Proving the equation for 2D determinant

[liengen]

Homework Statement

So, I'm in a 2D Euclidean space V. The two vectors v1 and v2 forms the orthonormal basis set for V.

Consider the transformation T(v1) = a*v1 + b*v2, T(v2) = c*v1 + d*v2

Show, using scalar products (not vector products), that the area scale factor of the transformation T is ad-bc.

Homework Equations

The Attempt at a Solution

The "area scale factor" is simply the determinant in this case. So all I have to do is to show that det([a b;c d]) = ad-bc.

A 2D transformation maps the orthonormal basis set from a unit square to a parallelogram. The area of this parallelogram must then be the area scale factor (since the area of a unit square is 1).

Extending this to 3D and then use the cross product is the simple answer (by definition, the |axb| where a and b are 3D vectors is the area of the parallelogram spanned by a and b)

However, my task is to solve this using the dot product only. Hence, I'm not allowed to extend to 3D. I just cannot see how I'm supposed to do this :-/

 

[Dick]

You can find the area of the parallelogram by multiplying length of base times height. It's easy to get the length of the base using the dot product. Can't you figure out how to get the height using the dot product? Hint: rotate the vector you want to call the 'base' by 90 degrees.

 

[liengen]

Yes, I've tried that, but I'm definitely doing something wrong along the way, since my result is wrong...

Let's say the parallelogram is spanned by v3 and v4 and my base is v3.

According to my calculations, the height is: sqrt(|v4|^2 - dot(v3,v4)^2)
That is, v4 is the hypotenuse and dot(v3,v4) and height are the two other sides.

Then the area is: |v3|*sqrt(|v4|^2 - dot(v3,v4)^2)

Which certainly not equals the equation for the 2D determinant... :-/

 

[Dick]

Yes, I've tried that, but I'm definitely doing something wrong along the way, since my result is wrong...

Let's say the parallelogram is spanned by v3 and v4 and my base is v3.

According to my calculations, the height is: sqrt(|v4|^2 - dot(v3,v4)^2)
That is, v4 is the hypotenuse and dot(v3,v4) and height are the two other sides.

Then the area is: |v3|*sqrt(|v4|^2 - dot(v3,v4)^2)

Which certainly not equals the equation for the 2D determinant... :-/

I think sqrt(|v4|^2-dot(v3,v4)^2) is only works to get the height if v3 and v4 happen to be perpendicular. Like I said before rotate the vector v4 by 90 degrees and look at the dot product of that with v3.

 

[liengen]

I see ^^. Or actually not.

Dot product between rotated vector and the basis:
[-b a].[c d] = -bc + ad = ad - bc (which is the area)

Generally speaking, let's say we have a parallelogram spanned by two vectors, say v1 and v2 and the rotated version of for example v1 is v3. Then, why do the dot product between v3 and v2 produce the area of the parallelogram?

This seems like pure magic for me. Any properties of the dot product that explain this?

 

[Dick]

I see ^^. Or actually not.

Dot product between rotated vector and the basis:
[-b a].[c d] = -bc + ad = ad - bc (which is the area)

Generally speaking, let's say we have a parallelogram spanned by two vectors, say v1 and v2 and the rotated version of for example v1 is v3. Then, why do the dot product between v3 and v2 produce the area of the parallelogram?

This seems like pure magic for me. Any properties of the dot product that explain this?

The height of the parallelogram is the projection of v4 onto the direction perpendicular to v3. To get the projected length you take a dot product of v3 with a unit normal vector pointed perpendicular to v4. Actually I'm starting to see where your other formula came from. Can you show |v3|^2*|v4|^2-dot(v3,v4)^2=(ad-bc)^2? Can you explain why that's true geometrically?

 

[liengen]

Ah, sorry, I was a little to fast earlier. When I expanded the dot product and got the actual result, I thought I was finished.

So, the rotated unit v3 vector is, v3rot = [-b a]/sqrt(b^2+a^2)
Now we can do the scalar projection from v4 onto this, and we get the height of the parallelogram: h = dot(v4,v3rot) = (ad-bc)/sqrt(b^2+a^2).

Then the area of the parallelogram is: A=base*height = |v3|*(ad-bc)/sqrt(b^2+a^2)
= sqrt(b^2+a^2)*((ad-bc)/sqrt(b^2+a^2)) = ad-bc

That's just beautiful.

When it comes to the other equation there, I see that its the same as the area:
|v3|^2*|v4|^2-dot(v3,v4)^2 = (a^2+b^2)(c^2+d^2)-(ac+bd)^2 = (ad-bc)^2
However, I cannot see this geometrically.

Anyway, thanks a lot man. I think I had the answer in front of me from the beginning, it's just sometimes hard to actually see it ^^

 

[Dick]

Ah, sorry, I was a little to fast earlier. When I expanded the dot product and got the actual result, I thought I was finished.

So, the rotated unit v3 vector is, v3rot = [-b a]/sqrt(b^2+a^2)
Now we can do the scalar projection from v4 onto this, and we get the height of the parallelogram: h = dot(v4,v3rot) = (ad-bc)/sqrt(b^2+a^2).

Then the area of the parallelogram is: A=base*height = |v3|*(ad-bc)/sqrt(b^2+a^2)
= sqrt(b^2+a^2)*((ad-bc)/sqrt(b^2+a^2)) = ad-bc

That's just beautiful.

When it comes to the other equation there, I see that its the same as the area:
|v3|^2*|v4|^2-dot(v3,v4)^2 = (a^2+b^2)(c^2+d^2)-(ac+bd)^2 = (ad-bc)^2
However, I cannot see this geometrically.

Anyway, thanks a lot man. I think I had the answer in front of me from the beginning, it's just sometimes hard to actually see it ^^

For the other approach you know |v3|*|v4|*sin(theta), where theta is the angle between the vectors, is the area you want. Use that dot(v3,v4)=|v3|*|v4|*cos(theta).

 

[liengen]

Ah, I see. By using the identity sin(theta)=(+-)sqrt(1-cos^2(theta)):

A = |v3|*|v4|*sin(theta)
A = |v3|*|v4|*(+-)sqrt(1-cos^2(theta))
A^2 = |v3|^2*|v4|^2*(1-cos^2(theta)) = |v3|^2*|v4|^2 - dot(v3,v4)^2

Again, thanks a lot man :-D

 

[cubzar]

Also, since the transform of a parallelogram with a vertex at the origin has area sclae factor ad-bc, the same applies for any triangle with a vertex at the origin, because the triangle can be added to its reflection in the side not touching the origin to make a parallelogram. Any polygon containing the origin can be split into triangles with the origin as a vertex and the areas summed. A polygon not containing the origin can be adjoined to another polygon, the whole expressed as a sum of triangles, then the adjoined polygon split up and its area subtracted. A shape which is not straight-sided can be approximated with an arbitrarily large number of triangles, and the number of triangles allowed to tend to infinity.
Therefore any 2D shape in a euclidean geometry has an area scale factor of ad-bc for the transformation.