Linear Algebra 35 | Rank-Nullity Theorem

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(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)

[@user-zr4ns3hu6y]

What a nice explanation! Thx!

[@brightsideofmaths]

You're welcome! And thanks for your support :)

[@tnuts92]

Nice thanks ! Please which software are you using to make these nice interactive presentations ?

[@brightsideofmaths]

Xournal :)

[@blakemcalevey-scurr1454]

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?

[@arturo3511]

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!

[@brightsideofmaths]

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 :)

[@sameerkhnl1]

@5:15 You say that dim(Ker(A)) = 1. But since Ker(A)= [1,1,1]^T, then should its dimension not be 3?

[@brightsideofmaths]

No, it's one :)

[@VictorHugo-xn9jz]

There's a more intuitive proof which involves showing that the null space and range are orthogonal to each other.

[@brightsideofmaths]

Don't forget: both subspaces live in different spaces.

[@VictorHugo-xn9jz]

@@brightsideofmaths oops, haha, I’m really sorry, I completely confused row space and column space in my proof. Thank you for the hint