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albedo

A few questions regarding Determinants

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albedo    1
Hi,
I know how to compute determinants and I'm familiar with the geometrical meaning of determinant
as the scaling factor of a unit (point/square/cube/hypercube)'s area/volume by applying a linear transformation (using a matrix).
However, I have several questions:
  1. Let's say I define determinant to have the above meaning. How can one
    derive the formula for computing determinant following just the visual/geometrical meaning?
  2. Let's say I have an arbitrary closed 2D polytope [latex]P[/latex] and I transform all of its vertices by a matrix [latex]\mathbf{A}[/latex].
    Is [latex]\det\left(\mathbf{A}\right)[/latex] the scaling factor of polytope's [latex]P[/latex] area after the transformation, i.e. [latex]\det\left(\mathbf A\right) = \frac{P\text{'s area after transform}}{P\text{'s area before transform}}[/latex]?
  3. Imagine I have an open 2D polytope [latex]\overline P[/latex] (which clearly doesn't have any area).
    How does [latex]\det\left(\mathbf{A}\right)[/latex] relate with the transformed polytope [latex]\mathbf {A}\overline P[/latex]?
  4. Suppose there's a vector [latex]\mathbf x[/latex]. What does [latex]\det\left(\mathbf{A}\right)[/latex] say about the transformed vector [latex]\mathbf {Ax}[/latex]?
I'd be glad to get answer to any one of these questions. Thanks.

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ajb    1567

2. is correct... but can be stated more carefully and a little more generally

 

Let us work in [math]\mathbb{R}^n[/math] and let us pick a basis [math](e_{1}, e_{2}, \cdots , e_{n})[/math].

 

The general motion of a volume here is via the wedge product - totally antisymmetric product of vectors.

 

[math] Vol(e) = | e_{1}\wedge e_{2} \wedge \cdots \wedge e_{n} | [/math]

 

gives the volume in the basis given above. The absolute value is with respect to the obvious Euclidean norm - we could do something more general here, but not for now.

 

Now let us take some linear transformation - which we know can always be written as a matrix. As a linear operator we have

 

[math]A(e_{1}\wedge e_{2} \wedge \cdots \wedge e_{n}) = A(e_{1})\wedge A( e_{2}) \wedge \cdots \wedge A( e_{n})[/math].

 

As we are working with n vectors in an n-dimensional space and the product is totally antisymmetric the Volume is an element of a one dimensional vector space. Any changes in this volume can always be written as the volume multiplied by some scalar.

 

Now lets build this linear oeperator - so [math]A(e_{1}) = e_{1}'[/math] etc. Then we see

 

[math]A(e_{1}\wedge e_{2} \wedge \cdots \wedge e_{n}) = \alpha (e_{1}\wedge e_{2} \wedge \cdots \wedge e_{n}) = e_{1}'\wedge e_{2}' \wedge \cdots \wedge e_{n}'[/math]

 

This scalar [math]\alpha[/math] is the determinant of the linear transformation A.

 

So lets do this for the 2d case. Let us take some linear operator that I write as a matrix

 

[math]A = \left(\begin{array}{ll}a & b \\c & d\end{array}\right)[/math]

 

Then look at its action on an arbitary vector and feed this into the wedge product; we obtain

 

[math](ax + by) \wedge (cx +dy) = ac \: x \wedge x + ad \: x \wedge y + bc \: y \wedge x + bd \: y\wedge y[/math].

 

Now remember that the wedge product is totally antisymmetric - which just means antisymmetric when we only have two vectors. So [math]x \wedge x = y \wedge y =0[/math] and we are left with

 

[math](ax + by) \wedge (cx +dy) = (ad - bc) \: x \wedge y[/math],

 

which is what we wanted. You could try the same thing yourself in dimension 3. In higher the same sort of thing works.

 

This also then gives you a solution to question 1. You can use this as a geometric definition of the determinant of a linear transformation.

 

I hope that helps a little :)

Edited by ajb
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albedo    1

Hello ajb,

 

thank you for your time and efforts, this definitely helps!

I was really hopeless since I asked this question on several forums without any success – and now I finally got a response.

 

Thank you.

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ajb    1567

Thank you.

You are welcome.

 

With your question 4. I am not sure there is some general answer. However, you might want to think about eigensystems - so solutions to

 

[math]A \underline{x} = \lambda \: \underline{x}[/math]

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albedo    1

You are welcome.

 

With your question 4. I am not sure there is some general answer. However, you might want to think about eigensystems - so solutions to

 

[math]A \underline{x} = \lambda \: \underline{x}[/math]

 

Thanks for tip, (unfortunately) I'm familiar with eigen(vectors/values).

 

But I have another determinants-related question (which could help me "intuitize" :) the broader meaning of determinant):

Just to state some facts: AFAIK, determinant is used for:

  1. To solve sets of linear equations (AFAIK this is why it was first invented - by Seki in Japan).
  2. To compute the volume distortion of parallelepiped (AFAIK this meaning come later - introduced by Lagrange).

A matrix can represent:

  1. Set of linear equations (row-wise).
  2. The basis vectors of a coordinate system (column-wise).

The question: how does the two meanings of matrix (row and column) relate?

I.e. let the matrix [latex]\mathbf A[/latex] represent a set of linear equations. What coordinate system does the matrix represent (i.e. what is the meaning of matrix [latex]\mathbf A[/latex] column-wise)?

 

Let's say I know the above meaning. Then I guess I could connect the "Seki" and "Lagrange" meanings of determinant - for which I don't see any connection now.

I.e. I could solve a set of linear equations (represented by a matrix) graphicaly using the knowledge of the volume of parallelepiped formed by the matrix's basis vectors.

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ajb    1567

I am not sure there is a great answer here - I an not sure what you are really looking for. However,

 

[math] \left( \begin{array}{ll} a & b \\ c & d\end{array} \right) \left(\begin{array}{l}1 \\0 \end{array} \right) = \left(\begin{array}{l}a \\b \end{array} \right)[/math]

 

and so on... not sure that is any help though.

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