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This annoys me the most when using the cross product. How is it that the determinant of the matrice is the magnitude for the new vector?

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This annoys me the most when using the cross product. How is it that the determinant of the matrice is the magnitude for the new vector?

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If the matrix is considered an operator or linear transformation, the determinant is then the "scale factor" which describes how the operand transforms in size (vector norm, for example)

- #3

matt grime

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In general it is a little more tricky to prove, and but can be explained thusly:

given a vector space V there is another vector space [itex]\Lambda^2V[/itex] (which I shall for typesetting reasons label /\^2(V)) called the alternating or antisymmetric square. If e_i is a basis of V then /\^2(V) then has as basis the symbols e_i/\e_j subject to the relation that e_i/\e_j=-e_j/\e_i. That is swapping over the arguments changes sign, hence the name alternating. There are other rules too such as the symbol (e_i+e_j)/\e_k = e_i/\e_k + e_j/\e_k.

Now, I can repeat this construction and define /\^n(V) by taking as a basis the 'n-fold wedge', so this has basis the symbols where we have n terms not two, but with the same rules that if we swap over two arguments we change the sign.

What happens if n=dimV? I claim that the resutling space /\^n(V) is one dimensional. To see this notice that if any wedge has a repeated element then it is identified with the zero vector (in /\^2 where it is easier to write e/\f=-f/\e, so if I let e=f, then e/\e=-e/\e, so e/\e=0). What this means is that because of the rule about swapping things round, given e_i/\e_j/\../\e_k I can swap them round so that the subscripts are all increasing, and because if I repeat a subscript I get zero then for non-zero terms these subscripts must be strictly increasing. If n=dimV how can choose a strictly increasing set of length n from the numbers 1,..,n? In exactly one way, so there is exactly one basis vector.

With n=dimV, what /\^n represents is volume. If you take n vectors, write them in terms of the basis elements, then wedge them together the resulting number (1-d vector spaces are jsut numbers remember) is the volume of the object that the vectors span (we're normalizing things so that we consider the basis vectors of V to have length 1). It is easiest to justify this by saying: consider some box defined by n vectors, if you double all the lengths of all the vectors you multiply the volume by 2^n, which is exactly waht should happen to an n dimensional volume.

So, back to determinants. Given a linear map M acting on V I can make it act on /\^n(V) by just letting M act on each of the components of a wedge of n things at the same time. When n=dim(V) M now just acts by multiplication by a number. That number is what we define to be det(M). Using the properties of wedges you can also now work out a formula for the determinant of any nxn matrix, and it gives precisely the things you know in the 2x2 and 3x3 case. Here are the determinants written out for small matrices

M= m_{ij} for 1<=i,j<=2

det(M)=[itex] m_{11}m_{22} - m_{12}m_{21}[\itex]

now for i,j>=3

det(M)=

[tex]m_{11}m_{22}m_{33}- m_{12}m_{21}m_{33}-m_{13}m_{22}m_{31} - m_{11}m_{23}m_{32} + m_{12}m_{23}m_{31}+m_{13}m_{21}m_{32}+m_{13}m_{22}m_{31}[/tex]

I've deliberately written in that form so that you can try to spot a pattern: the subscripts all go [itex]m_{1i}m_{2j}m_{3k}[/tex] and {i,j,k}={1,2,3}, the sign in front tells me how many swaps I made in the order of 1,2,3 to get i,j,k, i.e. to get from 1,2,3 to 2,1,3 you need one swap, 1,2,3 to 2,3,1 needs two swaps. Obviously 'the number of swaps' is not well defined, but what is well defined is the parity, no matter how you get from 1,2,3 to 2,1,3 you will always use an odd number, and from 1,2,3 to 2,3,1 an even number.

These formulae follow from the rules about things in /\^n. Try it for the 2x2 case and see what happens: If we take the usual 2x2 matrix with entries a,b,c,d, where does it send e_1? To ae_1+ce_2, and e_2 goes to be_1+de_2. Now what is

(ae_1+ce_2)/\(be_1+de_2)?

We can distribute over the brackets:

abe_1/\e_1 + ade_1/\e_2 + cbe_2/\e_1+cde_2/\e_2.

Remember repeated indices mean the symbol is equal to zero:

ade_1/\e_2+bce_2/\e_1

and we swap signs if we swap arguments:

ade_1/\e_2 - bc e_1/\e_2 = (ad-bc)e_1/\e_2

as required.

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mathwonk

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e.g. the area of the parallelogram spanned by (a,b) and (c,d) is the absolute value of ad-bc.

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matt grime

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Beware of LaTeX error in my post that I cannot now correct.

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arildno

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Sculptured:

1. The first thing to remember is:

How is the number called the determinant of the matrix DEFINED?

Not knowing your book, I expect your book defines it to be the number given by an unambiguous computation process

(i.e resolving the "how to compute the determinant" issue)

Consider this to furnish you with the "core concept" of what the determinant is. That core concept is very simple to grasp:

The determinant is just some number you get out of a particular computation process.

It is VERY important that you understand that we are at perfect liberty in maths to define whatever objects we want as long as the definition of it is clear!

This is a totally separate issue from the issue WHY should we bother about some object defined in this or that way!

2. It can be PROVEN that this number has certain PROPERTIES, that has been excellently explained to you already.

These proofs then, enriches your rather trite core concept of the determinant by pointing out non-intuitive features of that number, and really furnish us with the justification of why it was important to bother about the "determinant" in the first place.

Pay therefore particular care to such proofs; be sure you understand why they are correct!

1. The first thing to remember is:

How is the number called the determinant of the matrix DEFINED?

Not knowing your book, I expect your book defines it to be the number given by an unambiguous computation process

(i.e resolving the "how to compute the determinant" issue)

Consider this to furnish you with the "core concept" of what the determinant is. That core concept is very simple to grasp:

The determinant is just some number you get out of a particular computation process.

It is VERY important that you understand that we are at perfect liberty in maths to define whatever objects we want as long as the definition of it is clear!

This is a totally separate issue from the issue WHY should we bother about some object defined in this or that way!

2. It can be PROVEN that this number has certain PROPERTIES, that has been excellently explained to you already.

These proofs then, enriches your rather trite core concept of the determinant by pointing out non-intuitive features of that number, and really furnish us with the justification of why it was important to bother about the "determinant" in the first place.

Pay therefore particular care to such proofs; be sure you understand why they are correct!

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arildno said:Sculptured:

1. The first thing to remember is:

How is the number called the determinant of the matrix DEFINED?

Not knowing your book, I expect your book defines it to be the number given by an unambiguous computation process

(i.e resolving the "how to compute the determinant" issue)

Consider this to furnish you with the "core concept" of what the determinant is. That core concept is very simple to grasp:

The determinant is just some number you get out of a particular computation process.

It is VERY important that you understand that we are at perfect liberty in maths to define whatever objects we want as long as the definition of it is clear!

This is a totally separate issue from the issue WHY should we bother about some object defined in this or that way!

2. It can be PROVEN that this number has certain PROPERTIES, that has been excellently explained to you already.

These proofs then, enriches your rather trite core concept of the determinant by pointing out non-intuitive features of that number, and really furnish us with the justification of why it was important to bother about the "determinant" in the first place.

Pay therefore particular care to such proofs; be sure you understand why they are correct!

Although the book (Elementary Linear Algebra by Larason/Edwards/Falvo) gives the definition of a determinant of a 2x2 matrix in the first section dealing with the determinant, it leaves me with nothing. After all, my teacher seems to already assume that because we have taken precalculus before, we can already use the determinant to check for the consistency of the linear system. The biggest problem with this is I can't see why this simple calculation can tell us something that important about the system. I can at least begin to have an idea what this calculation means from the way mathwonk put it. Otherwise, sorry to say, but the use of concepts and terms such as vector spaces is something I have yet to reach. Looks like I will have to come back to this once I finish my course.

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jbusc said:

If the matrix is considered an operator or linear transformation, the determinant is then the "scale factor" which describes how the operand transforms in size (vector norm, for example)

So, if our matrix is the Hamiltonian, then det(H) tells you...something about the effect of this linear tranformation on a vector?

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matt grime

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Einstein Mcfly said:So, if our matrix is the Hamiltonian, then det(H) tells you...something about the effect of this linear tranformation on a vector?

The only thing you can say is that if the matrix has non-zero determinant the image of some (non-zero) vector is not zero. Apart from that the determinant tells you nothing about the action of a matrix on a vector (ignore the 1x1 case), and there is no need to introduce the notion that the matrix is a hamiltonian (in whatever sense you mean that).

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matt grime said:The only thing you can say is that if the matrix has non-zero determinant the image of some (non-zero) vector is not zero. Apart from that the determinant tells you nothing about the action of a matrix on a vector (ignore the 1x1 case), and there is no need to introduce the notion that the matrix is a hamiltonian (in whatever sense you mean that).

Well, we know that the determinant of a Hermetian operator is constant with respect to basis change, but others have been saying that the value itself has some meaning (I'd imagine only relative to the determinant of some other matrix). So if det(A)=a and det(B)=b and a>b, what does that tell you about A and B relative to one another or relative to their effects on some vector? Thanks for the response, by the way.

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matt grime

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The value of the determinant has a meaning, but that meaning doesn't specify the action of the matrix on *one* vector. It tells you about the action of the matrix on volumes - spaces enclosed by n linearly independent vectors in R^n.

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matt grime said:

The value of the determinant has a meaning, but that meaning doesn't specify the action of the matrix on *one* vector. It tells you about the action of the matrix on volumes - spaces enclosed by n linearly independent vectors in R^n.

Alright, so the determinant of an operator tells you about its action on volumes of basis in R^n; so if det(A)=a and det(B)=b, and a>b, what does that tell you about the operators A and B relative to each other? If we switch from basis {x} to basis {x'}, how does a, b, and (a-b) vary, and what does this say about A and B in the {x} relative to the {x'} basis?

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matt grime

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Determinants are basis invariant: that is an easy exercise.

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matt grime said:Determinants are basis invariant: that is an easy exercise.

Yeah, I knew that.

So does that the value of the determinant of one operator tell you relative to another?

- #15

matt grime

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That they have different effects on volumes.

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matt grime said:That they have different effects on volumes.

:rofl:

Okay, you seem to be sick of this line of questioning. Thanks for your reponses.

Anyone else have a thought on what operator A having a "larger effect on volumes" has to say about it relative to B if is has a "smaller effect on volumes"? That is, does the determinant tell you anything INTERESTING about the operator relative to another operator?

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matt grime

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matt grime said:

My question is what that means physically. Does this property of the determinate of an operator tell you anything interesting about the PHYSICS. I understand what you’ve said in terms of the geometrical interpretation in R^n. What I want to know is what the physical significance is of this property, if any. Thanks.

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You can find a general definition for a determinant http://en.wikipedia.org/wiki/Determinant#General_definition_and_computation".

I didn't see someone else mention this way of looking at it but it seems pleasant to me since it is the definition.

I didn't see someone else mention this way of looking at it but it seems pleasant to me since it is the definition.

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matt grime

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omagdon: that is what I wrote, but without demonstrating how to find this unique alternating form.

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Thanks for the answer.matt grime said:Einstein Mclfy: That would entirely depend on what they matrix was. In general it is the eigenvalues that are important, and you cannot read off those from the determinant - you need more information.

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Try looking at it in this way...

Remember the following rules:

A row of a matrix can have all of its entries be multiplied by a number. The effect that this has on the determinant of this matrix is that the determinant gets multiplied by the number.

If you do this on all rows of the matrix with the same number, you need to multiply the determinant by the number to the Nth power. (if the matrix is NxN)

Now, let the determinant of an NxN matrix [tex]L[/tex] be some non-zero number given by [tex]\lambda[/tex].

[tex]det(L)=\lambda[/tex]

Now, take the Nth root of lambda.

[tex]\sqrt[N]{\lambda}[/tex]

Multiply each of the rows (there are N of them) in [tex]L[/tex] by [tex]\frac{1}{\sqrt[N]{\lambda}}[/tex]. Recall that this is the same as multiplying the*determinant* by [tex]\frac{1}{\lambda}[/tex], it is also equivalent to multiplying the entire *matrix* by [tex]\frac{1}{\sqrt[N]{\lambda}}[/tex]. Let's give this new matrix a name, M. So,

[tex]M=\frac{1}{\sqrt[N]{\lambda}}L[/tex]

The determinant of this new matrix M is ONE (by design).

Now, let's look at the difference between what L and M do to a vector [tex]\vec{v}[/tex].

[tex]L\vec{v}=\vec{l}[/tex]

[tex]M\vec{v}=\vec{m}[/tex]

But, we know that [tex]M=\frac{1}{\sqrt[N]{\lambda}}L[/tex] so the last equation becomes

[tex]\frac{1}{\sqrt[N]{\lambda}}L\vec{v}=\vec{m}[/tex]

Let's multiply by [tex]\sqrt[N]{\lambda}[/tex] to get

[tex]L\vec{v}=\sqrt[N]{\lambda}\vec{m}[/tex]

When we compare the last equation with [tex]L\vec{v}=\vec{l}[/tex], we can see that

[tex]\vec{l}=\sqrt[N]{\lambda}\vec{m}[/tex]

Meaning that the action of L and M on a vector [tex]\vec{v}[/tex] are simply related by multiplication of a numerical factor. If you divide the value of the determinant of an NxN matrix by a certain number b, it multiplies the vector it is acting on by the Nth root of 1/b. To show that explicitly, solve for [tex]\vec{m}[/tex] in the equation

[tex]\vec{l}=\sqrt[N]{\lambda}\vec{m}[/tex]

to get

[tex]\frac{1}{\sqrt[N]{\lambda}}\vec{l}=\vec{m}[/tex]

This means M acting on [tex]\vec{v}[/tex] gives

[tex]M\vec{v}=\frac{1}{\sqrt[N]{\lambda}}\vec{l}[/tex]

The action of M is shrunk by comparison with L's action, by a factor of the 1/Nth root of lambda. You could also look at it in a reciprocal way, solving the other way around and getting this equation

[tex]L\vec{v}=\sqrt[N]{\lambda}\vec{m}[/tex]

Or that L's action on v is stretched by [tex]\sqrt[N]{\lambda}[/tex] when compared to the action of M. This simple relationship is only true for a special case however, remember that M is basically the same as L, just changed by a scale factor. If the matrices are completely different, the volume (read: all basis vectors) needs to be looked at, not just a single vector. In other words, one matrix could transform stuff with a determinant of 1 and another different transformation could transform with determinant 1. But one could be stretching vectors in the x direction while shrinking in the y in a perfect way to keep the volume the same while the other could be stretching in the z and shrinking in the x and you wouldn't be able to tell from the determinant.

Remember the following rules:

A row of a matrix can have all of its entries be multiplied by a number. The effect that this has on the determinant of this matrix is that the determinant gets multiplied by the number.

If you do this on all rows of the matrix with the same number, you need to multiply the determinant by the number to the Nth power. (if the matrix is NxN)

Now, let the determinant of an NxN matrix [tex]L[/tex] be some non-zero number given by [tex]\lambda[/tex].

[tex]det(L)=\lambda[/tex]

Now, take the Nth root of lambda.

[tex]\sqrt[N]{\lambda}[/tex]

Multiply each of the rows (there are N of them) in [tex]L[/tex] by [tex]\frac{1}{\sqrt[N]{\lambda}}[/tex]. Recall that this is the same as multiplying the

[tex]M=\frac{1}{\sqrt[N]{\lambda}}L[/tex]

The determinant of this new matrix M is ONE (by design).

Now, let's look at the difference between what L and M do to a vector [tex]\vec{v}[/tex].

[tex]L\vec{v}=\vec{l}[/tex]

[tex]M\vec{v}=\vec{m}[/tex]

But, we know that [tex]M=\frac{1}{\sqrt[N]{\lambda}}L[/tex] so the last equation becomes

[tex]\frac{1}{\sqrt[N]{\lambda}}L\vec{v}=\vec{m}[/tex]

Let's multiply by [tex]\sqrt[N]{\lambda}[/tex] to get

[tex]L\vec{v}=\sqrt[N]{\lambda}\vec{m}[/tex]

When we compare the last equation with [tex]L\vec{v}=\vec{l}[/tex], we can see that

[tex]\vec{l}=\sqrt[N]{\lambda}\vec{m}[/tex]

Meaning that the action of L and M on a vector [tex]\vec{v}[/tex] are simply related by multiplication of a numerical factor. If you divide the value of the determinant of an NxN matrix by a certain number b, it multiplies the vector it is acting on by the Nth root of 1/b. To show that explicitly, solve for [tex]\vec{m}[/tex] in the equation

[tex]\vec{l}=\sqrt[N]{\lambda}\vec{m}[/tex]

to get

[tex]\frac{1}{\sqrt[N]{\lambda}}\vec{l}=\vec{m}[/tex]

This means M acting on [tex]\vec{v}[/tex] gives

[tex]M\vec{v}=\frac{1}{\sqrt[N]{\lambda}}\vec{l}[/tex]

The action of M is shrunk by comparison with L's action, by a factor of the 1/Nth root of lambda. You could also look at it in a reciprocal way, solving the other way around and getting this equation

[tex]L\vec{v}=\sqrt[N]{\lambda}\vec{m}[/tex]

Or that L's action on v is stretched by [tex]\sqrt[N]{\lambda}[/tex] when compared to the action of M. This simple relationship is only true for a special case however, remember that M is basically the same as L, just changed by a scale factor. If the matrices are completely different, the volume (read: all basis vectors) needs to be looked at, not just a single vector. In other words, one matrix could transform stuff with a determinant of 1 and another different transformation could transform with determinant 1. But one could be stretching vectors in the x direction while shrinking in the y in a perfect way to keep the volume the same while the other could be stretching in the z and shrinking in the x and you wouldn't be able to tell from the determinant.

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- #23

lavinia

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This annoys me the most when using the cross product. How is it that the determinant of the matrice is the magnitude for the new vector?

The rows of the matrix span an n dimensional parallelapiped. Its volume equals the absolute value of the determinant.

The proof is an exercise in Euclidean geometry.

I think - but am not 100% sure - that algebraically, the determinant may be defined as the unique multilinear function of the rows of a matrix that assigns the value 1 to the identity matrix and changes sign if two adjacent rows are switched.

These two algebraic properties can be used to deduce the formula for the determinant.

- #24

mathwonk

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well volumes tell you also something about uniqueness of solvability of equations. I.e. a linear operator sends some non zero vector to zero, and hence send some two vectors to the same value, if and only if it sends some block of non zero volume to a lower dimensional block, i.e., to a block of zero volume. And this happens if and only if the determinant is zero. Looked at another way, isomorphisms can be recognized by having non zero determinant. So the most interesting distinction provided by the determinant is between operators that multiply volumes by non zero scalars as opposed to ones that essentially kill volumes altogether, i.e. that multiply them by zero. But it is also interesting to distinguish operators that multiply volumes by positive numbers as opposed to ones that multiply by negative numbers, as the latter change orientation i space.

But I myself think that volume is intrinsically interesting, after that's what integrals compute. If you look at the change of variable formula for a higher dimensional integral, you will see a determinant. I was about to refer you to my free linear algebra book on my web page, but they seem to have removed the link to my page since I retired. oops. I guess I no longer exist in cyberspace.

Here is a link http://www.math.uga.edu/vigre/STP.html [Broken]

where on pages 63-68 of "roy smith's notes for math 4050" I give a complete treatment of determinants including proofs of uniqueness.

try this:

www.math.uga.edu/~roy

wow when was THAT picture taken? Actually I still own that shirt so sometime in the last 40 years.

But I myself think that volume is intrinsically interesting, after that's what integrals compute. If you look at the change of variable formula for a higher dimensional integral, you will see a determinant. I was about to refer you to my free linear algebra book on my web page, but they seem to have removed the link to my page since I retired. oops. I guess I no longer exist in cyberspace.

Here is a link http://www.math.uga.edu/vigre/STP.html [Broken]

where on pages 63-68 of "roy smith's notes for math 4050" I give a complete treatment of determinants including proofs of uniqueness.

try this:

www.math.uga.edu/~roy

wow when was THAT picture taken? Actually I still own that shirt so sometime in the last 40 years.

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