Any linear map between finite-dimensional vector spaces may be represented by a matrix, and conversely. Matrix-matrix multiplication corresponds to map composition, and matrix-vector multiplication corresponds to map application.

This provides a nice way of understanding matrices and the origin of the rules for their calculation. Having attained this perspective is a huge boon to me, because I've always resented that matrix math is basically just taught as a collection of arbitrary manipulations of symbols (multiply the entries! now add them! don't ask why!).

But is this understanding always meaningful?

For in general, just because you've found an interpretation for a mathematical concept doesn't mean it will always apply. For instance, the trig functions can be thought of as simply being the coordinates on a unit circle, but this fails to explain why they show up as solutions to differential equations having nothing to do with circles (or indeed geometry) - to explain this we require a more abstract understanding of the trig functions.

Thus, in all non-trivial uses of matrices in pure and applied mathematics, can we meaningfully say that the matrices involved represent, at least implicitly, linear maps? For instance, we can think of solving a system of equations as simply trying to find a vector which, when a given map is applied to it, yields a specific image.

By non-trivial, I mean that the particular use of matrices involves, at minimum, matrix multiplication and/or determinants or other elementary aspects of the calculus of matrices. For instance, you could call a spreadsheet a matrix, but unless you intend to do anything matrix-like with it apart from draw it as a grid, you may as well just call it a set of numbers.

Note: this question may be soft, but I feel it can be given a clean answer. Either you can come up with a situation where the linear-map interpretation doesn't apply, or you can't.

  • $\begingroup$ Until you make a choice of basis, a matrix is just a matrix. $\endgroup$ Mar 12, 2013 at 21:30
  • $\begingroup$ But in the context of a particular problem, it often implicitly represents a linear map. $\endgroup$
    – Jack M
    Mar 12, 2013 at 21:31
  • 1
    $\begingroup$ Well in that case it probably is best understood as a linear map! (The key is understood as - some people make the error of thinking that a matrix really is a linear map.) $\endgroup$ Mar 12, 2013 at 21:41
  • 4
    $\begingroup$ A matrix can represent a quadratic form as well. It doesn't have to represent a linear map. $\endgroup$
    – user1551
    Mar 12, 2013 at 22:19
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    $\begingroup$ In my opinion a great question. +1 $\endgroup$
    – Alexander Gruber
    Mar 13, 2013 at 0:59

4 Answers 4


I think that when you see a matrix multiplication, more likely than not there will be a composition or application of a linear map going on. However I've seen very remarkable determinants being evaluated (in combinatorics notably) where it is really very hard to bring any linear operation into the picture. Sometimes arranging expressions into a matrix is just a convenient notational means to create a very complicated combination of those expressions simply by applying $\det$.

Here is an example. For a partition $\lambda=(\lambda_0,\lambda_1,\ldots,\lambda_k)$ (with therefore $\lambda_0\geq\lambda_1\geq\cdots\geq\lambda_k\geq0$) one can define the Schur polynomial $s_\lambda\in\mathbf Z[X_0,X_1,\ldots,X_k]$ by $$ s_\lambda = \frac{\begin{vmatrix} X_0^{\lambda_0+k}&X_0^{\lambda_1+k-1}&\cdots&X_0^{\lambda_{k-1}+1}&X_0^{\lambda_k}\\ X_1^{\lambda_0+k}&X_1^{\lambda_1+k-1}&\cdots&X_1^{\lambda_{k-1}+1}&X_1^{\lambda_k}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ X_{k-1}^{\lambda_0+k}k&X_{k-1}^{\lambda_1+k-1}&\cdots&X_{k-1}^{\lambda_{k-1}+1}&X_{k-1}^{\lambda_k}\\ X_k^{\lambda_0+k}&X_k^{\lambda_1+k-1}&\cdots&X_k^{\lambda_{k-1}+1}&X_k^{\lambda_k}\\ \end{vmatrix}} {\begin{vmatrix} X_0^k&X_0^{k-1}&\cdots&X_0^1&1\\ X_1^k&X_1^{k-1}&\cdots&X_1^1&1\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ X_{k-1}^k&X_{k-1}^{k-1}&\cdots&X_{k-1}^1&1\\ X_k^k&X_k^{k-1}&\cdots&X_k^1&1\\ \end{vmatrix}} = \frac{\det\bigl((X_i^{\lambda_j+k-j})_{i,j=0}^k\bigr)}{\prod_{0\leq i<j\leq k}(X_i-X_j)}. $$ Here I would have a hard time bringing a linear map into the picture in any useful manner.


Note that a Matrix is not a linear function, it is just a matrix. With a basis given we can interpret it as a linear function. For example there are matrixes that represents graph (like the vertex or so called adjacency matrix which don't represent linear maps).

You have set of vertices $\{P_i\}_{1\leq i \leq n}$ and the matrix $M$ is a $n \times n$ with $$m_{ij} = \left\{\begin{array}{rl} 1 & P_i \to P_j\\ 0& \text{else} \\ \end{array}\right.$$

  • $\begingroup$ I see. Well it's certainly very interesting that adjacency matrices can be meaningfully operated on with matrix multiplication. But perhaps there is a way to imagine them as linear maps over some sort of space somehow representing the graph? Maybe there's a higher, more abstract view point to account for both linear maps and adjacency matrices? $\endgroup$
    – Jack M
    Mar 12, 2013 at 21:47
  • $\begingroup$ What actually happens when one tries taking the determinant, inverse, or product of one or more of these graph matrices, or applying it as a map to a vector? Are you sure it doesn't yield anything interesting? $\endgroup$ Mar 12, 2013 at 21:48
  • $\begingroup$ Wikipedia provides some examples en.wikipedia.org/wiki/Adjacency_matrix#Properties , and I'm stretching my mind trying to interpret nodes as vectors and so on, but nothing concrete yet. Certainly food for thought (though I should probably finish my linalg textbook before I get too caught up with it). $\endgroup$
    – Jack M
    Mar 12, 2013 at 21:57
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    $\begingroup$ Well, it does. You can denote $m_{ij}$ as the number of paths of length $1$ between vertices $i$ and $j$, then $M^k$ is the number of paths of length $k$ and so on. If that does confuse you, it is easier seen with probabilities, e.g. see Markov chains. $\endgroup$
    – dtldarek
    Mar 12, 2013 at 22:39

The power of the linear map point of view is that it doesn't depend on the choice of a basis. On the other hand, for problems which inherently involve a fixed basis, typically matrices are better suited.

One example is the Gauss algorithm. I don't know a single linear algebra book presenting it for linear maps. The matrices are more intuitive here.

The reason is the following: You cannot express the Gauss algorithm it in terms of linear maps alone, you additionally need some fixed basis of the vector space. (the elementary transformations, the reduced row echelon form etc. depend on a basis). Using matrices, that is not necessary, since there is already a distinguished basis of the vector space (the canonical one).

  • $\begingroup$ For Gauss's method, isn't this just a notational issue? I mean, what would presenting Gauss's method "for linear maps" even look like? The fact that we use matrix notation for that purpose doesn't mean that the matrix in question can't, conceptually, represent a linear map. $\endgroup$
    – Jack M
    Mar 12, 2013 at 22:02
  • $\begingroup$ Of course you can formulate it for linear maps. But note that the normal form in the end depends on a chosen basis. So you can't formulate it with linear maps alone, you additionally need a fixed basis. The matrices already come with a fixed basis. $\endgroup$
    – azimut
    Mar 12, 2013 at 22:04
  • $\begingroup$ The Gaussian elimination algorithm is useful for computing the kernel and the image spaces of a linear map, or solving an equation $L(x)=y$ involving the map $L$. However by nature the algorithm operates on entries of a matrix, so you cannot apply the algorithm without first expressing the linear map in a basis. $\endgroup$ Mar 12, 2013 at 22:04
  • $\begingroup$ Okay, I'm confused about how the linear map represented by a matrix depends on a choice of basis, so at least now I know what I need to study next. $\endgroup$
    – Jack M
    Mar 12, 2013 at 22:07
  • $\begingroup$ @JackM Do you know what similar matrices are? $\endgroup$
    – EuYu
    Mar 12, 2013 at 22:09

The characteristic polynomial of a matrix with entries in $\mathbb{Z}$ is a monic polynomial in $\mathbb{Z}[X]$. Conversely, given a monic polynomial in $\mathbb{Z}[X]$, its companion matrix is an example of a matrix with entries in $\mathbb{Z}$ and with the given polynomial as characteristic polynomial. This implies that the set of algebraic integers, i.e. the set of numbers which are a root of some polynomial in $\mathbb{Z}[X]$, is also the set of eigenvalues of matrices with entries in $\mathbb{Z}$.

This provides a constructive proof of the fact that the set of algebraic integers is closed under addition and multiplication. Indeed, let $x$ and $y$ be eigenvalues of the matrices $A$ and $B$ respectively, both with entries in $\mathbb{Z}$ and with respective sizes $n$ and $m$. Then

  • the matrix $A \otimes B$ has entries in $\mathbb{Z}$ and has $xy$ as an eigenvalue.
  • the matrix $A \otimes I_m + I_n \otimes B$ has entries in $\mathbb{Z}$ and $x+y$ as an eigenvalue.

Of course matrix addition, Kronecker products and eigenvalues have a meaning in terms of linear maps, but I think it is debatable whether this translation gives extra meaning to this particular problem. When moving away from fields to commutative rings (and possible even further to noncommutative rings) as entries of a matrix ring, the view of linear maps loses its geometric meaning, I think.


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