A good question from the Khan site (w/ an answer): @ 20:30, when we are trying to find the equation of the plane that contains all the vectors of C(A) : Sal says that we can use the definition : C(A) = {b | A*x = b} where x is a vector of R^4, b a vector of R^3. Then he says : (A*x = b) solve the system represented by the matrix [A | b] But I don't think it is true, because finding x,y,z such as the system represented by [A | b] is true is equivalent to A*[(vector(1 1 1)) = b So I don't understand why Sal is right. ANSWER: We have defined b = A x, meaning that the elements of b are whatever the elements of A x are. For concreteness, say x = [x1,x2,x3,x4], and A is the matrix defined in the video. Thus, A x = [x1+x2+x3+x4 , 2*x1+x2+4*x3+3*x4 , 3*x1+4*x2+x3+2*x4]. Thus we get b = [x1+x2+x3+x4 , 2*x1+x2+4*x3+3*x4 , 3*x1+4*x2+x3+2*x4]. For compactness, and to translate to our geometric interpretation, let us re-label b as b=[x,y,z]. This means that x=x1+x2+x3+x4, y=2*x1+x2+4*x3+3*x4 and z=3*x1+4*x2+x3+2*x4. This is now a system of three linear equations, which may perfectly well be represented by the augmented matrix [A | b]. x1, x2, x3 and x4 are hidden in this view, because they are the variables you are solving for, while the augmented matrix shows only their coefficients.
6:43 , can someone explain why , here is taken as a position vector. I have a doubt because the two and in the column space only seems to define direction vectors. Since to form a plane I assume we need two direction vectors and a position vector. I Never mind I understand it as I write. Since this is a plane that passes through origin, can also be viewed as a vector space.... Bingo
10:56 how is 5x-y-z C(A). As far as I've understood, column space is the span of column vectors, i.e all the linear combinations of column vectors and while calculating the equation of plane, i guess we are doing n(vector)*(linear combination). How can this give the column space?
13:25 I don't understand this part, why is Sal emphasizing that we know the 2 basis vectors lie on the plane because the 0 vector is in the span? I thought when you write span([1 2 3], [1 1 4]) we are already assuming these are position vectors? Which would mean the vectors start at the point (0,0,0)?
the 2 vectors [1,2,3] and [1,1,4] are on the plane AND are position vectors at the same time), (the plane passes through the origin (0,0,0) too) edit: of course as I understand.
I have read many comments in this series and other series on this site. It seems many are self learning mathematics. Others, it seems, are badly let done by their lecturer and/or text. Below is a search that points to resources that I find useful. Perhaps others willl also. Search this How to Become a Pure Mathematician (or Statistician): a List of Undergraduate and Basic Graduate Textbooks and Lecture Notes
A good question from the Khan site (w/ an answer):
@ 20:30, when we are trying to find the equation of the plane that contains all the vectors of C(A) :
Sal says that we can use the definition : C(A) = {b | A*x = b} where x is a vector of R^4, b a vector of R^3.
Then he says :
(A*x = b) solve the system represented by the matrix [A | b]
But I don't think it is true, because finding x,y,z such as the system represented by [A | b] is true is equivalent to A*[(vector(1 1 1)) = b
So I don't understand why Sal is right.
ANSWER:
We have defined b = A x, meaning that the elements of b are whatever the elements of A x are.
For concreteness, say x = [x1,x2,x3,x4], and A is the matrix defined in the video. Thus, A x = [x1+x2+x3+x4 , 2*x1+x2+4*x3+3*x4 , 3*x1+4*x2+x3+2*x4]. Thus we get b = [x1+x2+x3+x4 , 2*x1+x2+4*x3+3*x4 , 3*x1+4*x2+x3+2*x4]. For compactness, and to translate to our geometric interpretation, let us re-label b as b=[x,y,z]. This means that x=x1+x2+x3+x4, y=2*x1+x2+4*x3+3*x4 and z=3*x1+4*x2+x3+2*x4.
This is now a system of three linear equations, which may perfectly well be represented by the augmented matrix [A | b]. x1, x2, x3 and x4 are hidden in this view, because they are the variables you are solving for, while the augmented matrix shows only their coefficients.
Column Space is the same thing as image. Null Space is the kernel.
The way Sal explain all the terms may become very lengthy and confusing at times, but once you get the logic, it means u get it.
At 19:02 can it not also be the case that the 1st and 2nd row of that augmented matrix will further limit the possible values of x, y & z?
6:43 , can someone explain why , here is taken as a position vector. I have a doubt because the two and in the column space only seems to define direction vectors. Since to form a plane I assume we need two direction vectors and a position vector.
I Never mind I understand it as I write. Since this is a plane that passes through origin, can also be viewed as a vector space.... Bingo
This was a source of much confusioin for me too. In general, if the plane does not pass through the origin, then this approach will not work.
You are such a great teacher!
20:18
Wow!
10:56 how is 5x-y-z C(A). As far as I've understood, column space is the span of column vectors, i.e all the linear combinations of column vectors and while calculating the equation of plane, i guess we are doing n(vector)*(linear combination). How can this give the column space?
He just found the orthogonal vector so that he could put in the equation, n(orthogonal to column space) .(dot) linear combination = 0
Mathematics is always stunning... This just stunned me!!!
I have a 3rd approach. It involves Wolfram alpha..... 😂😂😂😂 Great video Sal, the different approaches are quite illuminating!
13:25 I don't understand this part, why is Sal emphasizing that we know the 2 basis vectors lie on the plane because the 0 vector is in the span? I thought when you write span([1 2 3], [1 1 4]) we are already assuming these are position vectors? Which would mean the vectors start at the point (0,0,0)?
the 2 vectors [1,2,3] and [1,1,4] are on the plane AND are position vectors at the same time), (the plane passes through the origin (0,0,0) too)
edit: of course as I understand.
I have read many comments in this series and other series on this site.
It seems many are self learning mathematics. Others, it seems, are badly let done by their lecturer and/or text.
Below is a search that points to resources that I find useful.
Perhaps others willl also.
Search this
How to Become a Pure Mathematician (or Statistician):
a List of Undergraduate and Basic Graduate Textbooks and Lecture Notes
Thank you so much!!! This is what I needed.
The coordinate system at 5:08 is not possible
한국어 번역해주신분 정말정말 감사합니다!!!!!!!
can you not take the span of other two matrix equals to C(A)
2x-y-z+3x=0 how come this works. How come this also works for 4th dimensional systems?
What do you say about NCC
watching all your vids x1.75, now if I hear you talking at normal speed it seems like you are talking on slowmo
The English captions on this video are not in English. Please fix.
Sal you're not drawing the axes right! Remember your right hand rule... tsk! :)
is column space the kernel or image?
NoobTastic1234 Image
thank you, but id like it if the pointer was easier to see