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# Vector transformations

Introduction to the notion of vector transformations. Created by Sal Khan.

## Want to join the conversation?

• •   A tuple is a list, with some specific number of elements. In first/second year algebra, you mainly use two and three tuples to express points in 2 and 3 space; the Cartesian and three dimensional spaces. In linear algebra, the idea of these tuples is abstracted to n-component lists. Tuples are also always ordered; the three tuple (1,2,3) is different than (2,3,1).
• Since R^3 has infinitely more values then R^2 how can every point in R^3 only map to one point in R^2? • I have noticed that when talking about Vectors Sal generally denotes them as a column matrix like he did at when describing vector x?
IS there a reason to do so or could your represent them as a row vector also? • In vector mapping can we map from one dimensional space? If yes, then how will the relation defined by the function be since in this case the codomain must be of 2-tuples and above. • I've always thought of vectors as merely n-tuples (as Sal states at around ). However, some books (and some parts of Khan Academy, such as the "Vector dot and cross products" playlist videos) make an effort to differentiate between points and the position vectors used to represent those points (for instance, the point "P(x1,x2,x3)" vs. the position vector "<x1,x2,x3>"), which has confused me because I've always thought of both points and vectors as tuples--i.e. as one and the same.

Anyway, could someone help clarify this point of confusion? In addition, could someone please explain why it's necessary to distinguish between points & position vectors and not merely represent both as n-tuples?

I apologize for the long-windedness of this question... • Don't apologize! There's nothing wrong with a non-TL/DR mentality :)
First of all, both are tuples. You basically say A implies C, B implies C, thus mustn't it be that A implies B? No. It could be, but for instance in this particular example it is not true. A series of points that belong to eachother, say P(x1, x2, x3), is a tuple. A position vector is a vector with its tail in O(0,0,0) and its head in P(x1, x2, x3). However, since it is a vector, we need to consider the associated unit vectors with each component in the position vector <x1, x2, x3>. This means that we can express the position vector as an N-tuple, where N is the amount of components of the point P and its position vecotr, but this N-tuple does not consist merely of x1, x2 and x3 like the point itself, it consists of 2-tuples including each component and their respective unit vector.
In other words, to express the position vector of point P(x1, x2, x3), we can use this notation: ((x1, i-hat), (x2, j-hat), (x3, k-hat)).
We can conclude that you represent both as N-tuples, but the point consists of N scalars, and the vector consists of N 2-tuples. This is due to the fact that vectors have a direction, which we can see through the fact that now every component is indeed a component vector of the position vector, thanks to the usage of unit vectors.

Hopefully my lack of theoretical knowledge (I simply thought of this all myself, so mistakes in reasoning might make this entire answer flawed), will not keep you from understanding the analysis. Anyone reading this, be sure to point out mistakes if I'm wrong and it appears that vectors by some definition cannot correctly be expressed as N-tuples of 2-tuples.
• Is a function in the imaginary plane a vector valued function? You are mapping from a 2D space (real and complex part) to another 2D space (another real and complex part). • The definition of a vector valued function tells us that the codomain of that function must have more than one dimension. This means that it does not matter what the domain of our function is: we only care what that domain maps to, in other words the codomain. Since a codomain in 2D space simply means codomain: R^2, where 2 is strictly bigger than 1, we can conclude that the function you described is in fact a vector valued function.
Right?
No, because we made the assumption that our function codomain is a plane of real numbers, which they are not. A function in the imaginary plane has a codomain of Z^1, which means that the function is in fact a scalar valued function (but NOT real valued function).

Eventually it all depends on whether or not you tell your function to be real-valued or not: you could simply take Cartesian space and say that the y-axis has no unit while your x-axis has the unit "i", the imaginary number. Basically now we see that simply said, complex numbers can be expressed as 2-tuples of real numbers, given that one of those 2-tuple numbers has a unit "i".

Thus it is ambivalent to say that your function is vector valued or scalar valued.
It is ambivalent to say that it is both real-valued and vector valued, or both complex-valued and scalar valued.
• the position vectors of T=(5,2)and S=(1,-3)and P =(7,13). Find the values of the scalars of a and b such that aT+bS=P • Bascally you can set it up like a system of equations (though as you go through linear algebra you will be getting systems and turning them into vectors.)

5a + 1b = 7
2a - 3b = 13

I will solve for s in the first equation

b = 7 - 5a
2a - 3b = 13

Then plug in s int he second one

b = 7 - 5a
2a - 3(7 - 5a) = 13 = 2a - 21 + 15a = 17a -21

Then solve for a in the second
b = 7 - 5a
a = 2

Then back substitute
b = 7 - 5(2) = -3
a = 2

And that should do it.

If you have learned reduced row echelon form you can turn the vectors into a 2 by 3 matrix and put it into rref
• Why was f(x1, x2, x3) in R^3 while (x1, 2x2, 3x3) is in R^2?

By the way, I didn't know how to make some of the numbers subscripts so I just put regular numbers. Eg. x1, x2, x3.

Thank you! :)   