# How did the idea determinants come up?

1. Oct 8, 2012

### Avichal

Its easy to come with the idea of matrices. Its just a representation of data.

But how did the concept of determinants come up? The way we expand determinants with alternate plus and minus sign and then multiplying with the co-factors - how did that come up?

2. Oct 8, 2012

3. Oct 8, 2012

### HallsofIvy

Staff Emeritus
Go ahead and look at the Wikipedia reference.

But I think it started from a simple observation. The matrix equation:
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}u \\ v\end{bmatrix}$$
is equivalent to the system of equations
ax+ by= u, cx+ dy= v.

One of the first things we learn about such equations is to solve them by elimination. multiply the first equation by c, to get acx+ bcy= cu, and the second equation by a to get acx+ ady= av. Subtracting the two equations eliminates x: (bc- ad)y= cu- av. We can solve for y by dividing through by bc- ad. And whether or not we can solve that equation depends upon whether or not bc- ad is non-negative. That is, of course, the determinant of the original matrix:
$$\left|\begin{array}{cc}a & b \\ c & d\end{array}\right|= ad- bc$$

If we were to do the same thing for three or four equations in three or four unknowns, we would find that we would eventally need to divide by some number so that whether or not the system had a solution depended upon whether or not that number is 0. And that number is the determinant of the coefficient matrix.

4. Oct 8, 2012

### Avichal

Oh okay so it all started with whether a set linear equations have a solution or not.
And then onwards people started discovering that determinants can be used to solve linear equations itself - not just check them.
Thanks - I saw the Wikipedia article before coming here but still I thought it lacked something. The observation you mentioned made it much clearer

5. Oct 8, 2012

### A. Bahat

This is all much clearer from the basis-free point of view. A matrix is just a way of representing a linear transformation T:V--->W between vector spaces. We naturally want to know when this transformation is invertible, i.e. when the corresponding system of equations Av=w has a unique solution. Thinking of this geometrically we see this can't happen when T maps the unit cube onto a lower-dimensional subspace, i.e. when det(T), which is just the signed volume of the image of the cube under T, vanishes (unless w=0 of course).

Last edited: Oct 8, 2012
6. Oct 9, 2012

### dumbQuestion

Thank you so much for this explanation. I had never stopped and thought of it like this and that makes so much sense

7. Oct 10, 2012

### mathwonk

that wikipedia article seems to have been written by someone who scanned the historical literature without reading it.

8. Oct 10, 2012

### alan2

Everyone who has ever studied determinants has asked the exact same question. This is one of my favorite papers. Dr. Axler also has a very popular linear algebra text based in which determinants appear. As a computational tool, determinants are useful. Beyond that, they tend to confuse.

http://www.axler.net/DwD.pdf

9. Oct 10, 2012

### mathwonk

i have enormous respect for axler, but one should not take such ridiculous statements as the title of his paper seriously. this sort of thing does enormous harm in my opinion. since math is not only about proving existence theorems, but sometimes also about actually producing the items known to exist. and in computations, determinants are very useful.

moreover, it is of great interest in many settings to actually find a polynomial which defines something of significance. E.g. in the theory of a Riemann surface of genus g, the classical theorem of Riemann himself computes the degree of the singularity of the variety of line bundles of degree g-1, at agiven line bundle L, to be the dimension of the space of sections of L.

The modern approach via the theory of determinantal varieties, produces also the lowest degree Taylor polynomial of this variety, as a dterminant. Indeed the famous book, Geometry of algebraic curves, by Arbarello, Cornalba, Grifiths, Harris,

https://www.amazon.com/Geometry-Algebraic-Curves-mathematischen-Wissenschaften/dp/0387909974

is largely a detailed treatment of the modern theory of varities defined by classifying maps to the space of matrices, with especial regard to the subvarieties defined by vanishing of determinants of various orders. see chapter 2, e.g.

another deep study is contained in :

https://www.amazon.com/s/ref=nb_sb_...-keywords=birgit+iversen,+linear+determinants

To disparage determinants to beginners does a definite disservice, since those who say such things are usually themselves very familiar with determinants and are fully able to use them when needed, but the poor student who believes their half true propaganda against them, unfortunately may fail to master them himself.

Such headlines are just meant as provocative discussion causing statements, not to taken at face value at all.

I.e. by all means learn as much about determinants as possible, but then listen , even if skeptically, to the arguments of those who point out how to avoid them at certain times.

indeed one should understand the geometric meaning of the vanishing of determinants but one should definitely be aware of the use of determinants for computing polynomials that make these geometric statements quantifiable.

in general never listen seriously to anyone who tells you you do not need to learn some fundamental and time honored subject, or do so at you peril. if something fundamental is confusing or puzzling, do not take that as evidence to ignore it but quite the opposite, as evidence that it will likely repay continued thought.

forgive me for this stump speech, but i have lived a long time and suffered greatly from such smart alecky stuff as axler is putting out there. one can learn a lot from him as he is very very smart. but try to ignore him when he says you do not need to learn something that he himself, and everyone else, has learned very well.

there is nothing shocking about proving most result without using determinants. in the 4 documents on my webpage that treat linear algebra, 3 of them, notes for linear algebra, math 8000, and math 4050, mention determinants only as an after thought, in appendices. the math 845 part 2c notes use them along with other techniques for computing normal forms of matrices, but treat them in detail again only in an appendix, not even included on the website.

but i am leaving them aside there for the same reason axler does, namely they are a lot of tedious work. it is easier to avoid this work than to do it in detail. still to use them, one only needs to know a minimal amount of facts about them, not all the nitty gritty details, as is clear in my 843-4-5 notes. but to ignore them completely or worse, to claim that they are somehow bad, is irresponsible.

Last edited by a moderator: May 6, 2017
10. Oct 12, 2012

### mathwonk

to be more specific, determinants are fundamental in understanding area and volume, i.e. change of variables formulas for integration, differential forms, formulas for surface integrals, as well as families of linear equations, families of matrices, behavior of sections of vector bundles, abel - jacobi maps for riemann surfaces, ... etc etc...

In the famous book Intersection theory, by Fulton, check out chapter 14 on degeneracy loci [i.e. loci where certain determinants vanish], or just type in "determinant" in the search box and see what you get.

https://www.amazon.com/Intersection...&sr=1-1&keywords=william+fulton,+intersection

Last edited: Oct 12, 2012
11. Oct 12, 2012

### mathwonk

as far as the original question goes, namely where do the alternating plus and minus signs come from, draw a parallelogram in the plane spanned by two vectors emanating from (0,0) and ending at (a,b) and (c,d).

Then draw the big rectangle enclosing this parallelogram, with diagonal from (0,0) to (a+c,b+d). When you compute the area of the parallelogram, by starting from the area of the big rectangle, and subtracting off the areas of the smaller rectangles and right triangles inside the big rectangle, but outside the parallelogram, you will see that the determinantal area formula for the parallelogram arises by a process of adding and subtracting the areas of rectangles.

this explains the plus and minus signs. i.e. any slanted figure can be obtained by adding and subtracting figures with right angle sides, hence the volume can be computed by adding and subtracting the products which compute those simpler volumes.

not that the coordinates of a vector express that vector in terms of its perpendicular projections on various axes. the problem is to use those projections to compute the volume of the slanted block spanned by several vectors, in terms of those projected measurements.

an interesting phenomenon of this type is the pythagorean formula for areas. i.e. if we have a parallelogram in space and project it onto the three axis planes, the sum of the squares of the projected areas will equal the square of the original area. I.e. the area spanned by two vectors in 3 space, the rows of a 2 by 3 matrix, can also be computed as a determinant, or rather in terms of the three 2by2 sub determinants.\
determinants are rather beautiful and amazing.

Last edited: Oct 12, 2012
12. Oct 12, 2012

### Studiot

I am sorry to hear that you have suffered at all.

In substance and spirit I am pretty much with you that there is more to determinants than just the solution of a set of linear equations.

Old time engineering books, from fluids mechanics to electricity theory to elasticity to... used the following phrase frequently:-
'This is nothing more than the determinant', because determinants pervade engineering, perhaps because of this one fact not so far noted.

A determinant is a (real) number. In fact it is a map from the vector space to the reals.

Engineers like(real) numbers.

A Bahat one small correction

Not all matrices represent transformations, linear or otherwise. They too have other lives.
Perhaps you meant square matrices?

13. Oct 12, 2012

### A. Bahat

I'm not quite sure what you mean by this. Every m x n matrix A over a field k determines a linear transformation T:k^n--->k^m, namely left-multiplication by A. Conversely, if we are given a linear transformation T:V--->W and bases of V and W (i.e. isomorphisms V≈k^n and W≈k^m) there is some matrix associated with T in these bases.

Now, sometimes matrices are used in contexts independent of linear maps (I have in mind more analytic topics like stochastic matrices). But this doesn't change the fact that every matrix gives a linear map and every linear map gives a matrix once a basis is chosen.

14. Oct 12, 2012

### Studiot

Perhaps you have not met the incidence matrix when applied to a colour table or other dataset?

15. Oct 12, 2012

### A. Bahat

Like I said, we don't always have to think of matrices as linear transformations, but the fact is that a matrix always gives us a linear transformation whether we want one or not! Besides, adjacency matrices and incidence matrices in graph theory are useful precisely because they allow us to use the power of linear algebra to study combinatorial problems.

16. Oct 12, 2012

### Muphrid

Something relevant to the idea of determinants is the notion of exterior algebra--the use of vectors to build higher-dimensional objects called bivectors and trivectors and so on. A linear operator on a vector can be extended to act on such obects by considering its action on each of the individual vectors that make up those objects.

The highest-dimensional objects in such a space naturally represent volumes (or their analogues). These objects form a 1-dimensional vector space--after all, in 3d dimensions, there is only 1 unique unit volume, and all other volumes are just scalar multiples of that volume. The action of a linear operator on the unit volume thus must return some multiple of the unit volume--or, it must return zero. The unit volume is often an eigenvector of a linear operator, and the associated eigenvalue has a special name: the determinant.

17. Oct 12, 2012

### mathwonk

well 'suffered" is maybe too strong. more like, "if i had only known then what i know now" type of existential suffering.

yes indeed, exterior algebras are exactly the kind of thing one needs in many situations.

the book, "a geometric approach to differential forms" by david bachmann is an excellent place to learn to understand the geometric meaning of determinants, i.re. of alternating functions (pluses and minuses).

the last part of my math 845 notes on my website also give some careful treatment of exterior algebras and their properties.

the most important formula is on page 56 of these notes:

http://www.math.uga.edu/%7Eroy/845-3.pdf [Broken]

the key formula is given a shorter, more memorable, but less explicit statement in the remark at the top of page 61.

Last edited by a moderator: May 6, 2017
18. Oct 12, 2012

### mathwonk

the geometric s-wedge construction takes a vector space and constructs the associated vector space of s-fold blocks spanned by s tuples of vectors in the original space, and their linear combinations.

this is a functor, so every linear map between two vector spaces induces a linear map between their s-fold exterior products. moreover if the space is n dimensional, its nth exterior product has dimension one.

thus in case of a finite dimensional vector space of dimension n, and an endomorphism of this space, the induced map from the nth exterior product is a map from a one dimensional space to itslf. such a map is defined by multiplication by a scalar, the determinant of the original map.

thus the determinant can be considered as a map

det: Hom(V,V)-->Hom(^(dimV)(V), ^(dimV)(V))) ≈ R.

19. Oct 12, 2012

### A. Bahat

Very nice notes, mathwonk.

This is where I disagree with Axler. On the one hand, I'll admit his book has some interesting determinant-free proofs. But determinants 'done right' (i.e. via exterior algebra), in addition to being computationally essential, help conceptually organize a lot of properties of linear maps. Long live determinants!

20. Oct 12, 2012

### Studiot

What transformation can be applied to the colour 'red' ?
A ball, for instance, is either red or it isn't. There is no transformation involved or available.

This is taking the discussion off topic. I will happily expand upon non transformation matrices if you would like to start a discussion thread.