# Understanding the core idea behind determinants

1. Dec 26, 2007

### O.J.

I'm really putting some effort into understanding the core idea behind determinants. For a 2x2 matrix, I obviously saw how to derive the formula for the determinant (using AA^-1=I). The question is, how did they define the |A| for higher order matrices? I'm reading a textbook on linear algebra and it uses permutations in its definition of the determinant? What was the motivation for such a definition? How did they know that such a definition can be applied to all nxn matrices???

2. Dec 26, 2007

### HallsofIvy

The most fundamental definition- for all n by n square matrices is this: form all possible products $a_{1i_1}a_{2i_2}\cdot\cdot\cdot a_{ni_n}$ taking exactly one number from each row and column in the matrix: there are exactly n! such products. Since no row or column is duplicated, each $i_1, i_2, \cdot\cdot\cdot, i_n$ is a permutation of $1, 2, \cdot\cdot\cdot, n$. Multiply each product by 1 or -1 depending on whether the permutation is even or odd. Add them all together. I suspect that is the definition that "uses permutations" that your book gives. As to the motivation, it was a lot like other such general definitions. People working with systems of 2 or 3 equations saw that such numbers could be used to write general solutions (Cayley's formula for example), found some formal way of writing that worked for both 2 by 2 and 3 by 3 systems and extended it to n by n. As for "How did they know that such a definition can be applied to all nxn matrices???" the whole point is that obviously we can do that for all n by n matrices. If you are asking "how did they know that such a determinant would solve n equations in n unknowns?" it is really based on how the determinant of a matrix is related to the inverse of the matrix.

3. Dec 26, 2007

### mathwonk

if you solve a general system of n linear equations in n unknowns, the determinant is the denominator of all the solutions.

4. Dec 26, 2007

### O.J.

Mathwonk... could you provide an example please?
And ivy.. I know they can use it to DO all nxn matrices, question is: How did they ARRIVE at such a definition that can solve all nxn matrices...

5. Dec 26, 2007

### slider142

Go ahead and write out a system of 2 equations in 2 unknowns and a system of three equations in 3 unknowns and solve them symbolically (no numbers, just use coefficients a11, a12, etc.) using methods that do not involve determinants. You'll see the solutions all have the same denominator. Try to find the pattern.

6. Dec 26, 2007

### O.J.

Why don't they explain this in our math texts? Are there books that are dedicated to explaining such things?

7. Dec 26, 2007

### mathwonk

perhaps you are reading the wrong texts. consult my thread "who wants to be a mathematician" for book recommendations, in the first 40 or so posts,

e.g. courant, differential and integral calculus, vol. 2, pages 24-25. on determinants.

Last edited: Dec 27, 2007
8. Dec 27, 2007

### Ben Niehoff

If you consider the columns (or rows) of an nxn matrix to be n-dimensional vectors, then the determinant of the matrix gives the "signed volume" of the n-dimensional parallelepiped defined by those vectors. In a sense, then, it gives the "magnitude" of the matrix, multiplied by a sign that indicates whether the orientation of the vectors is "right-handed" or "left-handed" with respect to the underlying coordinate system.

If any one of the vectors lies in the same (n-1)-dimensional space spanned by the remaining (n-1) vectors, then the whole parallelepiped is "squashed" to zero volume (consider the three-dimensional case, where you have three vectors all lying in the same plane). A matrix composed of such vectors, then, represents a degenerate system of equations, where the equations are not all independent (and thus, permit infinitely many solutions). One can consider the matrix equation

$$\mathbf{AX} = \mathbf{B}$$

as a single, abstract entity, with the solution

$$\mathbf{X} = \mathbf{A^{-1}B}$$

In this case, the determinant of A can again be regarded as a sort of "magnitude". If it is zero, then there are infinitely many X which solve the equation, in an analogous sense to solving the one-dimensional equation

$$0x = b$$

Furthermore, the expression

$$\mathbf{A^{-1}B}$$

yields Cramer's Rule when written out in terms of $\det \mathbf{A}$ and the cofactors of A.

9. Dec 27, 2007

### mathwonk

cramers rule is a more detaield way of saying ther denominator of a solution is the determinant. i.e. it also gives the numerators.

10. Jan 4, 2008

### musicheck

"The determinant of a matrix is an (oriented) volume of the parallelepiped whose edges are its columns. If the students are told this secret (which is carefully hidden in the purified algebraic education), then the whole theory of determinants becomes a clear chapter of the theory of poly-linear forms. If determinants are defined otherwise, then any sensible person will forever hate all the determinants, Jacobians and the implicit function theorem."

Given this geometric interpretation, it makes sense that permutations are related to determinants: permuting two column vectors in a matrix is just changing the order in which you are writing the list of column vectors spanning the parallelapiped.

11. Jan 6, 2008

### Ben Niehoff

Quite true, that. I absolutely hated determinants until I realized the geometric meaning of them. Until then, they were just pointless tedium, and my disinterest certainly hindered my understanding of linear algebra.

I blame this partly on the fact that determinants are often taught using a procedure on 3x3 matrices that does not generalize to higher-order matrices (i.e., the "diagonals" procedure). Students learn a cookie-cutter shortcut, without being taught what they are actually calculating (what else is new?).

12. Jan 7, 2008

### HallsofIvy

How long ago was this? Calculation of determinants are now (and for at least 30 years in my experience) taught using "expansion by rows" or "row reduction" both of which extend immediately to higher dimensions.

13. Jan 8, 2008

### musicheck

I'm currently in the middle of a year long vector calc + linear algebra class, and we saw the diagonals procedure. We have not defined n by n determinants yet, though I think our book defines it axiomatically (the unique multilinear map with some list of properties).

14. Jan 8, 2008

### Count Iblis

You can learn the theory on determinants in an hour or so. The two most useful properties of the determinant are:

det(AB) = det(A) det(B),

det[exp(A)] = exp[Tr(A)]

If you are done studying the elementary stuff, you should http://arxiv.org/abs/math/9902004" [Broken]

Last edited by a moderator: May 3, 2017
15. Jan 8, 2008

### Count Iblis

http://en.wikipedia.org/wiki/Dodgson_condensation" [Broken]

Last edited by a moderator: May 3, 2017
16. Jan 9, 2008

### HallsofIvy

Probably apocryphal story: Queen Victoria so enjoyed the "Alice" books that she invited Louis Carrol (aka the Reverend Charles Dodgson) to court and requested that he send her a copy of his next book. He did- a treatise on matrices including the method for calculating determinants cited above.

17. Jan 9, 2008

### The_Ouroboros

cool :D

Last edited by a moderator: May 3, 2017
18. Jan 15, 2008

### WWGD

Take a look at Borwein and Borowski's Dictionary of Mathematics. It is nicely
written in that sense, i.e, many of the definitions are made within a context.
I never leave home without it. Try also looking at some of Ian Stewart's books,
which I think do a nice job in the same respect. I think Stewart treads well the
line in writing popular books in mathematics and his writing is neither too technical
nor oversimplified for the most part.