This seems to be the same thing as the case of an inhomogeneous linear Dif. Eq. where the solution is the general homogeneous solution plus a particular solution to the inhomogeneous equation. I always find it remarkable when the same thing occurs in different branches of mathematics.
If I pick p-q as my homogeneous solution then could I say t(p-q)+p is the non-homogeneous solution of the line M? Or must I set q-p because the line is oriented from p to q?
For an mxn matrix A, If Ax = 0 has only the trivial solution then Ax = b has a unique solution for each b is an element of the set R^m. Could you explain why this is False apparently that's the answer and I cannot figure out why? Am I trying to devise an example to explain it but I keep getting stuck because if there are pivot positions in every row and Ax = 0 is trivial(no free variables) then Ax = b would be a unique solution. I tried thinking about maybe if Ax = b had more columns then rows but then it wouldn't be trivial.
For anyone reading this 4 years later, I believe Elias had an error when reading the answers. If Ax = 0 is trivial, then Ax = b is necessarily unique. Literally just solved a similar problem in my own linear book, and it illustrates why it has to be necessarily unique. Maybe there's something about "for each b is an element of the set R^m", I don't really understand. Just saying.
Good explanation, however it feels lacking. You derived the conclusion that Ax = b is built from [homogeneous vector] + [translation], but this is super confusing since you derived that conclusion by row-reducing a nonhomogenous system. I.E: You explain how to get to M, by starting from 0 -- but in your parametric example, you start from Ax = b, you start from M to find 0 and the translation! This is super confusing, since it doesn't explain WHY after row-reducing a nonhomogeneous system, we end up with a homogeneous vector! I don't understand why inherently row-reducing provides us that? The only thing I can think of, is the simplification of: let c and d be solutions to Ax = b, then: A(c-b) = A(c) - A(b) = b - b = 0. Let v = (c-b) As such, A(v) = 0. And every linear transformation of v will also solve this equation, i.e: A(sv) = sA(v) = s0 = 0 So the general solution of the homogeneous system, A(s*v), correlates with the linear transformation of s*c and s*b, two vectors on the same line (m). But that also doesn't explain why row-reducing a nonhomogeneous system, provides us directly with a [homogeneous vector] + translation !
![[Linear Algebra] Linear Independence](http://i.ytimg.com/vi/XI2kYIxhe-o/mqdefault.jpg)
![[Linear Algebra] Linear Independence](/img/tr.png)
![[Linear Algebra] Homogeneous Linear Systems and Parametric Form](http://i.ytimg.com/vi/VNApFaUOJ9o/mqdefault.jpg)
you saved my life by making these tutorials for linear algebra and discrete maths. thanks!
Thanks so much for this!! It’s makes so much more sense when you explain it instead of a textbook 😂😂
your explanations are so intuitive, thank you very much!
This seems to be the same thing as the case of an inhomogeneous linear Dif. Eq. where the solution is the general homogeneous solution plus a particular solution to the inhomogeneous equation. I always find it remarkable when the same thing occurs in different branches of mathematics.
They happen to be connected since the differential operator happens to be a linear operator. Beautifully connecting linear algebra to calculus.
you explained it with the visuals so well! Thank you Trev!
Thank you so much!
What software do you use for your videos? The hand drawing looks so natural.
Really? How does it take 60 seconds to go through all of his other videos and read all of the comments or watch them all? Genuinely curious.
Benargee you cant do that in 60 seconds? Cmon man
Yeah that Matt dude sounds like a jackass
did anyone ever figure this out???
this is gold
Thanks sir ..
If I pick p-q as my homogeneous solution then could I say t(p-q)+p is the non-homogeneous solution of the line M? Or must I set q-p because the line is oriented from p to q?
how do you quickly find the 'translation'
thanx a lot. sir can u tell me from where to practice such questions to have confidence.??
For an mxn matrix A, If Ax = 0 has only the trivial solution then Ax = b has a unique solution for each b is an element of the set R^m. Could you explain why this is False apparently that's the answer and I cannot figure out why? Am I trying to devise an example to explain it but I keep getting stuck because if there are pivot positions in every row and Ax = 0 is trivial(no free variables) then Ax = b would be a unique solution. I tried thinking about maybe if Ax = b had more columns then rows but then it wouldn't be trivial.
For anyone reading this 4 years later, I believe Elias had an error when reading the answers. If Ax = 0 is trivial, then Ax = b is necessarily unique. Literally just solved a similar problem in my own linear book, and it illustrates why it has to be necessarily unique.
Maybe there's something about "for each b is an element of the set R^m", I don't really understand. Just saying.
can u do a class on DCT,DST, projections
can there be a non homogeneous equation with a trivial solution?
in my solution book,in some cases they use the same variable for two different free variables. for example x3=a and x4=a? why is this?
typo? take a screenshot so we can see it
Good explanation, however it feels lacking. You derived the conclusion that Ax = b is built from [homogeneous vector] + [translation], but this is super confusing since you derived that conclusion by row-reducing a nonhomogenous system.
I.E: You explain how to get to M, by starting from 0 -- but in your parametric example, you start from Ax = b, you start from M to find 0 and the translation! This is super confusing, since it doesn't explain WHY after row-reducing a nonhomogeneous system, we end up with a homogeneous vector! I don't understand why inherently row-reducing provides us that?
The only thing I can think of, is the simplification of:
let c and d be solutions to Ax = b, then:
A(c-b) = A(c) - A(b) = b - b = 0.
Let v = (c-b)
As such, A(v) = 0. And every linear transformation of v will also solve this equation, i.e: A(sv) = sA(v) = s0 = 0
So the general solution of the homogeneous system, A(s*v), correlates with the linear transformation of s*c and s*b, two vectors on the same line (m).
But that also doesn't explain why row-reducing a nonhomogeneous system, provides us directly with a [homogeneous vector] + translation !
how about t(q-p)+p is it works
What about the closest solution?