Matrix transformations and effects on the unit square

[fogvajarash]

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

 

[Simon Bridge]

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

 

[fogvajarash]

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.

 

[Simon Bridge]

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.

 

[CellCoree]

Dear student,

Thank you for bringing up this question about matrix transformations and their effects on the unit square. The relationship between the area of the unit square and the determinant of the transformation matrix is an important concept in linear algebra.

To answer your question, we need to understand the concept of determinant and how it relates to matrix transformations. The determinant of a square matrix A is a scalar value that represents the scaling factor of the transformation induced by A. In other words, it tells us how much the area of a shape changes under the transformation A.

When we apply a matrix transformation A to the unit square, the resulting shape is a parallelogram with sides given by the transformed unit vectors T(i) and T(j). The area of this parallelogram is given by the magnitude of the cross product between T(i) and T(j), which is represented by ||T(i) x T(j)||. This is where the term [[Ai x Aj]] comes from in your notes.

Now, since T(i) and T(j) are unit vectors, they are perpendicular to each other and the angle between them is 90 degrees. This means that the sine of the angle θ between them is 1, and therefore we can simplify the expression to just abs(det(A)).

In summary, the reason why the area of the unit square under a matrix transformation is equal to abs(det(A)) is because the determinant of A represents the scaling factor of the transformation, and the area of the resulting parallelogram is given by the magnitude of the cross product between the transformed unit vectors, which can be simplified to abs(det(A)) in this case.

I hope this explanation helps clarify your understanding. If you have any further questions, please don't hesitate to ask.

Best regards,