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### Course: Linear algebra>Unit 2

Lesson 6: More determinant depth

# Determinant as scaling factor

Viewing the determinant of the transformation matrix as a scaling factor of regions. Created by Sal Khan.

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• Does this break down once you go beyond 2x2?
• can you guys talk about scales or scal factor or scale factor of a model im having trouble with it
• It reminds me of a derivative in the case of linear functions from the reals to reals or R -> R case. The dy/dx = slope part gives the same feel as Area(T(region))/Area(region) = |det(A)|, where A is T's matrix. Is their a relationship between determinants and derivatives?
• The analogy you're drawing is a little tricky.
Derivatives act not on numbers or vectors but on functions. Thus, the derivative doesn't act on the set of real numbers but instead it maps the set of all possible functions to the set of all possible functions.
The derivative gives the rate of change of a function which in general won't be a linear function, while the determinant gives the change in area/volume/hypervolume of a linear transformation.
The derivative itself is a linear operator (obeying the rules of linearity), while the determinant is not, so this is quite a strong difference between them.
The derivative also generally gives more information than the determinant. From the derivative of a function, you can reproduce the original function up to an added constant. From the determinant, you have almost no capacity to reproduce the original linear operator.
As such, the similarity you've found between the determinant and derivative appears to be quite superficial.

For fun, since the derivative is a linear operator (albeit in the space of functions not numbers), and one where the domain and codomain are equal (meaning the corresponding matrix is square), then it should have a determinant. It's fairly easy to calculate. We can see that dy/dx is not one-to-one, because all constants get mapped to the same value (zero), so dy/dx is not invertible, so the determinant of the derivative is zero.
• Sorry this may be due to my aged brain...75, but. I was confused until I did the second pass at Sal's use of (for example) vectors a and b in the 2D diagram and then scalar symbols for a and b in the matrix. To Sal's great credit, at least for me, there are very, very, very few situations where the steps in Sals videos don't quite flow one to the next. I did understand after watching 2 times, and 3 for good luck. Great video otherwise with important conclusion. Signed: the curious old grouch.
• is there a way to do it even faster with out all the long work?
• Sure, calculate the original Area and multiply it by the absolute value of the determenant of the image, as specified in the video.
• What is the easiest way to find a scale factor?
• For a transformation T(x) = Ax in R^2, the scaling factor for the area associated with 2 vectors is det(A).
(1 vote)
• How do I find scale factor
• For a transformation T(x) = Ax (in R^2 at least) the scaling factor is det(A).
(1 vote)
• In this video, Sal plots the transformation using the matrix A on the domain R^2, but I am not sure how the points on the coordinate system have been identified. The column vector elements of the transformation matrix are mere alphabets denoting scalars, so I would presume that the plotting has been an approximation at its best, and arbitrary otherwise, unless someone else can clarify this point. Again, the points such as bK1 and dK2 or aK1 and cK2 in the plot do seem to specify a scaling with respect to the origin, but what is perhaps jarring is the fact that each of these specifications seem to be confined to only a single axis (i.e R^1) with no corresponding coordinate in R^2 (unless they are presumed to be 0), which by itself appears to be a contradiction of the domain space notion to me.