What Is the Tensor Product of a Vector with Itself?

If you tensor a vector with itself, the result is a second-order tensor, not a scalar. In a basis, you can calculate it by multiplying every component of the vector by every component again. This article explains the notation, works through a numerical example, and also covers what changes when the second factor is a different tensor.

What does v ⊗ v mean?

Let V be a vector space over a field, and let v be an element of V. The tensor product of v with itself is written

$$v \otimes v \in V \otimes V.$$

The tensor product is bilinear: scaling either input scales the result, and addition in either input distributes over the operation. It combines two vector inputs into an object in the tensor-product space. It does not require a length, angle, or inner product.

Choose a basis $e_1,\ldots,e_n$ for V and write the vector as

$$v = \sum_{i=1}^n v_i e_i.$$

Then

$$v \otimes v = \sum_{i=1}^n \sum_{j=1}^n v_i v_j (e_i \otimes e_j).$$

Thus the component with indices $i,j$ is $v_i v_j$. If you arrange these components in a square array, the result is the outer-product matrix $vv^{\mathsf T}$.

Worked example

Take $v=(2,-1)$ in $\mathbb{R}^2$. Its tensor product with itself has four components:

$$v \otimes v = \begin{pmatrix}2\\-1\end{pmatrix} \begin{pmatrix}2&-1\end{pmatrix} = \begin{pmatrix} 4&-2\\ -2&1 \end{pmatrix}.$$

For instance, the upper-right entry is $2\cdot(-1)=-2$. This matrix is symmetric, and its rank is one because the vector is nonzero. If instead $v=0$, then $v\otimes v=0$.

The matrix form depends on the chosen basis, but the tensor $v\otimes v$ is the underlying coordinate-independent object. In an orthonormal basis of a Euclidean space, the matrix also represents the linear map $x\mapsto \langle v,x\rangle v$. That interpretation uses an inner product to identify vectors with covectors; it is an extra structure, not part of the tensor product itself.

How it differs from a dot product

A dot product takes two vectors and, when an inner product is defined, returns a scalar. For the example above,

$$v\cdot v = 2^2+(-1)^2=5,$$

whereas $v\otimes v$ is a two-index tensor represented by a $2\times2$ matrix. These operations answer different questions: the dot product measures a scalar relationship, while the tensor product retains the component combinations as a larger object.

In three dimensions, the cross product is different again: it returns a vector and is a special operation on $\mathbb{R}^3$, not the general tensor product.

What if the other factor is a different tensor?

The tensor product need not use the same vector twice. For $v\in V$ and $w\in W$, the result is $v\otimes w\in V\otimes W$. If their coordinates are $v_i$ and $w_j$, its components are $v_iw_j$. For example, vectors with 2 and 3 components produce a $2\times3$ array of pairwise products in the corresponding bases.

More generally, if $T$ is a tensor with components $T_{j_1\ldots j_k}$, then $v\otimes T$ has components

$$(v\otimes T)_{i j_1\ldots j_k}=v_iT_{j_1\ldots j_k}.$$

This adds a vector index to the tensor. The order of the factors matters: $v\otimes T$ and $T\otimes v$ belong to spaces with their factors in different orders, though a natural reordering isomorphism can relate them. For tensors involving both vectors and covectors, keep track of each factor’s type as well as its order.

Common mistakes

• Confusing the outer product with the dot product: $v\otimes v$ has two indices; $v\cdot v$ is a scalar.
• Thinking the result is always a matrix: A matrix is a coordinate representation in a chosen basis. Abstractly, the result lies in $V\otimes V$.
• Assuming an inner product is required: The tensor product is defined for vector spaces without one. An inner product is needed only for interpretations that identify vectors and covectors or measure angles and lengths.
• Assuming tensor products are limited to vectors: You can tensor vectors with other tensors; the result has the combined factors and indices.

The standard linear-algebra reading of “a vector tensor-multiplied by itself” is therefore its tensor square $v\otimes v$. If “another tensor” means a separate factor, form the tensor product with that tensor and retain the order and type of the factors.