Matrix transformations and effects on the unit squareby fogvajarash Tags: algebra, cross product, effects, linear transf, matrix, transformations, unit square, vectors 

#1
Nov613, 07:51 PM

P: 126

I was looking over my notes today, and I realized that there was a point that isn't pretty clear.
If we have the image under T (being T a matrix transformation induced by a matrix A) of the unit square, then its area should be abs(det(A)). Why is this though? I was looking at the proof and I saw that the transformation was defined as T(i) = Ti = (a c 0) and T(j) = Tj = (b d 0) [in this case i and j are the unit vectors]. However, the area of the parallelogram made between these vectors is stated to be as [[Ai x Aj]] (without the sinθ being θ the angle these vectors make). Or is it because as if they are unit vectors, they are 90 degrees to each other? Thank you very much. Edit: This link (around page 1) should help if i'm not clear with my explanations http://www.math.mun.ca/~mkondra/linalg2/la2set7.pdf 



#2
Nov613, 10:06 PM

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PF Gold
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You don't need the trig function to evaluate a cross product.




#3
Nov713, 08:12 AM

P: 126

Oh darn, I finally saw my notes and checked that indeed it is just the vector product lv x wl (for the trig function, it involves the magnitude of both vectors or lal lbl sinθ, which is completely different from the vector product). So this means that we can as well find the area of that parallelogram using the sine of the angle and the two magnitudes?
Thank you Simon. 



#4
Nov713, 08:39 PM

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Matrix transformations and effects on the unit square
The advantage of the vector form is that it gives you the direction of the cross product as well.
Consider if ##\vec{v} = (v\cos\phi,v\sin\phi,0)^t## and ##\vec{u}=(u\cos\theta, u\sin\theta, 0)^t## Where ##\theta## is the angle ##\vec{u}## makes to the x axis and ##\phi## is the angle ##\vec v## makes to the x axis. Then $$\vec{u}\times\vec{v} = \left \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k}\\ u\cos\theta & u\sin\theta & 0\\ v\cos\phi & v\cos\phi & 0\end{array}\right = uv\sin(\phi\theta)\hat{k}$$ ... which result required trig identities. Notice that ## \theta \phi ## is the angle between the vectors  so the equation you are used to is just for the magnitude. It gets trickier when you go to more than three dimensions. 


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