Nullspace of a matrix

Finding a basis for Nul(A), the nullspace/kernel of A, by row-reducing
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[@eamon_concannon]

I really appreciate your teaching style. It is such a refreshing change from most math videos!

[@holys6348]

prof there should be a -6 at around 5:43 for the first column 5th component of the column.

[@ablation791]

Also noticed this.

[@josh3460]

Thought I was losing my mind lol EDIT: He fixed it

[@CrypticPulsar]

You bring a breath of fresh air to a difficult subject! Thank you!❤

[@lightspd714]

Dr. Peyam, your teaching style is awesome. Love the energy!

[@drpeyam]

Thank you!!

[@eamon_concannon]

On the independence of the solution vectors, I think it becomes clear when we consider that the free variables (y and t here) always appear on their own in the solution vector ( (2y+6t y -2t t) here). (The pivot variables x and z appear as linear combinations of the free variables in the solution vector). We can write the solution vector as a linear combination of vectors by letting y=0, t=1 and then y=1, t=0 guaranteeing that the vectors are independent.

[@khbye2411]

Thank you for the video!! I just learnt about the nullspace, range and rank-nullity theorem in the more general context of linear transformation functions. Before that, we learnt about the rank-nullity theorem in matrices only...I think I finally got the link to the special case where we look only at matrices Our notes defined nullspace as N(T)=the set of all the vectors v in vector space V (the domain), such that when you apply T to the v (T refers to a linear transformation, with domain V and codomain W), T(v)=the zero vector in the codomain W Nullspace of a matrix, is when you let vector space V (domain) be the set of all column matrices, size n*1 (this is the 'home' of the solution x in Ax=0 !!), vector space W (codomain) be the set of matrices, size m*1 v be the column matrix x (the solution to Ax=0), T(v) be the linear transformation function which maps from x-->Ax (the rule of T), where A is a m*n matrix that's left-multiplied onto the x and the zero vector in codomain W is the zero matrix :)

[@jasonbroadway8027]

I enjoyed the video, but at 5:43 you might have made an error.

[@jonathanb.burnett3204]

This was a nice video. Educational and puts a smile on my face.

[@ChitranshSagarMTAIE]

Excellent Teaching Style , Superb Sir :)

[@newtonnewtonnewton1587]

Wonderful thanks alot doctor peyam

[@jasonbroadway8027]

You fixed it! Nevermind!

[@WarzoneMasters]

just amazing thanks a lot sir

[@chanokplaisub3463]

The great one!

[@saudahmad5624]

Thanks Dr. Peyam

[@rexyancey2181]

bro is cracked at matrices.

[@rolfdoets]

In Holland we also say NUL for 0, thnx for your lecture

[@anuragbasu1813]

You teach awesome👍👍👍👍

[@dhunt6618]

I'm trying to figure out all the relations of spaces, is the following correct? space time ⊃ outer space ⊃vector space ⊃ space ⊃ (nasa ∪ esa ∪ jaxa) space exploration ⊃ spacex ⊃ sub space ⊃ personal space ⊃ null space ⊃ monopoly board space (iI just got the 'go to jail, go directly to jail, do not pass go, do not collect $200' card, darn it!) Thanks for your videos enjoying both mathematics and silliness!

[@nuraisyahbintiomarupm6124]

that singing hahahhahahha you will always be my basis~

[@user-bu8mg7uq3s]

thank you

[@gvantsasakaruli9900]

When you started the geometric interpretation i thought somebody was going to draw time!!!

[@StewieGriffin]

The first 2 columns is multiplied by -2

[@yashj8238]

What does it mean to be a variable to be free?

[@dinhkhoa3665]

Hi Dr. Peyam, can you prove why the construction of the Null space will always be a linearly independent set? My textbook does not give a proof on this one.

[@eamon_concannon]

I think it becomes clear when we consider that the free variables (y and t here) always appear on their own in the solution vector ( (2y+6t y -2t t) here). (The pivot variables x and z appear as linear combinations of the free variables in the solution vector). We can write the solution vector as a linear combination of vectors by letting y=0, t=1 and then y=1, t=0 guaranteeing that the vectors are independent.

[@joshdilworth3692]

How do you know by looking at the matrix in RREF that y and t are going to be free variables? is it their lack of a pivot?

[@drpeyam]

Yes basically; the free variables are in the non-pivot columns

[@joshdilworth3692]

@@drpeyam thank you for your quick reply! Your attention and care that you show towards your followers is why I (we) respect you so much 😁. I (we) cannot wait for your upcoming videos.

[@deeptochatterjee532]

How is this comment from a week ago but the video was just uploaded?

[@drpeyam]

It’s been unlisted for a week (so technically available), but officially released today

[@federicopagano6590]

And in english null and VoId menans without Value

[@vijayank1170]

Dr. Peyam Can you please make a video explaining.. Pointwise and uniform convergence of series , explaining their differences and their relation to continuity with examples.. It would be really helpful.. Because I am having a hard time understanding the idea behind them.

[@drpeyam]

There’s a video on covering compactness and uniform continuity. Not quite what you want but a good start

[@foreachepsilon]

vijayan K my understanding is: uniform convergence is a substype of pointwise convergence. A series of functions is pointwise convergent if any given x can result in a convergent series (in other words, pick a value for x, like five, and see if the series converges; you can pick a different epsilon depending on the x). Uniform convergence says you can't pick any special x when showing the series converges; there has to be one epsilon that fits every choice of x. It's been a few years so I might be off.

[@vijayank1170]

@@drpeyam Dr peyam can you suggest a good read..

[@foreachepsilon]

www.personal.psu.edu/auw4/M401-notes1.pdf

[@foreachepsilon]

math.byu.edu/~bakker/M341/Lectures/Lec28.pdf

[@joluju2375]

I don't get it. The definition is so simple, I thought we'd just solve for (x,y,z,t) vectors. But then strange things arose, an unknown notation with a vertical bar, undefined terms like pivot, modifying rows with a recipe I never heard of ... Pretty sure the result is the same with a naive method, but where did the method you used come from ? Is there a video somewhere I should see ? Please :)

[@drpeyam]

Yeah! Check out my Systems of Equations (Lay Chapter 1) playlist. It’s called Gaussian Elimination

[@joluju2375]

@@drpeyam I got it now, thanks ! I feel stupid, it's what I had been taught very long ago, but with a fancy layout and wording. I didn't recognize the lady at first because of the makeup, but she's the same.

[@sugarfrosted2005]

I always thought it was odd that linear algebra uses different terminology that most of algebra. Null Space rather than the Kernel. I suspect it's because linear algebra developed before abstract algebra.

[@drpeyam]

I think it’s to distinguish it from linear transformations, although they’re the same

[@AndDiracisHisProphet]

in german it is also "Kern" in linear algebra and not "Nullraum" or something

[@arthurlbn]

The Ker(T) is a null space?

[@drpeyam]

Yes same thing

[@cameronspalding9792]

@5:43 I think you made a mistake

[@yashj8238]

ma man

[@pawanshkl60]

In Hindi it's "Lul" means zero. So same it is 😀😀😀😀

[@myliserta]

At 5:20 , R1-3R2 ->R1 should be [-1, -2, 0, -6] not [-1, -2, 0, 0]

[@drpeyam]

That’s mentioned in the comments already

[@myliserta]

@@drpeyam Sorry, I could not find it. Anyway, congratulations for the great explanation!

[@thenewdimension9832]

You will always be my Baby😂😂😂😂😂😂

[@KANA-rd8bz]

in polish we say "JĄDRO" which also means ... testicle.😂😂😂😂😂😂😂 "Find the testicle of A"

[@linguafranca7834]

Ohkkiii😂

[@Pradowpradow]

Haha in french we use the word Ker(A) and not Nul :D

[@drpeyam]

We use Ker for linear transformations

[@Pradowpradow]

@@drpeyam ain't it what we are currently studying?

[@foreachepsilon]

A is a matrix. T(x) = Ax is its linear transformation.

[@drpeyam]

Yes, but it’s just to distinguish kernels for matrices from kernels of general linear transformations (on infinite dimensional vector spaces)

[@foreachepsilon]

In my study, with the above definition I have, we would have ker(T) = null(A).

[@yhmah6729]

Nul ! Nul ! Nul ! Germain !