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A few questions regarding Determinants


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#1 albedo

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Posted 16 July 2016 - 02:48 AM

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 P and I transform all of its vertices by a matrix \mathbf{A}.
    Is \det\left(\mathbf{A}\right) the scaling factor of polytope's P area after the transformation, i.e. \det\left(\mathbf A\right) = \frac{P\text{'s area after transform}}{P\text{'s area before transform}}?
     
  3. Imagine I have an open 2D polytope \overline P (which clearly doesn't have any area).
    How does \det\left(\mathbf{A}\right) relate with the transformed polytope \mathbf {A}\overline P?
     
  4. Suppose there's a vector \mathbf x. What does \det\left(\mathbf{A}\right) say about the transformed vector \mathbf {Ax}?
 
I'd be glad to get answer to any one of these questions. Thanks.

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#2 ajb

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Posted 16 July 2016 - 08:30 AM

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

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

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

 Vol(e) = | e_{1}\wedge e_{2} \wedge \cdots \wedge e_{n} |

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

A(e_{1}\wedge e_{2} \wedge \cdots \wedge e_{n}) = A(e_{1})\wedge A( e_{2}) \wedge \cdots \wedge A( e_{n}).

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 A(e_{1}) = e_{1}' etc. Then we see

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}'

This scalar \alpha 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

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

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

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

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

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

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, 16 July 2016 - 08:41 AM.

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#3 albedo

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Posted 16 July 2016 - 09:27 AM

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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#4 ajb

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Posted 16 July 2016 - 09:48 AM

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

A \underline{x} = \lambda \: \underline{x}
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#5 albedo

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Posted 16 July 2016 - 07:59 PM

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

A \underline{x} = \lambda \: \underline{x}

 

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 \mathbf A represent a set of linear equations. What coordinate system does the matrix represent (i.e. what is the meaning of matrix \mathbf A 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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#6 ajb

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Posted 17 July 2016 - 07:32 AM

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

 \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)

and so on... not sure that is any help though.
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#7 uncool

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Posted 7 August 2016 - 01:01 AM

Think you mixed up a and c there, ajb. 


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