n, the dot product is invariant under a coordinate transformation based on an orthogonal matrix. This corresponds to the following two conditions:

• the new basis is again orthonormal (i.e., it is orthonormal expressed in the old one)
• the new base vectors have the same length as the old ones (i.e., unit length in terms of the old basis)

The dot product in physics

In physics, magnitude is a scalar in the physical sense, i.e. a physical quantity independent of the coordinate system, expressed as the product of a numerical value and a physical unit, not just a number. The dot product is also a scalar in this sense, given by the formula, independent of the coordinate system. The formula in terms of coordinates is evaluated with not just numbers, but numbers times units. Therefore, although it relies on the basis being orthonormal, it does not depend on scaling.

Example:

• Mechanical work is the dot product of force and displacement.

Properties

This is a very useful identity (also known as Lagrange's formula) involving the dot- and cross-products. It is written as

a × (b × c) = b(a · c) − c(a · b),

which is easier to remember as “BAC minus CAB”, keeping in mind which vectors are dotted together. This formula is commonly used to simplify vector calculations in physics.

Derivatives

If a and b are functions, then the derivative of a • b is a' • b + a • b'.

Matrix representation

An inner product can be represented as a matrix. For example, given two vectors

$\mathbf {a} ={\begin{bmatrix}a_{u}\\a_{v}\\a_{w}\end{bmatrix}},\qquad \mathbf {b} ={\begin{bmatrix}b_{u}\\b_{v}\\b_{w}\end{bmatrix}}$

with respect to the basis set $\mathrm {S}$

$\mathrm {S} =\{\mathbf {u} ,\mathbf {v} ,\mathbf {w} \}=\left\{{\begin{bmatrix}u_{1}\\u_{2}\\u_{3}\end{bmatrix}},{\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}},{\begin{bmatrix}w_{1}\\w_{2}\\w_{3}\end{bmatrix}}\right\}$

any inner product can be represented as follows:

$\langle \mathbf {a} \,,\mathbf {b} \rangle =\mathbf {a} ^{T}\cdot \mathbf {M} \cdot \mathbf {b}$

where M is the 3x3 matrix representation of the inner product. Given the matrix of the inner product through S called $\mathbf {C} _{S}$, M can be calculated by solving the following system of equations.

$\mathbf {C} _{S}={\begin{bmatrix}\langle \mathbf {u,u} \rangle &\langle \mathbf {u,v} \rangle &\langle \mathbf {u,w} \rangle \\\langle \mathbf {v,u} \rangle &\langle \mathbf {v,v} \rangle &\langle \mathbf {v,w} \rangle \\\langle \mathbf {w,u} \rangle &\langle \mathbf {w,v} \rangle &\langle \mathbf {w,w} \rangle \end{bmatrix}}={\begin{bmatrix}\mathbf {u} ^{T}\cdot \mathbf {M} \cdot \mathbf {u} &\mathbf {u} ^{T}\cdot \mathbf {M} \cdot \mathbf {v} &\mathbf {u} ^{T}\cdot \mathbf {M} \cdot \mathbf {w} \\\mathbf {v} ^{T}\cdot \mathbf {M} \cdot \mathbf {u} &\mathbf {v} ^{T}\cdot \mathbf {M} \cdot \mathbf {v} &\mathbf {v} ^{T}\cdot \mathbf {M} \cdot \mathbf {w} \\\mathbf {w} ^{T}\cdot \mathbf {M} \cdot \mathbf {u} &\mathbf {w} ^{T}\cdot \mathbf {M} \cdot \mathbf {v} &\mathbf {w} ^{T}\cdot \mathbf {M} \cdot \mathbf {w} \end{bmatrix}}$

Example

Given a basis set

$\mathrm {S} =\{\mathbf {u} ,\mathbf {v} ,\mathbf {w} \}=\left\{{\begin{bmatrix}1\\0\\0\end{bmatrix}},{\begin{bmatrix}1\\1\\0\end{bmatrix}},{\begin{bmatrix}1\\1\\1\end{bmatrix}}\right\}$

and a matrix of the inner product through $\mathrm {S}$

$\mathbf {C} _{S}={\begin{bmatrix}5&2&0\\2&6&2\\0&2&7\end{bmatrix}}$

we can set each element of C_S equal to the inner product of two of the basis vectors as follows

$\mathbf {C} _{S}[i,j]=\langle \mathrm {S} [i],\mathrm {S} [j]\rangle$

$\mathbf {C} _{S}[0,0]=5=\langle \mathbf {u,u} \rangle ={\begin{bmatrix}1&0&0\end{bmatrix}}\cdot \mathrm {M} \cdot {\begin{bmatrix}1\\0\\0\end{bmatrix}}$

$\mathbf {C} _{S}[0,1]=2=\langle \mathbf {u,v} \rangle ={\begin{bmatrix}1&0&0\end{bmatrix}}\cdot \mathrm {M} \cdot {\begin{bmatrix}1\\1\\0\end{bmatrix}}$

$\cdots$

which gives nine equations and nine unknowns. Solving these equations yields

$\mathbf {M} ={\begin{bmatrix}5&-3&-2\\-3&7&-2\\-2&-2&9\end{bmatrix}}$

Generalization

The inner product generalizes the dot product to abstract vector spaces and is normally denoted by <a , b>. Due to the geometric interpretation of the dot product the norm ||a|| of a vector a in such an inner product space is defined as

$\|\mathbf {a} \|={\sqrt {\langle \mathbf {a} \,,\mathbf {a} \rangle }}$

such that it generalizes length, and the angle θ between two vectors a and b by

$\cos {\theta }={\frac {\langle \mathbf {a} \,,\mathbf {b} \rangle }{\|\mathbf {a} \|\,\|\mathbf {b} \|}}.$

In particular, two vectors are considered orthogonal if their dot product is zero

$\mathbf {a} \cdot \mathbf {b} =0.$

The Frobenius inner product defines an inner product on matrices as though they are two vectors, summing up the products of corresponding components.

Proof of the geometric interpretation

Note: This proof is shown for 3-dimensional vectors, but is readily extendable to n-dimensional vectors.

Consider a vector

$\mathbf {v} =v_{1}\mathbf {i} +v_{2}\mathbf {j} +v_{3}\mathbf {k} .\,$

Repeated application of the Pythagorean theorem yields for its length v

$v^{2}=v_{1}^{2}+v_{2}^{2}+v_{3}^{2}.\,$

But this is the same as

$\mathbf {v} \cdot \mathbf {v} =v_{1}^{2}+v_{2}^{2}+v_{3}^{2},\,$

so we conclude that taking the dot product of a vector v with itself yields the squared length of the vector.

Lemma 1

$\mathbf {v} \cdot \mathbf {v} =v^{2}.\,$

Now consider two vectors a and b extending from the origin, separated by an angle θ. A third vector c may be defined as

$\mathbf {c} \ {\stackrel {\mathrm {def} }{=}}\ \mathbf {a} -\mathbf {b} .\,$

creating a triangle with sides a, b, and c. According to the law of cosines, we have

$c^{2}=a^{2}+b^{2}-2ab\cos \theta .\,$

Substituting dot products for the squared lengths according to Lemma 1, we get

$\mathbf {c} \cdot \mathbf {c} =\mathbf {a} \cdot \mathbf {a} +\mathbf {b} \cdot \mathbf {b} -2ab\cos \theta .\,$                   (1)

But as c ≡ a − b, we also have

$\mathbf {c} \cdot \mathbf {c} =(\mathbf {a} -\mathbf {b} )\cdot (\mathbf {a} -\mathbf {b} )\,$,

which, according to the distributive law, expands to

$\mathbf {c} \cdot \mathbf {c} =\mathbf {a} \cdot \mathbf {a} +\mathbf {b} \cdot \mathbf {b} -2(\mathbf {a} \cdot \mathbf {b} ).\,$                     (2)

Merging the two c • c equations, (1) and (2), we obtain

$\mathbf {a} \cdot \mathbf {a} +\mathbf {b} \cdot \mathbf {b} -2(\mathbf {a} \cdot \mathbf {b} )=\mathbf {a} \cdot \mathbf {a} +\mathbf {b} \cdot \mathbf {b} -2ab\cos \theta .\,$

Subtracting a • a + b • b from both sides and dividing by −2 leaves

$\mathbf {a} \cdot \mathbf {b} =ab\cos \theta .\,$

Q.E.D.

See also

• Cross product
• Matrix multiplication
• Cauchy–Schwarz inequality

External links

• Weisstein, Eric W. "Dot product". MathWorld.
• Java demonstration of dot product
• Another Java demonstration of dot product
• Explanation of dot product including with complex vectors
• "Dot Product" by Bruce Torrence, The Wolfram Demonstrations Project, 2007.