Currently, I'm taking matrix theory, and our textbook is Strang's Linear Algebra. Besides matrix theory, which all engineers must take, there exists linear algebra I and II for math majors. What is the difference,if any, between matrix theory and linear algebra?


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    $\begingroup$ Likely the version of the course called "linear algebra" is proof-based and gets deeper into the conceptual content, whereas "matrix theory" probably focuses on applications. It's a matter of emphasis, really. $\endgroup$ Jan 13, 2010 at 17:20
  • $\begingroup$ The other difference I've seen is that matrix theory usually concentrates on the theory of real complex matrices. Linear algebra cares about those, but also rational canonical forms, etc... $\endgroup$ Apr 1, 2010 at 14:52

8 Answers 8


Let me elaborate a little on what Steve Huntsman is talking about. A matrix is just a list of numbers, and you're allowed to add and multiply matrices by combining those numbers in a certain way. When you talk about matrices, you're allowed to talk about things like the entry in the 3rd row and 4th column, and so forth. In this setting, matrices are useful for representing things like transition probabilities in a Markov chain, where each entry indicates the probability of transitioning from one state to another. You can do lots of interesting numerical things with matrices, and these interesting numerical things are very important because matrices show up a lot in engineering and the sciences.

In linear algebra, however, you instead talk about linear transformations, which are not (I cannot emphasize this enough) a list of numbers, although sometimes it is convenient to use a particular matrix to write down a linear transformation. The difference between a linear transformation and a matrix is not easy to grasp the first time you see it, and most people would be fine with conflating the two points of view. However, when you're given a linear transformation, you're not allowed to ask for things like the entry in its 3rd row and 4th column because questions like these depend on a choice of basis. Instead, you're only allowed to ask for things that don't depend on the basis, such as the rank, the trace, the determinant, or the set of eigenvalues. This point of view may seem unnecessarily restrictive, but it is fundamental to a deeper understanding of pure mathematics.

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    $\begingroup$ While it is true that people doing "Matrix Theory" often spend a lot of time with a choice of basis, it's important to note that this is frequently in pursuit of quantities that are invariant of choice of basis. $\endgroup$
    – Dan Piponi
    Jan 13, 2010 at 23:44
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    $\begingroup$ An even more basic question but in the same line is "What is the difference of a vector and a row (collum) matrix".Vectors are mathematical object living in a linear space or vector space (which satisfy certain properties). Choosing a special set of vectors called a base, we can decompose every vector in the vector space into a kind of sum of vectors in this base. Thus every vector in a code, and this is the row (collum) matrix. The next step is to look at the homomorphisms (maps) between linear spaces. Choosing the base of the domain and the range we can represent the homomorphism by a matrix $\endgroup$
    – abcdxyz
    Jan 14, 2010 at 15:00
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    $\begingroup$ Even worse, matrices depend on a choice of an ordered basis. $\endgroup$ Mar 30, 2010 at 22:30
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    $\begingroup$ Belated comment: Depends on what you call a matrix, Harry. If $X$ and $Y$ are sets and $K[X]$ and $K[Y]$ are their free K-vector spaces then a linear map $\colon K[X] \to K[Y]$ is the same as a map of sets $X \to K^Y = X \times Y \to K$. I'd argue this is what a matrix really is and that ordering is an artifact of trying to write something in linear order on a piece of paper. $\endgroup$ Aug 4, 2010 at 5:36

A counter-quotation to the one from Dieudonné:

We share a philosophy about linear algebra: we think basis-free, we write basis-free, but when the chips are down we close the office door and compute with matrices like fury.

(Irving Kaplansky, writing of himself and Paul Halmos)

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    $\begingroup$ I totally agree with this one. Thanks for sharing. $\endgroup$
    – user1855
    Apr 2, 2010 at 2:27
  • $\begingroup$ +1! I confess that I like this quote much more than the other one (Dieudonné's), which, at least to me, appears a little arrogant. In my opinion, 'abstract' is not automatically 'better.' There are cases when one needs a concrete and efficient computation [Or, are all the algorithms implemented in Matlab just not smart enough because they use matrices? :) ] $\endgroup$
    – user2734
    Apr 2, 2010 at 7:48
  • $\begingroup$ To echo the comment above: Kaplansky's quotation is that much more appropriate for people who code in low-level or numerical languages. It's possible to do a heck of a lot of symbolic calculation in such settings through the judicious use of integral matrices (here "integral" should be considered broadly). $\endgroup$ Apr 23, 2010 at 16:26

Let me quote without further comment from Dieudonné's "Foundations of Modern Analysis, Vol. 1".

There is hardly any theory which is more elementary [than linear algebra], in spite of the fact that generations of professors and textbook writers have obscured its simplicity by preposterous calculations with matrices.

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    $\begingroup$ You should add this to the great quotes in mathematics thread. $\endgroup$ Mar 30, 2010 at 22:30
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    $\begingroup$ It is ironic that a textbook on analysis would make such an outrageous claim on the trivially of another field: the analytic parts of linear algebra are truly deep and quite actively researched. See, for example, Loewner's classification of matrix-monotone functions, or most any paper in quantum Shannon theory. Additionally, the entire field of quantum information theory (QIT) is essentially the study of unitary and self-adjoint operators on tensor products of Hilbert spaces, and a large majority the interesting questions in QIT retain 99% of their interest in the finite-dimensional case. $\endgroup$
    – Jon
    Apr 4, 2010 at 18:06
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    $\begingroup$ I don't think that there are many who can claim to have a better understanding of that things than Dieudonné. So instead of trying so hard to misunderstand him, try to find a meaning in his comment. $\endgroup$
    – Tile
    Apr 4, 2010 at 18:53
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    $\begingroup$ It should also be pointed out that the "analytic parts of linear algebra" are more properly thought of as linear analysis, or in the case of operator monotone functions and calculations with the c.b. norm, even as non-linear analysis. I think castigating Dieudonné for this quote is taking unnecessary umbrage. $\endgroup$
    – Yemon Choi
    Apr 4, 2010 at 19:25
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    $\begingroup$ Of course Dieudonne meant "elementary" as in "simple and foundational", here not using "simple" to mean easy, but simple in the sense of structural complexity. It's not an arrogant statement about how easy he thinks linear algebra is, but rather a castigation of those "generations of professors and textbook writers" who turned an elegant subject into a jumbled mess. $\endgroup$ Jun 10, 2010 at 8:45

The difference is that in matrix theory you have chosen a particular basis.


Although some years ago I would have agreed with the above comments about the relationship between Linear Algebra and Matrix Theory, I DO NOT agree any more!

See, for example Bhatia's "Matrix Analysis" GTM book. For example, doubly-(sub)stochastic matrices arise naturally in the classification of unitarily-invariant norms. They also naturally appear in the study of quantum entanglement, which really has nothing to do with a basis. (In both instances, all sorts of NONarbitrary bases come into play, mainly after the spectral theorem gets applied.)

Doubly-stochastic matrices turn out to be useful to give concise proofs of basis-independent inequalities, such as the non-commutative Holder inequality:

tr |AB| $\le$ $||A||_p$ $||B||_q$

with 1/p+1/q=1, $|A|=(A^*A)^{1/2}$, and $||A||_p = (tr |A|^p)^{1/p}$

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    $\begingroup$ Doubly-stochastic matrices (in one interpretation, anyway) describe transition probabilities of some Markov chain where all the transitions are reversible. The relevant vector space is the free vector space over the states of the chain. Maybe this interpretation isn't directly relevant to the application you're thinking of, but there should be some connection. $\endgroup$ Apr 4, 2010 at 18:25
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    $\begingroup$ In the application to the Holder inequality, one uses the fact that if U is a unitary operator, then replacing the matrix elements of U by the squares of their absolute values yields a doubly-stochastic matrix. $\endgroup$
    – Jon
    Apr 7, 2010 at 1:33

Matrix theory is the specialization of linear algebra to the case of finite dimensional vector spaces and doing explicit manipulations after fixing a basis. More precisely: The algebra of $n \times n$ matrices with coefficients in a field $F$ is isomorphic to the algebra of $F$-linear homomorphisms from an $n$-dimensional vector space $V$ over $F$, to itself. And the choice of such an isomorphism is precisely the choice of a basis for $V$.

Sometimes you need concrete computations for which you use the matrix viewpoint. But for conceptual understanding, application to wider contexts and for overall mathematical elegance, the abstract approach of vector spaces and linear transformations is better.

In this second approach you can take over linear algebra to more general settings such as modules over rings(PIDs for instance), functional analysis, homological algebra, representation theory, etc.. All these topics have linear algebra at their heart, or, rather, "is" indeed linear algebra..


I'm with Jon. Matrices don't always appear as linear transformations. Yes, you can look at them as linear transformations, but there are times when it's better not to and study them for their own right. Jon already gave one example. Another example is the theory of positive (semi)definite matrices. They appear naturally as covariance matrices of random vectors. The notions like schur complements appear naturally in a course in matrix theory, but probably not in linear algebra.

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    $\begingroup$ Covariance matrices are essentially inner products, aren't they? That's just thinking of matrices as tensors of type (0, 2) instead of as tensors of type (1, 1). I think the theory of linear algebra is really good at clarifying the distinction between this type of matrix and the "usual" type of matrix; for example it gets to the heart of when similarity is relevant vs. when conjugation is relevant. So I don't think this is a good example. $\endgroup$ Apr 4, 2010 at 18:20

My opinion: matrix theory mostly deals with matrix of a paticular kind , or a few relevant ones. But linear algebra cares about the general, underlying structrue.


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