# Are tensors vectors?

Take the definition of a tensor as a multilinear map $$T: (V)^n \times (V^*)^m \rightarrow \mathbb R$$. One may then define the addition of two tensors as the sum of their outputs on their input spaces (assuming of course the tensors have the same rank), and scalar multiplication as the output scaled by the factor used (if $$T$$ maps its input to $$b$$, then $$aT$$ maps its input to $$ab$$).

As one can easily check, tensors would then satisfy all of the vector space axioms, so then a tensor could be considered a specific type of vector. This bothered me, though, because vectors are often defined as objects with magnitude and direction in less formal contexts, and tensors can often be informally considered to have many different directions.

This leads me to the writing of this question. What distinguishes a tensor from a classical, "ordinary" vector? How can one derive the definition of a vector as something with magnitude and direction from the definition of a vector space, and where does this definition break down for tensors (if anywhere)?

• Well, I’m a purely-algebraically-oriented guy, but as far as I’m concerned, a vector is an element of a vector space. Period. No talk of magnitude and direction. Jan 15, 2021 at 2:21
• The definition of vector space does not inherently give the notion of length or direction. You additionally need an "inner product" to define length. Usually in $\Bbb{R}^n$ you take the ordinary "dot product", and this lets you talk about length.
– Nick
Jan 15, 2021 at 2:22
• One cannot. A vector is an element of a vector space and that's that. The concept of "something that has a magnitude and direction" is usually reserved for elements of $\mathbb{R}^n$ and this description implicitly relies on $\mathbb{R}^n$ being more than just a vector space, namely an inner product space, so it does not generalize. Talking about something with "magnitude and direction" is informal anyhow and not a proper mathematical definition. Jan 15, 2021 at 2:22
• That then raises the question of why a vector space with tensors as its elements can't be endowed with an inner product. Jan 15, 2021 at 2:26
• Responding simply to the title, "no". Jan 15, 2021 at 3:37

What distinguishes a tensor from a vector relies on understanding what you're talking about. Strictly speaking, tensors of a fixed rank form a vector space (over $$\mathbf R$$, say), and thus "tensors are vectors" for pure mathematicians who don't work in anything related to physics or differential geometry.

But nobody means anything like that when they bring up the issues of tensors vs. vectors. This is why "tensors are not vectors" for physicists, and it explains why you are getting the answers "yes" and "no" in the comments.

Look at how tensors are built. You start with a vector space $$V$$ and take a tensor product of tensor powers of $$V$$ and the dual space $$V^*$$, say $$V^{\otimes n} \otimes (V^*)^{\otimes m}$$. Elements of $$V^{\otimes n} \otimes (V^*)^{\otimes m}$$ are called tensors, elements of $$V$$ are called vectors, and elements of $$V^*$$ are called covectors. In this setting, where the word vector means "element of $$V$$", the objects in $$V^{\otimes n} \otimes (V^*)^{\otimes m}$$ are not vectors (unless $$n = 1$$ and $$m = 0$$). Vector is just a synonym for "element of $$V$$". That's all there is to it.

You ask in a comment why tensors can't be given an inner product. That's incorrect: they can, provided $$V$$ is given an inner product first. If $$V$$ and $$W$$ are both real vector spaces and they are each given an inner product $$\langle \cdot,\cdot\rangle_V$$ and $$\langle \cdot,\cdot\rangle_W$$, then there is a unique inner product on $$V \otimes W$$ for which $$\langle v \otimes w,v'\otimes w'\rangle = \langle v,v'\rangle_V\langle w,w'\rangle_W$$ on elementary tensors $$v \otimes w$$ and $$v' \otimes w'$$ in $$V \otimes W$$. Check that. That is, figure out why there exists exactly one inner product on $$V \otimes W$$ having the above behavior on pairs of elementary tensors in $$V \otimes W$$.

By iterating this construction, when $$V$$ has an inner product $$\langle \cdot,\cdot\rangle_V$$, each tensor power $$V^{\otimes n}$$ has a unique inner product that looks like $$\langle v_1 \otimes \cdots \otimes v_n,v_1'\otimes \cdots \otimes v_n'\rangle = \langle v_1,v_1'\rangle_V\cdots \langle v_n,v_n'\rangle_V$$ on pairs of elementary tensors $$v_1 \otimes \cdots \otimes v_n$$ and $$v_1'\otimes \cdots \otimes v_n'$$.

When $$V$$ is finite-dimensional, an inner product $$\langle \cdot,\cdot\rangle_V$$ on $$V$$ gives us an isomorphism $$V \to V^*$$ by $$v \mapsto \langle\cdot,v\rangle_V$$. Using this, we can transport the inner product on $$V$$ over to an inner product on $$V^*$$: for $$\varphi$$ and $$\varphi'$$ in $$V^*$$, simply declare $$\langle \varphi,\varphi'\rangle_{V^*} := \langle v,v'\rangle_V$$ where $$\varphi = \langle \cdot,v\rangle_V$$ and $$\varphi' = \langle \cdot,v'\rangle_V$$. Now you can use this inner product on $$V^*$$ to define an inner product on $$(V^*)^{\otimes m}$$ by the method described above for creating an inner product on tensor powers of $$V$$ from an inner product on $$V$$ (just replace $$V$$ by $$V^*$$ everywhere). And then you can use the inner products on $$V^{\otimes n}$$ and $$(V^*)^{\otimes m}$$ to define an inner product on $$V^{\otimes n} \otimes (V^*)^{\otimes m}$$ by the method described up above for putting an inner product on $$V \otimes W$$ when $$V$$ and $$W$$ each have an inner product. Thus, when a finite-dimensional real vector space $$V$$ has an inner product, every space of tensors $$V^{\otimes n} \otimes (V^*)^{\otimes m}$$ for fixed $$m$$ and $$n$$ gets an inner product from the inner product on $$V$$. Concretely, for elementary tensors $$t = v_1 \otimes \cdots \otimes v_n \otimes \varphi_1 \otimes \cdots \otimes \varphi_m$$ and $$t' = v_1' \otimes \cdots \otimes v_n' \otimes \varphi_1' \otimes \cdots \otimes \varphi_m'$$ in $$V^{\otimes n} \otimes (V^*)^{\otimes m}$$, $$\langle t,t'\rangle = \prod_{i=1}^n \langle v_i,v_i'\rangle_V \prod_{j=1}^m \langle w_j,w_j'\rangle_V,$$ where $$\varphi_j = \langle \cdot,w_j\rangle_V$$ and $$\varphi_j' = \langle \cdot,w_j'\rangle_V$$.

• When considering infinite-dimensional inner-product spaces, $V^*$ typically means the continuous dual (well, this is also true, trivially, in the finite-dimensional case). So the restriction to finite-dimensional $V$ in the second half of your post is not really necessary (given that you note that $V^*$ is understood to be the continuous dual). Sep 19, 2022 at 1:02
• Also, there is a typo: "to defined" should be "to define" Sep 19, 2022 at 1:04
• Well, there is a small caveat in the infinite-dimensional case: if $V$ is incomplete, then it is only isomorphic to a dense subspace of $V^*$. But the inner product can be uniquely extended (or alternatively, one can complete $V$ and use the natural isomorphism $\hat V\cong V^*$), so it does not really matter. Sep 19, 2022 at 1:19

This is basically contained in KCd's answer but it's worth saying explicitly: what almost no one will say explicitly about tensors is that they are a fundamentally relative notion, and what they are defined relative to is a "base" vector space $$V$$. There is in an important sense no such thing as a "tensor" in isolation; the grammar of "tensor" is "a tensor of type $$(m, n)$$ with respect to a vector space $$V$$," namely an element of the tensor product $$V^{\otimes m} \otimes (V^{\ast})^{\otimes n}$$. Often the vector space $$V$$ is understood from context and left implicit.

This can be formalized category-theoretically using the language of functors, which here amounts to making precise what people mean when they talk about tensors transforming in such-and-such a way under change of coordinates - they mean coordinates of the base vector space $$V$$.

It is true that every tensor product of vector spaces is another vector space, and hence every tensor is technically a tensor of type $$(1, 0)$$ (in other words, a vector) with respect to this tensor product of vector spaces. But this is irrelevant; when we talk about tensors we have in mind a particular choice of base vector space $$V$$, and in that context "vector" means an element of $$V$$, as KCd says.

(So, in other words, the word "vector" is overloaded; it refers to several related but actually somewhat different concepts. This sort of thing is unfortunately common.)

• In the usual convention, a tensor of type $(m,n)$ should be an element of $V^{\otimes m}\otimes (V^*)^{\otimes n}$, no? Sep 19, 2022 at 0:08
• Yes, that's a typo, thanks. Sep 19, 2022 at 4:42