A few questions regarding Determinants

[albedo]

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.

[ajb]

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 July 16, 2016 by ajb

[albedo]

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.

[ajb]

  On 7/16/2016 at 9:27 AM, albedo said:

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

 

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

[albedo]

  On 7/16/2016 at 9:48 AM, ajb said:

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.

[ajb]

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.

[uncool]

Think you mixed up a and c there, ajb.