Lost student

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rolo91
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Lost student

Postby rolo91 » Mon Aug 06, 2012 10:55 am UTC

First things first: Hi to everyone, as this is my first post in these forums :D Also, excuse my English if it's not quite correct, it's not my native language.

I live in Spain, and this last course I started studying a degree in mathematics and statistics (that's how it's called here, I don't know which is the equivalent title in the American system)

The reason I'm starting this thread is that I'm still finishing my freshman year, and athough I really like the subjects, I feel somewhat lost.

That feeling is related to almost anything related to the field: For example, I still don't know what are the job opportunities related to this degree or what can I reallistically expect to achieve with the knowledge I'll have. Basically,the course began and we started studying the tecnichal aspects of algebra, analysis and so on without stopping for a second to understand why, what is this knowledge useful for, etc.

Don't get me wrong, I'm not the kind of guy who needs an inmediate practical use for something in order to have interest about it. I'm not complaining about the knowledge I'm given being "useless"; My problem is that I feel that I'm not really learning, as I seem to be incapable of grasping the ideas behind the technical procedures.

I'll try to be more clear with an example: Right now, I'm studying linear algebra. I've learned how to calculate the determinant of a matrix through Laplace's method. However, I honestly have no idea about what a determinant represents, why knowing how to calculate it is important, or what was it developed for. I just know how to calculate them.

It's pretty bad to me, as I feel I'm just learning mechanichal methods. And that's not quite what I expected, honestly.

So the purpose of this thread is:

1- To know if there are other people here who has had these kind of feelings. I want to know your experiences in the matter.

And, 2 - I'm trying to fix the mess in my head :P How can I understand better these ideas and concepts that my books and teachers don't explain (or that I'm supposed to know, somehow, but I don't)?

Thanks to every of you for reading this pseudo-rant, every response would be appreciated :D

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Re: Lost student

Postby eSOANEM » Wed Aug 08, 2012 9:55 am UTC

Hi. I can't offer any first-hand experience (I'm starting uni next year, I'll be studying natural sciences but will end up with a degree in theoretical physics) but I can pass on the advice I was given by my school. Also, the advice I was given was about the UK so may not translate so well to Spain so take all this with a pinch of salt.

rolo91 wrote:That feeling is related to almost anything related to the field: For example, I still don't know what are the job opportunities related to this degree or what can I reallistically expect to achieve with the knowledge I'll have. Basically,the course began and we started studying the tecnichal aspects of algebra, analysis and so on without stopping for a second to understand why, what is this knowledge useful for, etc.


Degrees in mathematics (particularly at the more applied end of the spectrum such as those including statistics) are very highly sort after by employers. Whatever degree someone picks to study, chances are that what they learnt will not be directly applicable (assuming we're only talking about traditional academic subjects i.e. the arts and sciences). What is relevant to employers and is relevant to how good an employee you can be is how you have been trained to think.

People with mathematical degrees have been trained to think in a very precise and analytical way and this is what employers want. Now, there are some areas of the jobs market which take a larger proportion of people with mathematical degrees than others and these are mainly in financial services (including accounting) but there are very few areas where a maths degree will look worse than another degree and there are lots where it will look a lot better.

Still, your university will probably have a careers service who will probably be able to offer more precise and relevant information (as I said, I was given this advice from my teachers in the UK so it may not apply in Spain so much). I suggest very strongly that if you're having trouble seeing where this degree will lead you, you talk to them.

rolo91 wrote:I'll try to be more clear with an example: Right now, I'm studying linear algebra. I've learned how to calculate the determinant of a matrix through Laplace's method. However, I honestly have no idea about what a determinant represents, why knowing how to calculate it is important, or what was it developed for. I just know how to calculate them.

It's pretty bad to me, as I feel I'm just learning mechanichal methods. And that's not quite what I expected, honestly.


Unfortunately, this is something you're probably going to be stuck with. Often you need to learn the mechanism before you can learn how to use it. In this case, a determinant represents (in some sense) the magnitude of the matrix in that, if we have a matrix M and a vector v then |Mv|=det(M)*|v|. It tells you the scaling factor of the transformation represented by the matrix.

Edit: I got this a bit wrong. Rather, if we have a 3x3 matrix M and some region of space R and define |x| as the volume of the region x then |MR|=det(M)*R. More generally, this is true for any nxn matrix and |x| being the n-volume of the region x rather than the 3-volume.

It is also used in the calculation of inverses of matrices which are themselves used for solving systems of linear equations which in turn can be used to describe many physical phenomena.

It is then also used for finding eigenvalues of a matrix, the eigenvalues being the solutions to the characteristic equation det(M-λI)=0 where M is the matrix, λ the eigenvalue and I the identity matrix. Eigenvalues then are fundamental to quantum mechanics.

The problem is, until you learn how to calculate a determinant, it is somewhat hard to teach these applications. Once you know how to calculate them, by giving a few simple transformations to calculate the determinant of, the student can discover for themself that it represents the scale factor and so, in some sense, the size. Once you know this, it becomes clear that the inverse of a matrix M should be some matrix with determinant 1/det(M), that the matrix is simply the transpose of the matrix of cofactors all multiplied by 1/det(M) is not obvious so that is another method which must be taught.

So, often it is necessary to teach the method before the application. This is clearly not desirable as it makes people feel that they don't understand it properly but, in some cases at least, it is unavoidable.

rolo91 wrote:And, 2 - I'm trying to fix the mess in my head :P How can I understand better these ideas and concepts that my books and teachers don't explain (or that I'm supposed to know, somehow, but I don't)?


If you're not sure about a concept, I've found the fora here really great. I've never posted here asking for help and not ended up understanding it more. You might also find Khan Academy helps with things like linear algebra. Sal has a very good way of explaining concepts and likes to introduce an intuition about what something is before going through the mechanics of it and it sounds like this is something you really want. I think most of his videos have been translated into Spanish too.
Last edited by eSOANEM on Thu Aug 09, 2012 7:59 am UTC, edited 1 time in total.
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z4lis
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Re: Lost student

Postby z4lis » Wed Aug 08, 2012 12:39 pm UTC

Ah, determinants. They're almost magical. Given some matrix, you do some awful computation, and somehow the single number that comes out gives you so much information about the matrix itself, the (arguably) most important piece is whether or not the matrix is invertible, by Cramer's Rule. As eSOANEM says, knowing whether or not a matrix is invertible has applications nearly everywhere. It lets you compute eigenvalues. It lets you determine whether or not you can locally solve a nonlinear system (see this). I believe it also lets you get a good estimate on how nicely a computer will be able to invert your matrix, which is important since solving linear systems is so important in applied math.

But to reinforce what eSOANEM says, it's very difficult to really convey how important determinants are without just showing you how to compute a few, first. I know I didn't quite grasp the depth and importance of linear algebra the first time I saw it, but as you use the tools you realize how powerful they really are.
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

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Re: Lost student

Postby MartianInvader » Wed Aug 08, 2012 10:53 pm UTC

eSOANEM wrote:Unfortunately, this is something you're probably going to be stuck with. Often you need to learn the mechanism before you can learn how to use it. In this case, a determinant represents (in some sense) the magnitude of the matrix in that, if we have a matrix M and a vector v then |Mv|=det(M)*|v|. It tells you the scaling factor of the transformation represented by the matrix.

This, um, isn't true. It does sort of represent a "scaling factor" of the matrix, but it's really easy to find M and v such that |Mv| =/= det(M) *|v| (for example, a non-invertible matrix doesn't have to be the zero matrix). In fact, such v exist for any M that isn't of the form aI.
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

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Re: Lost student

Postby eSOANEM » Thu Aug 09, 2012 7:55 am UTC

MartianInvader wrote:
eSOANEM wrote:Unfortunately, this is something you're probably going to be stuck with. Often you need to learn the mechanism before you can learn how to use it. In this case, a determinant represents (in some sense) the magnitude of the matrix in that, if we have a matrix M and a vector v then |Mv|=det(M)*|v|. It tells you the scaling factor of the transformation represented by the matrix.

This, um, isn't true. It does sort of represent a "scaling factor" of the matrix, but it's really easy to find M and v such that |Mv| =/= det(M) *|v| (for example, a non-invertible matrix doesn't have to be the zero matrix). In fact, such v exist for any M that isn't of the form aI.


You're quite right. If nothing else, my statement would require that all eigenvectors of a transformation had the same eigenvalue. Something which is clearly not the case.

I think I know where I went wrong. I just remembered it from my class as a "scale factor" but didn't stop to think about what was being scaled. A quick search tells me that it is the area/volume/4-volume/n-volume etc. scale factor for a 2x2/3x3/4x4/nxn etc. matrix.

I've added this correction to my original post.
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Re: Lost student

Postby f5r5e5d » Thu Aug 09, 2012 5:36 pm UTC

rolo91 wrote:First things first: Hi to everyone, as this is my first post in these forums :D Also, excuse my English if it's not quite correct, it's not my native language.

I live in Spain, and this last course I started studying a degree in mathematics and statistics (that's how it's called here, I don't know which is the equivalent title in the American system)

The reason I'm starting this thread is that I'm still finishing my freshman year, and athough I really like the subjects, I feel somewhat lost...

I'll try to be more clear with an example: Right now, I'm studying linear algebra. I've learned how to calculate the determinant of a matrix through Laplace's method. However, I honestly have no idea about what a determinant represents, why knowing how to calculate it is important, or what was it developed for. I just know how to calculate them.

It's pretty bad to me, as I feel I'm just learning mechanical methods. And that's not quite what I expected, honestly...



there are several issues - repetition, "mechanical" practice seems necessary to cause the brain to forge new links, make lasting memories, a condition for learning

some make analogies to language learning - a 'babytalk"/babel stage where words, pronunciation, grammar fragments need to be accumulated, at this stage you are certain you "don't know anything" about what you are putatively learning - and then a critical level is reached where enough structure is laid down to make a template for quickly assimilating higher order concepts - the subject "clicks", "jells", "snaps into focus"...

a commonplace observation is that you are really unlikely to be competent in a course' content until you have successfully completed the advanced courses that is was a prerequisite for - the "oh that's what that means", what this or that rule based procedure "really does" moments come more often in the advanced courses


but teaching, maybe most glaringly math teaching, hasn't appreciably adapted to modern computer math sw capabilities either in teaching or, possibly more significantly, in what should be taught when we can assume powerful math sw will be available in any technical job

there is literally a century of institutional inertia over what is appropriate early undergrad applied math

but the tech is changing at a ridiculous pace compared to the teaching innovation adoption rate – as a undergrad engineering/science student I could probably "out integrate" symbolic computer math programs at the time I graduated - the class entering next likely not – and that was 30 years ago

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Re: Lost student

Postby rolo91 » Sun Aug 12, 2012 6:26 pm UTC

Thanks for your responses, both about the determinant's use and about the general idea.

Yes, I'm aware that in some cases, and especially in abstract reasoning subjects, sometimes is necessary some practice in order to understand the advanced applications of a concept or idea. And I understand too, that most uses of the basic things I'm learning now will come when facing more advanced courses.

However, what I find lacking, more than an explanation about the different applications of each idea, is just some introductory explanations : For example, during my analysis classes, every time we faced a new theorem our teacher stopped to give us some background about the mathematician who developed that theorem, the problems he was trying to solve, how did he come up with the solution... It probably had to do with the fact that my teacher also teaches other advanced subject about math history, but it really helped to both make the classes interesting and also to understand better the theorem.

In comparison, every teacher and book I've had until now regarding algebra seems to not only ignore that background, but also it almost seems like the writers are being cryptic on purpose (I swear that the last book I took about algebra in my university's library was written in old Spanish -- just for the hell of it, it was written in 2002 :roll: ).

I don't know, I have the opinion that, if something's understandable, it's also possible to explain it in a more or less simple way, at least as an introduction to the formal explanation... Not that I'm looking for a "linear algebra for dummies" handbook, but I would love to see something more... appealing, may be the word. Like that famous quote... if you can't explain something to your grandmother, then you don't understand it well enough :P

I've seen the Khan Academy link, It seems something close to what I was looking for. If any of you has any books, courses or resources which you think fit what I'm looking for, I would love to see them :D

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Re: Lost student

Postby f5r5e5d » Mon Aug 13, 2012 4:19 am UTC

I am a working BSc level EE, not mathematically sophisticated - If you already have Abstract Algebra, Analysis courses you may be beyond my level already

but I have had to refresh, extend my Linear Algebra a few times - control theory is now fairly advanced/abstract Linear Algebra - at least by my undergrad EE/Science minimal math background

Gilbert Strang's book, MIT opencourseware has been multiply recommended when I ask the PhD's I climb with in the gym for pointers for an engineer

I also attended a ~20 hr, month long refresher lecture series taught from Axelrod's "Linear Algebra Done Right" - makes a point of not using "ugly" determinants in developing most of the theory - a curious stance

though I haven't worked through it Treil's "Linear Algebra Done Wrong" book pdf I found on the web looked fine in a quick skim too - determinants are real handy to the applied engineer's uses for linear algebra

I have Ian MacDondald's "Linear and Geometric Algebra" more because I am a Geometric Algebra fanboy (as distinct from a serious student really working at it)really doesn't get very far into either subject

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Re: Lost student

Postby Yakk » Mon Aug 13, 2012 7:46 pm UTC

So part of the problem is that you are experiencing acceleration. They want to be able to bring you up to competence in a mere 4 years, and they aren't stopping to smell the flowers along the way.

And the people teaching you are not holding your hand. They are throwing information at you as fast as they figure you can absorb it, and leave it up to you to swallow or drown.

...

Linear algebra is the study of a particular set of transformations that behave nicely. In particular, the study of vector spaces over scalars, where you can add vectors and multiply by scalars, and transformations that play nice with said adding of vectors and multiplying of scalars. As it happens, in the "finite dimensional" case, you can represent said linear transformations in the form of matrices (up to a choice of basis for your vector space's dimension), and compute using matrices, and everything works out really nicely.

Once you have learned this toolkit, there are many uses, from abusing them to do affine transformations for computer graphics, to connectivity calculations (which, among other things, is the seed of the google pagerank algorithm), to working out if a particular car design will shake itself to bits when going over a bumpy road (mmm, eigenvalues).

I've heard it argued that almost any problem we find tractable is tractable because we reduced it to linear algebra.

Calculus itself is the game of finding and manipulating linear approximations to functions. We turn the problem of curves into a tractable one by ... reducing it to linear algebra.

However, linear algebra is all of the above applications with the details boiled out. By abstracting the commonality, they hope to teach you techniques that are useful all over the place. However, the problems that linear algebra is useful for end up requiring that you be competent at linear algebra in the first place to understand them easily! (once reduced to a linear algebra problem, the problem becomes easy...) Sort of a catch-22, no?

...

The determinate, while weird, is unique.

It is the unique function on n x n matrices for which:
1) det(I) = 1
2) det(k M) = k^n det(M) for matrix M and scalar k (or, it is linear in each column of the matrix, and linear in each row).
3) det(M) = 0 <-> the matrix cannot be inverted. Alternatively, det(M) = 0 iff the rows of M are linearly dependent on each other (one can be expressed as a linear combination of the others), or ditto columns.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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Re: Lost student

Postby rolo91 » Wed Aug 15, 2012 1:00 am UTC

Yakk wrote:So part of the problem is that you are experiencing acceleration. They want to be able to bring you up to competence in a mere 4 years, and they aren't stopping to smell the flowers along the way.

And the people teaching you are not holding your hand. They are throwing information at you as fast as they figure you can absorb it, and leave it up to you to swallow or drown.



I don't know exactly why, but that words were quite motivational. Thanks.


The determinate, while weird, is unique.

It is the unique function on n x n matrices for which:
1) det(I) = 1
2) det(k M) = k^n det(M) for matrix M and scalar k (or, it is linear in each column of the matrix, and linear in each row).
3) det(M) = 0 <-> the matrix cannot be inverted. Alternatively, det(M) = 0 iff the rows of M are linearly dependent on each other (one can be expressed as a linear combination of the others), or ditto columns.


Yes, I know the definition and its relation with the matrix rank. What bothers me about determinants is precisely its formal deffinition. When I started learning, every concept seemed to appear as obvious once you got into it. But the determinants' deffinition seems to me like pulled out of somebody's sleeve. I know it works, I know it has a lot of applications, it just seems to appear in the books/lessons out of thin air, like a deux ex machina :|

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Re: Lost student

Postby Yakk » Wed Aug 15, 2012 2:14 am UTC

You could take a look at 2x2 determinates, and work out how it corresponds to 2-volume (ie, area) transformations of parallelograms? Then convince yourself it is the only such operation that works...
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

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Re: Lost student

Postby Sizik » Wed Aug 15, 2012 2:49 am UTC

In my university math class, a sequence of four courses appropriately* titled "Advanced Calculus", we derived the formula for determinants from its properties. Good thing I still have my notes!
Spoiler:
1. Axioms
detn is a function on nxn matrices that gives a number. We want it to have properties that will make |det M| = the volumn of the (n-dimensional) parallelepiped created using the n row vectors in M.
We'll use mj to denote the vector appearing as the jth row of M, and write detn(m1, ... , mn) instead of det(M).
Axiom 1: detn(m1, ... , mj-1, t*mj, mj+1, ... , mn)
= t*detn(m1, ... , mn).
(Scaling one of the edges of the parallelepiped scales the volume by the same amount.)

Axiom 2: detn(m1, ... , mj-1, A+B, mj+1, ... , mn)
= detn(m1, ... , mj-1, A, mj+1, ... , mn)
+ detn(m1, ... , mj-1, B, mj+1, ... , mn)
(If you split the parallelepiped in two along one of the edges, the volumes of the resulting smaller parallelepipeds sum to that of the larger one.)

Axiom 3: detn(M) = 0 if there are two rows that are equal.
(if two edges are equal, than the parallelepiped doesn't have any n-volume)

Axiom 4: detn(I) = 1
(All edges have length one, and thus the figure has volume 1).

2. Row operations

Our first step is to see what happens to the determinant when you do various operations on the rows.
(using det(A, B) to mean the determinant of a matrix with arbitrary rows A and B)

1. Multiply a row by a constant => multiply the determinant by the constant.
(Axiom 1.)

2. Adding a multiple of a row to another row => determinant unchanged.
( Proof: det(A+tB, B) = det(A, B) + det(tB, B) = det(A, B) + t*det(B, B) = det(A, B). )

3. Swapping two rows => determinant multiplied by -1.
( Proof: det(A, B) = det(A+B, B) = det(A+B, B-(A+B)) = det(A+B, -A) = det(B, -A) = -det(B, A). )

4. If {mj} is linearly dependent, then detn(m1, ... , mn) = 0.
( Linearly dependent means that one of the vectors is a multiple of another => Axioms 1 and 3.)

At this point, you can calculate any determinant by simply doing Gauss-Jordan elimination and keeping track of what the determinant gets multiplied by.

3. Cofactor Expansion

Suppose you wanted to compute
[math]\det{A} = \begin{vmatrix}
a_{1,1} & \cdots & a_{1,n} \\
\vdots & \ddots & \vdots \\
a_{m,1} & \cdots & a_{m,n}
\end{vmatrix}[/math]

You can view the top row as being a1,1e1 + a1,2e2 + ... + a1,nen, where e1 = [1,0,...,0], e2 = [0, 1, 0, ... 0] and so forth.
By breaking over sum and pulling out factors,
[math]\det{A} = a_{1,1} \begin{vmatrix} 1 & 0 & \cdots & 0 \\ a_{2,1} & \cdots \\ \vdots & \ddots \end{vmatrix} + a_{1,2} \begin{vmatrix} 0 & 1 & 0 & \cdots & 0 \\ a_{2,1} & \cdots \\ \vdots & \ddots \end{vmatrix} + \cdots + a_{1,n} \begin{vmatrix} 0 & \cdots & 0 & 1 \\ a_{2,1} & \cdots \\ \vdots & \ddots \end{vmatrix}[/math]
[math]= \sum_{j=1}^n a_{1,j} \begin{vmatrix} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ a_{2,1} & \cdots \\ \vdots & \ddots \end{vmatrix} = \sum_{j=1}^n a_{1,j} \begin{vmatrix} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ & & & 0 \\ & A & & \vdots & & A \\ & & & 0 \end{vmatrix}[/math]
where the 1's are in the jth column and the A's are just the rest of the matrix that was there before. (The last part is gotten by adding -ak,j times the first row to each row k, which doesn't change the determinant)

The matrix [math]\sum_{j=1}^n a_{1,j} \begin{bmatrix} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ & & & 0 \\ & A & & \vdots & & A \\ & & & 0 \end{bmatrix}[/math] is called the (1,k) cofactor of A. In general, replacing the jth row and kth column with 0's except for a 1 at (j,k) gives you the (j,k) cofactor.

4. Determinant of the cofactor
We want to find
[math]\begin{vmatrix} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ & & & 0 \\ & A & & \vdots & & A \\ & & & 0 \end{vmatrix}[/math]
where the 1 is in the kth column.
We can move the top row to the kth position by swapping rows (which multiplies the determinant by -1 each time):
[math]= (-1)^{k-1}\begin{vmatrix} a_{2,1} & \cdots & a_{2,k-1} & 0 \\ \vdots & \ddots & \vdots & \vdots \\ a_{k-1,1} & \cdots & a_{k-1,k-1} & 0 \\ 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ & & & 0 \\ & & & \vdots \\ & & & 0\end{vmatrix}[/math]
There is a trick to see the determinant is unchanged if you just delete the row and column with the 0's and 1's. Look at the above as a function of rows
[math]\begin{align}& (a_{2,1},\ldots,a_{2,k-1},a_{2,k+1},\ldots,a_{2,n})\\
& (a_{3,1},\ldots,a_{3,k-1},a_{3,k+1},\ldots,a_{3,n}) \\
& \text{no }k^{\text{th}}\text{ row} \rightarrow \vdots \\
& (a_{n,1},\ldots,a_{n,n})\end{align}[/math]
As a function of these n-1 rows, an nxn determinant with the special rows and columns added, it has linearity, is 0 if two rows are equal, and equals 1 if the rows are from In-1, so it is the (n-1)x(n-1) determinant.

Thus,
[imath]\begin{vmatrix} a_{1,1} & \cdots & a_{1,n} \\ \vdots & \ddots & \vdots \\ a_{m,1} & \cdots & a_{m,n} \end{vmatrix} = \sum_{j=1}^n a_{1,j} \begin{vmatrix} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ & & & 0 \\ & A & & \vdots & & A \\ & & & 0 \end{vmatrix} = \sum_{j=1}^n a_{1,j} (-1)^{j-1}\begin{vmatrix} a_{2,1} & \cdots & a_{2,j-1} & a_{2,j+1} & \cdots & a_{2,n} \\ \vdots & \ddots & & & & \vdots \\ a_{m,1} & \cdots & \cdots & \cdots & \cdots & a_{m,n} \end{vmatrix}[/imath]
which is how you calculate determinants.


*We started off by constructing the real numbers from scratch, and then derived integration without using limits.
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Re: Lost student

Postby rolo91 » Wed Aug 15, 2012 10:16 pm UTC

Sizik wrote:In my university math class, a sequence of four courses appropriately* titled "Advanced Calculus", we derived the formula for determinants from its properties. Good thing I still have my notes!
Spoiler:
1. Axioms
detn is a function on nxn matrices that gives a number. We want it to have properties that will make |det M| = the volumn of the (n-dimensional) parallelepiped created using the n row vectors in M.
We'll use mj to denote the vector appearing as the jth row of M, and write detn(m1, ... , mn) instead of det(M).
Axiom 1: detn(m1, ... , mj-1, t*mj, mj+1, ... , mn)
= t*detn(m1, ... , mn).
(Scaling one of the edges of the parallelepiped scales the volume by the same amount.)

Axiom 2: detn(m1, ... , mj-1, A+B, mj+1, ... , mn)
= detn(m1, ... , mj-1, A, mj+1, ... , mn)
+ detn(m1, ... , mj-1, B, mj+1, ... , mn)
(If you split the parallelepiped in two along one of the edges, the volumes of the resulting smaller parallelepipeds sum to that of the larger one.)

Axiom 3: detn(M) = 0 if there are two rows that are equal.
(if two edges are equal, than the parallelepiped doesn't have any n-volume)

Axiom 4: detn(I) = 1
(All edges have length one, and thus the figure has volume 1).

2. Row operations

Our first step is to see what happens to the determinant when you do various operations on the rows.
(using det(A, B) to mean the determinant of a matrix with arbitrary rows A and B)

1. Multiply a row by a constant => multiply the determinant by the constant.
(Axiom 1.)

2. Adding a multiple of a row to another row => determinant unchanged.
( Proof: det(A+tB, B) = det(A, B) + det(tB, B) = det(A, B) + t*det(B, B) = det(A, B). )

3. Swapping two rows => determinant multiplied by -1.
( Proof: det(A, B) = det(A+B, B) = det(A+B, B-(A+B)) = det(A+B, -A) = det(B, -A) = -det(B, A). )

4. If {mj} is linearly dependent, then detn(m1, ... , mn) = 0.
( Linearly dependent means that one of the vectors is a multiple of another => Axioms 1 and 3.)

At this point, you can calculate any determinant by simply doing Gauss-Jordan elimination and keeping track of what the determinant gets multiplied by.

3. Cofactor Expansion

Suppose you wanted to compute
[math]\det{A} = \begin{vmatrix}
a_{1,1} & \cdots & a_{1,n} \\
\vdots & \ddots & \vdots \\
a_{m,1} & \cdots & a_{m,n}
\end{vmatrix}[/math]

You can view the top row as being a1,1e1 + a1,2e2 + ... + a1,nen, where e1 = [1,0,...,0], e2 = [0, 1, 0, ... 0] and so forth.
By breaking over sum and pulling out factors,
[math]\det{A} = a_{1,1} \begin{vmatrix} 1 & 0 & \cdots & 0 \\ a_{2,1} & \cdots \\ \vdots & \ddots \end{vmatrix} + a_{1,2} \begin{vmatrix} 0 & 1 & 0 & \cdots & 0 \\ a_{2,1} & \cdots \\ \vdots & \ddots \end{vmatrix} + \cdots + a_{1,n} \begin{vmatrix} 0 & \cdots & 0 & 1 \\ a_{2,1} & \cdots \\ \vdots & \ddots \end{vmatrix}[/math]
[math]= \sum_{j=1}^n a_{1,j} \begin{vmatrix} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ a_{2,1} & \cdots \\ \vdots & \ddots \end{vmatrix} = \sum_{j=1}^n a_{1,j} \begin{vmatrix} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ & & & 0 \\ & A & & \vdots & & A \\ & & & 0 \end{vmatrix}[/math]
where the 1's are in the jth column and the A's are just the rest of the matrix that was there before. (The last part is gotten by adding -ak,j times the first row to each row k, which doesn't change the determinant)

The matrix [math]\sum_{j=1}^n a_{1,j} \begin{bmatrix} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ & & & 0 \\ & A & & \vdots & & A \\ & & & 0 \end{bmatrix}[/math] is called the (1,k) cofactor of A. In general, replacing the jth row and kth column with 0's except for a 1 at (j,k) gives you the (j,k) cofactor.

4. Determinant of the cofactor
We want to find
[math]\begin{vmatrix} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ & & & 0 \\ & A & & \vdots & & A \\ & & & 0 \end{vmatrix}[/math]
where the 1 is in the kth column.
We can move the top row to the kth position by swapping rows (which multiplies the determinant by -1 each time):
[math]= (-1)^{k-1}\begin{vmatrix} a_{2,1} & \cdots & a_{2,k-1} & 0 \\ \vdots & \ddots & \vdots & \vdots \\ a_{k-1,1} & \cdots & a_{k-1,k-1} & 0 \\ 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ & & & 0 \\ & & & \vdots \\ & & & 0\end{vmatrix}[/math]
There is a trick to see the determinant is unchanged if you just delete the row and column with the 0's and 1's. Look at the above as a function of rows
[math]\begin{align}& (a_{2,1},\ldots,a_{2,k-1},a_{2,k+1},\ldots,a_{2,n})\\
& (a_{3,1},\ldots,a_{3,k-1},a_{3,k+1},\ldots,a_{3,n}) \\
& \text{no }k^{\text{th}}\text{ row} \rightarrow \vdots \\
& (a_{n,1},\ldots,a_{n,n})\end{align}[/math]
As a function of these n-1 rows, an nxn determinant with the special rows and columns added, it has linearity, is 0 if two rows are equal, and equals 1 if the rows are from In-1, so it is the (n-1)x(n-1) determinant.

Thus,
[imath]\begin{vmatrix} a_{1,1} & \cdots & a_{1,n} \\ \vdots & \ddots & \vdots \\ a_{m,1} & \cdots & a_{m,n} \end{vmatrix} = \sum_{j=1}^n a_{1,j} \begin{vmatrix} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ & & & 0 \\ & A & & \vdots & & A \\ & & & 0 \end{vmatrix} = \sum_{j=1}^n a_{1,j} (-1)^{j-1}\begin{vmatrix} a_{2,1} & \cdots & a_{2,j-1} & a_{2,j+1} & \cdots & a_{2,n} \\ \vdots & \ddots & & & & \vdots \\ a_{m,1} & \cdots & \cdots & \cdots & \cdots & a_{m,n} \end{vmatrix}[/imath]
which is how you calculate determinants.


*We started off by constructing the real numbers from scratch, and then derived integration without using limits.


Wow, thank you! That was really clarifying. That kind of explanation is what I was looking for :D

That was a copypaste of your notes, or did you write all that for the post? If it's the second case, then a million thanks for taking the time. I think you could do quite well as a teacher (if you aren't already).

User avatar
Sizik
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Re: Lost student

Postby Sizik » Wed Aug 15, 2012 11:39 pm UTC

Copied from my handwritten notes.
she/they
gmalivuk wrote:
King Author wrote:If space (rather, distance) is an illusion, it'd be possible for one meta-me to experience both body's sensory inputs.
Yes. And if wishes were horses, wishing wells would fill up very quickly with drowned horses.


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