Super clear! To summarize or recap (with a slight abuse of notation) : Given an m x n matrix A with entries from the set of real numbers (more generally from a field F), a constant vector b in R^n (more generally F^n), and 0 in R^m (more generally F^m), we can define H = { x in R^n : Ax = 0 } and G = { x in R^n : Ax = b }. Then if G is nonempty, and you know in particular what an element of G is, which we will denote by the vector p, then it can be shown that G = { p } + H. The latter expression { p } + H is the set of vectors which are the sum of p and elements in H. if G is empty , then G is empty and we can say nothing further about G, and the linear system is inconsistent.
Hi, Firstly your lectures has been truly amazing, i am grateful for your sharing of knowledge. I would like to ask the following: Is the turquoise plane in 3:28 the null space of A?
One question Sir , why is it that in 3:16 , the sum of the two vectors are represented by the orange line when it is clear that the sum is the yellow line ?
Question that immediately comes to mind, can we use the ideas suggested in the video to prove that if Ax = b is consistent and has more than one solution then it has an infinite number of solutions. My proof attempt: Let p and p ' be distinct solutions to Ax = b. Then I claim there are an infinite number of solutions of the form p + t * ( p ' - p ) for any t in R . Then using linear properties of matrix arithmetic A ( p + t * ( p ' - p ) ) = Ap + A( t * (p ' - p ) ) = Ap + t * A( p ' - p ) = b + t * (Ap ' - Ap ) = b + t * ( b - b) = b + t * 0 = b + 0 = b. Since t can range over R, and R is infinite, there are an infinite number of solutions. Note that this does not exhaust all the solutions to Ax = b, only shows that if there are two distinct solutions there must be an infinite number of them. Geometrically this proof says that if the points p , p ' are in the affine subspace of solutions to Ax = b, then the line containing p, p' (a line contains an infinite number of points) is also contained in the affine subspace. Nice image of affine subspace en.wikipedia.org/wiki/Affine_space I don't know too much about affine subspaces but it's clear from the image that the solution space of Ax = b is not a vector space when b ≠ 0, but such a space still has some nice properties like closure under subtraction.
No, that's exactly what you do. We're not saying that (1) or (2) are *always* true. We're saying that *if* (1) is true, *then* (2) is true, and vice versa.
Greatest explanation I've ever heard. Now I got it!
Super clear!
To summarize or recap (with a slight abuse of notation) :
Given an m x n matrix A with entries from the set of real numbers (more generally from a field F),
a constant vector b in R^n (more generally F^n), and 0 in R^m (more generally F^m),
we can define H = { x in R^n : Ax = 0 } and G = { x in R^n : Ax = b }.
Then if G is nonempty, and you know in particular what an element of G is,
which we will denote by the vector p, then it can be shown that G = { p } + H.
The latter expression { p } + H is the set of vectors which are the sum of p and elements in H.
if G is empty , then G is empty and we can say nothing further about G, and the linear system is inconsistent.
The best lecture in Linear Algebra. Thank you!!
Thank you sir! These are truly great videos you have here. Much better than my professor.
Thank you very much, I learned a lot. I started from 1st video and will continue to the end.
great lectures.....this is so helpful ...thanks
Hi,
Firstly your lectures has been truly amazing, i am grateful for your sharing of knowledge.
I would like to ask the following:
Is the turquoise plane in 3:28 the null space of A?
One question Sir , why is it that in 3:16 , the sum of the two vectors are represented by the orange line when it is clear that the sum is the yellow line ?
The orange line represents *all* of the solutions to Ax=b. The yellow vector represents *one* of those solutions.
extremely helpful thank u!
Question that immediately comes to mind, can we use the ideas suggested in the video to prove that if Ax = b is consistent and has more than one solution then it has an infinite number of solutions.
My proof attempt:
Let p and p ' be distinct solutions to Ax = b. Then I claim there are an infinite number of solutions of the form p + t * ( p ' - p ) for any t in R .
Then using linear properties of matrix arithmetic
A ( p + t * ( p ' - p ) ) = Ap + A( t * (p ' - p ) ) = Ap + t * A( p ' - p ) = b + t * (Ap ' - Ap ) = b + t * ( b - b) = b + t * 0 = b + 0 = b.
Since t can range over R, and R is infinite, there are an infinite number of solutions.
Note that this does not exhaust all the solutions to Ax = b, only shows that if there are two distinct solutions there must be an infinite number of them.
Geometrically this proof says that if the points p , p ' are in the affine subspace of solutions to Ax = b,
then the line containing p, p' (a line contains an infinite number of points) is also contained in the affine subspace.
Nice image of affine subspace en.wikipedia.org/wiki/Affine_space
I don't know too much about affine subspaces but it's clear from the image that the solution space of Ax = b is not a vector space when b ≠ 0, but such a space still has some nice properties like closure under subtraction.
Great lecture!
i understand better now
thank you!!!!!
Thanks.
Good stuff.
But you can’t assume (2) to be true and in turn use that assumption to prove (1).....
No, that's exactly what you do. We're not saying that (1) or (2) are *always* true. We're saying that *if* (1) is true, *then* (2) is true, and vice versa.