What are the Differences Between a Matrix and a Tensor?

What is the difference between a matrix and a tensor? Or, what makes a tensor, a tensor? I know that a matrix is a table of values, right? But, a tensor?

• Continuing with your analogy, a matrix is just a two-dimensional table to organize information and a tensor is just its generalization. You can think of a tensor as a higher-dimensional way to organize information. So a matrix (5x5 for example) is a tensor of rank 2. And a tensor of rank 3 would be a "3D-matrix" like a 5x5x5 matrix. Jun 5, 2013 at 23:03
• I thought a rank n tensor is multi-linear function which takes n vectors and returns a vector of the same vector space? Jan 3, 2017 at 12:37
• One point of confusion is that in machine learning, people often use the term "tensor" when they really mean just "multidimensional array". I disagree that a tensor of rank 3 would be a "3D matrix", but admittedly it's not uncommon to hear the word "tensor" used in this way. stats.stackexchange.com/questions/198061/… Jan 4, 2017 at 10:53
• I think this video is much more useful then any of the answers. If you are still confused watch the video from beginning: youtu.be/f5liqUk0ZTw?t=7m27s Mar 7, 2018 at 5:29
• @Xitcod13: That video doesn't answer the question about the difference between a matrix and a tensor. Indeed, that video doesn't even explicitly mention matrices at all. Jan 16, 2020 at 14:20

Maybe to see the difference between rank 2 tensors and matrices, it is probably best to see a concrete example. Actually this is something which back then confused me very much in the linear algebra course (where we didn't learn about tensors, only about matrices).

As you may know, you can specify a linear transformation $$a$$ between vectors by a matrix. Let's call that matrix $$A$$. Now if you do a basis transformation, this can also be written as a linear transformation, so that if the vector in the old basis is $$v$$, the vector in the new basis is $$T^{-1}v$$ (where $$v$$ is a column vector). Now you can ask what matrix describes the transformation $$a$$ in the new basis. Well, it's the matrix $$T^{-1}AT$$.

Well, so far, so good. What I memorized back then is that under basis change a matrix transforms as $$T^{-1}AT$$.

But then, we learned about quadratic forms. Those are calculated using a matrix $$A$$ as $$u^TAv$$. Still, no problem, until we learned about how to do basis changes. Now, suddenly the matrix did not transform as $$T^{-1}AT$$, but rather as $$T^TAT$$. Which confused me like hell: how could one and the same object transform differently when used in different contexts?

Well, the solution is: because we are actually talking about different objects! In the first case, we are talking about a tensor that takes vectors to vectors. In the second case, we are talking about a tensor that takes two vectors into a scalar, or equivalently, which takes a vector to a covector.

Now both tensors have $$n^2$$ components, and therefore it is possible to write those components in a $$n\times n$$ matrix. And since all operations are linear resp. bilinear, the normal matrix-matrix and matrix-vector products together with transposition can be used to write the operations of the tensor. Only when looking at basis transformations, you see that both are, indeed, not the same, and the course did us (well, at least me) a disservice by not telling us that we are really looking at two different objects, and not just at two different uses of the same object, the matrix.

Indeed, speaking of a rank-2 tensor is not really accurate. The rank of a tensor has to be given by two numbers. The vector to vector mapping is given by a rank-(1,1) tensor, while the quadratic form is given by a rank-(0,2) tensor. There's also the type (2,0) which also corresponds to a matrix, but which maps two covectors to a number, and which again transforms differently.

The bottom line of this is:

• The components of a rank-2 tensor can be written in a matrix.
• The tensor is not that matrix, because different types of tensors can correspond to the same matrix.
• The differences between those tensor types are uncovered by the basis transformations (hence the physicist's definition: "A tensor is what transforms like a tensor").

Of course, another difference between matrices and tensors is that matrices are by definition two-index objects, while tensors can have any rank.

• This is a great answer, because it reveals the question to be wrong. In fact, a matrix is not even a matrix, much less a tensor. Jun 5, 2013 at 23:35
• I'm happy with the first sentence in your comment @RyanReich but utterly confused by: "a matrix is not even a matrix". Could you elaborate or point towards another source to explain this (unless I've taken it out of context?) Thanks.
– AJP
Sep 11, 2016 at 19:54
• @AJP It's been a while, but I believe what I meant by that was that a matrix (array of numbers) is different from a matrix (linear transformation = (1,1) tensor). The same array of numbers can represent several different basis-independent objects when a particular basis is chosen for them. Sep 11, 2016 at 20:00
• Proof basis change rule for quadratic form $q= \begin{bmatrix}x&y&z\end{bmatrix}\begin{bmatrix}\text A\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}$ Nov 16, 2018 at 13:41
• with $P$ being the change of basis matrix between $x,y,z$ and $u,v,w.$ \begin{align}q&= \begin{bmatrix}x&y&z\end{bmatrix}\begin{bmatrix}\text A\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\\ &=\begin{bmatrix}x\\y\\z\end{bmatrix}^\top\begin{bmatrix}\text A\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\\ &=\left(P\begin{bmatrix}u\\v\\w\end{bmatrix}\right)^\top \begin{bmatrix}\text A\end{bmatrix}P\begin{bmatrix}u\\v\\w\end{bmatrix}\\ &=\begin{bmatrix}u&v&w\end{bmatrix}P^\top\begin{bmatrix}\text A\end{bmatrix}P\begin{bmatrix}u\\v\\w\end{bmatrix} \end{align} Nov 16, 2018 at 13:52

Indeed there are some "confusions" some people do when talking about tensors. This happens mainly on Physics where tensors are usually described as "objects with components which transform in the right way". To really understand this matter, let's first remember that those objects belong to the realm of linear algebra. Even though they are used a lot in many branches of mathematics the area of mathematics devoted to the systematic study of those objects is really linear algebra.

So let's start with two vector spaces $$V,W$$ over some field of scalars $$\Bbb F$$. Now, let $$T : V \to W$$ be a linear transformation. I'll assume that you know that we can associate a matrix with $$T$$. Now, you might say: so linear transformations and matrices are all the same! And if you say that, you'll be wrong. The point is: one can associate a matrix with $$T$$ only when one fix some basis of $$V$$ and some basis of $$W$$. In that case we will get $$T$$ represented on those bases, but if we don't introduce those, $$T$$ will be $$T$$ and matrices will be matrices (rectangular arrays of numbers, or whatever definition you like).

Now, the construction of tensors is much more elaborate than just saying: "take a set of numbers, label by components, let they transform in the correct way, you get a tensor". In truth, this "definition" is a consequence of the actual definition. Indeed the actual definition of a tensor is meant to introduce what we call "Universal Property".

The point is that if we have a collection of $$p$$ vector spaces $$V_i$$ and another vector space $$W$$ we can form functions of several variables $$f: V_1\times \cdots \times V_p \to W$$. A function like this will be called multilinear if it's linear in each argument with the others held fixed. Now, since we know how to study linear transformations we ask ourselves: is there a construction of a vector space $$S$$ and one universal multilinear map $$T : V_1 \times \cdots \times V_p \to S$$ such that $$f = g \circ T$$ for some $$g : S \to W$$ linear and such that this holds for all $$f$$? If that's always possible we'll reduce the study of multilinear maps to the study of linear maps.

The happy part of the story is that this is always possible, the construction is well defined and $$S$$ is denoted $$V_1 \otimes \cdots \otimes V_p$$ and is called the tensor product of the vector spaces and the map $$T$$ is the tensor product of the vectors. An element $$t \in S$$ is called a tensor. Now it's possible to prove that if $$V_i$$ has dimension $$n_i$$ then the following relation holds:

$$\dim(V_1\otimes \cdots \otimes V_p)=\prod_{i=1}^p n_i$$

This means that $$S$$ has a basis with $$\prod_{i=1}^p n_i$$ elements. In that case, as we know from basic linear algebra, we can associate with every $$t \in S$$ its components in some basis. Now, those components are what people usually call "the tensor". Indeed, when you see in Physics people saying: "consider the tensor $$T^{\alpha \beta}$$" what they are really saying is "consider the tensor $$T$$ whose components in some basis understood by context are $$T^{\alpha \beta}$$".

So if we consider two vector spaces $$V_1$$ and $$V_2$$ with dimensions respectivly $$n$$ and $$m$$, by the result I've stated $$\dim(V_1 \otimes V_2)=nm$$, so for every tensor $$t \in V_1 \otimes V_2$$ one can associate a set of $$nm$$ scalars (the components of $$t$$), and we are obviously allowed to plug those values into a matrix $$M(t)$$ and so there's a correspondence of tensors of rank $$2$$ with matrices.

However, exactly as in the linear transformation case this correspondence is only possible when we have selected bases on the vector spaces we are dealing with. Finally, with every tensor it is possible to associate also a multilinear map. So tensors can be understood in their fully abstract and algebraic way as elements of the tensor product of vector spaces, and can also be understood as multilinear maps (this is better for intuition) and we can associate matrices to those.

So after all this hassle with linear algebra, the short answer to your question is: matrices are matrices, tensors of rank 2 are tensors of rank 2, however there's a correspondence between them whenever you fix a basis on the space of tensors.

My suggestion is that you read Kostrikin's "Linear Algebra and Geometry" chapter $$4$$ on multilinear algebra. This book is hard, but it's good to really get the ideas. Also, you can see about tensors (constructions in terms of multilinear maps) in good books of multivariable Analysis like "Calculus on Manifolds" by Michael Spivak or "Analysis on Manifolds" by James Munkres.

• I must be missing something, but can't you just set $S=W, g=\mathrm Id_W$? Apr 27, 2017 at 5:29
• The point is that we want a space $S$ constructed from the vector spaces $V_i$ such that we can use it for all $W$. In other words, given just $V_i$ we can build the pair $(S,g)$ and use once and for all for any $W$ and $f$. That is why it is calles universal property.
– Gold
Apr 27, 2017 at 14:48

As a place-holder answer waiting perhaps for clarification by the questioner's (and others') reaction: given that your context has a matrix be a table of values (which can be entirely reasonable)...

In that context, a "vector" is a list of values, a "matrix" is a table (or list of lists), the next item would be a list of tables (equivalently, a table of lists, or list of lists of lists), then a table of tables (equivalently, a list of tables of lists, or list of lists of tables...). And so on. All these are "tensors".

Unsurprisingly, there are many more sophisticated viewpoints that can be taken, but perhaps this bit of sloganeering is useful?

• In addition to the answer of celtschk, this makes tensors make some sense to me (and their different ranks) Jun 23, 2016 at 18:30
• So basically a tensor is an array of objects in programming. Tensor1 = array. Tensor2 = array of array. Tensor3 = array of array of array. Aug 28, 2017 at 20:59
• @Pacerier, yes, from a programming viewpoint that would be a reasonable starter-version of what a tensor is. But, as noted in my answer, in various mathematical contexts there is complication, due, in effect, to "collapsing" in the indexing scheme. Aug 28, 2017 at 21:37
• @paulgarrett Can we say objects with a rank of more than 2 are tensors? or a scalar is also a tensor? Dec 13, 2021 at 13:07
• Yes, we can (if we insist) say that a scalar is a tensor of rank $0$. And, yes, there are higher-rank tensors: sometimes that higher rank is visible in the number of subscripts and/or superscripts they carry. Such things arise in geometry (and, thereby, in general relativity). Dec 13, 2021 at 16:04

Tensors are objects whose transformation laws make them geometrically meaningful. Yes, I am a physicist, and to me, that is what a tensor is: there is a general idea that tensors are objects merely described using components with respect to some basis, and as coordinates change (and thus the associated basis changes), the tensor's components should transform accordingly. What those laws are follows, then, from the chain rule of multivariable calculus, nothing more.

What is a matrix? A representation of a linear map, also with respect to some basis. Thus, some tensors can be represented with matrices.

Why some? Well, contrary to what you may have heard, not all tensors are inherently linear maps. Yes, you can construct a linear map from any tensor, but that is not what the tensor is. From a vector, you can construct a linear map acting on a covector to produce a scalar; this is where the idea comes from, but it's misleading. Consider a different kind of quantity, representing an oriented plane. We'll call it a bivector: From a bivector, you can construct a linear map taking in a covector and returning a vector, or a linear map taking two covectors and returning a scalar.

That you can construct multiple maps from a bivector should indicate that the bivector is, in itself, neither of these maps but a more fundamental geometric object. Bivectors are represented using antisymmetric 2-index tesors, or antisymmetric matrices. In fact, you can form bivectors from two vectors. While you can make that fit with the mapping picture, it starts to feel incredibly arbitrary.

Some tensors are inherently linear maps, however, and all such maps can be written in terms of some basis as a matrix. Even the Riemann tensor, which has $\binom{n}{2}$ by $\binom{n}{2}$ components, can be written this way, even though it's usually considered a map of two vectors to two vectors, three vectors to one vector, four vectors to a scalar...I could go on.

But not all matrices represent information that is suitable for such geometric considerations.

• Your interpretation of bivectors as rank 2 antisymmetric tensors is very interesting. Where does it find its justification ? Could one construct something similar or analogous for "cobivectors", outer products of 2 covectors, or 2-forms ? This could help provide interesting visualizations of 2-forms. Nov 25, 2019 at 19:56

The shortest answer I can come up with is that a Tensor is described by a matrix (or rank 1 vector) but also the type of thing represented. Matrices have no such "type" associated with them. If you misapply linear algebra on inconsistently typed matrices the math yields mathematically valid garbage.

Intuitively you can't transform apples into peach pie. But you can transform apples into apple pie. Matrices have no intrinsic type associated with them so a linear algebra recipe to do the peach pie transform will produce garbage from the apples matrix.

A more mathematical example is that if you have a vector describing text terms in a document and a vector describing DNA codes, you cannot take the cosine of the normalized vectors (dot product) to see how "similar" they are. The dot product is mathematically valid but since they are from different types and represent different things the dot product is meaningless garbage. But if you do the same with 2 text term vectors you can make statements about how similar they are from the result, it is indeed not garbage.

What I'm calling "type" is more rigorously defined but the above gets the gist I think.

All matrices are not tensors, although all tensors of rank 2 are matrices.

Example

$$T = \begin{bmatrix} x & -y \\ x^2 & -y^2 \end{bmatrix}$$

This matrix $$T$$ is not tensor rank $$2$$. We test matrix $$T$$ to rotation matrix

$$A= \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix}$$

Now, expand tensor equation rank2,for example $$T'_{11}= \Sigma (A_{1i} * A_{1j} * T_{ij}) \quad (1)$$

Now, calculate $$T'_{11} = x' = x*\cos(\theta) + y * \sin(\theta) \quad (2)$$

You see (1) is unequal to (2), then we can conclude that the matrix $$T$$ isn't a tensor of rank 2.

Tensor must follow the conversion(transformation) rules, but matrices generally are not.

• Should you not sum over all indices in equation 2? So you would get x cos² -y (cos + sin) +x² (cos+ sin) -y² sin² (if I interpret your formula correctly, you sum over all i and j, right?) Feb 14 at 8:00
• Also, would that not just show that it is not a rank (2,0) tensor, but not that it is not a rank 2 tensor in general (like 1,1 or 0,2)? Feb 14 at 8:01

You may want to have a look at this web sites http://en.wikipedia.org/wiki/Tensor https://physics.stackexchange.com/questions/20437/are-matrices-and-second-rank-tensors-the-same-thing

"Tensors are generalizations of scalars (that have no indices), vectors (that have exactly one index), and matrices (that have exactly two indices) to an arbitrary number of indices. " [1]

You can't index into a scalar $s\in \mathbb{R}$. You can index into a vector $v\in \mathbb{R}^d$ with a single index $i$ as $v_i$. You can index into a matrix $M\in \mathbb{R}^{n \times m}$ with two indices $(i, j)$ as $M_{ij}$.

If I continue adding indices, you might imagine a matrix-like object $T\in\mathbb{R}^{n\times m \times p}$ you can index into with three indices $(i, j, k)$ as $T_{ijk}$. How about $4, 5, 6, ...$ indices? A tensor is such an object that allows any number of indices you want.

In this way scalars, vectors and matrices are the special case of tensors with 0, 1 and 2 indices.