ไม่สามารถเล่นวิดีโอนี้
ขออภัยในความไม่สะดวก
Linear Algebra 35 | Rank-Nullity Theorem
ฝัง
- เผยแพร่เมื่อ 4 ก.ย. 2024
- Support the channel on Steady: steadyhq.com/e...
Or support me via PayPal: paypal.me/brig...
Or via Ko-fi: ko-fi.com/theb...
Or via Patreon: / bsom
Or via other methods: thebrightsideo...
Watch the whole video series about Linear Algebra and download PDF versions and quizzes: tbsom.de/s/la
There is also a dark mode version of this video: • Linear Algebra 35 | Ra...
There is also a bright mode version of this video: • Linear Algebra 35 | Ra...
To find the TH-cam-Playlist, click here for the bright version: • Linear Algebra
And click here for the dark version of the playlist: • Linear Algebra [dark v...
Thanks to all supporters! They are mentioned in the credits of the video :)
This is my video series about Linear Algebra. I hope that it will help everyone who wants to learn about it.
#LinearAlgebra
#Vectors
#Matrices
#MachineLearning
#Eigenvalues
#Calculus
#Mathematics
(This explanation fits to lectures for students in their first year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)
Nice thanks ! Please which software are you using to make these nice interactive presentations ?
Xournal :)
What a nice explanation! Thx!
You're welcome! And thanks for your support :)
If the vectors ci are part of a basis for R^n, and that basis is linearly independent, doesn't that directly imply that ci are independent too? So the line of the proof labelled "basis of kernel" is redundant right? Is it just there for clarity, or am i missing something?
@5:15 You say that dim(Ker(A)) = 1. But since Ker(A)= [1,1,1]^T, then should its dimension not be 3?
No, it's one :)
In general if (v1,v2) are linearily independent then, T(v1), T(v2) are not necessarily linearily independent. Here we said c1, c2 linearily independent then also Ac1, Ac2 are linearily independent for a general map A. Why is it true here for all general linear maps A? thanks!
It's because we excluded the kernel: c_1 and c_2 are for spanning the range of A. The formal proof is at the end of this video :)
There's a more intuitive proof which involves showing that the null space and range are orthogonal to each other.
Don't forget: both subspaces live in different spaces.
@@brightsideofmaths oops, haha, I’m really sorry, I completely confused row space and column space in my proof. Thank you for the hint