I think if someone is actually putting chalk to board, working through an example is something very integral to the learning process. Most (not all) of my professors rely heavily on slideshows and it drives me up the wall.
Professor Ana Rita Pires, thank you for an outstanding video/lecture on Basis and Dimension in Introductory Linear Algebra. The example in this video is an excellent building block for understanding Bases and Dimensions. This is an error free video/lecture on TH-cam TV with Ana Rita Pires.
But the set of vectors is linearly depandant and doesn't form basis, because while checking depandancy or indepandancy of the set of vectors we have to take 4 unknowns, and the rank of the matrix formed by these vactors is 3, so here rank of the matrix < number of unknowns, means that system has non zero solution. therefore these vectors are linearly depandant and doesn't form basis.
She put the vectors into rows, not columns. From 6:19 where she shows how to do it writing the vectors as columns. There the columns with pivot variables ended up in the basis and the columns of the free variables didn't.
4:48 In which other cases, besides this case study, matrix can have a useless equation in its rows? How you can transpose a matrix and still get a basis which is also happened to be lighter? Is it because they are in the same field?
Okay so the dimension is 3 yet doesn't that mean it can be graphed on R 3 graph. How do you graph a vector with 5 components in 3d space? Someone tell me why I'm wrong with all this.
@@andybeyond9546 Ok so the only conclusion I've come to is that the dimension of the basis is simply defined as the cardinality of the number of vectors that are in that basis, in which here it is simply 3. However, we're confusing dimension of say R^2 , R ^3 as we know them with graphing just because their basis dimension is 3 in R^3, and 2 in R^2 doesn't mean that other vectors that have more than 3 components have to exist in R^3, to have a basis dimension of 3. Simply put 3 dimension space is different than thinking about dimension of a basis. Please correct me if anyone thinks otherwise.
This is not the method of getting basis vectors that I'm seeing everywhere else online and now I'm more confused. Everyone else seems to use parameters for the non pivot columns and turns the pivot columns into a set of basis vectors based on their parameter values or something. I'm more confused now.... :(
Thanks for solving by Column also for a while I was afraid that my answer gone wrong that I done by column way when you explained in last then I take a long breadth.
Dear Ana Rita Pires I am Ghani, from Indonesia. This tutorial video is very interesting to understand more for linear independence and basis. I just want to know the reason, why we need to prove that the vector space "linear independent" first before we call that vector space a basis? Or maybe is it related to "the uniqueness of solution"? Thank you.
Well I think it's just to avoid redundancy, the BASIS are the root vectors together making the set and including any dependent vector will just mean including a linear combination of the root vector which has no importance at all. Hope it helped.
wow thank you very much your way of teaching is awesome. i have linear paper on Thursday i.e tomorrow and i so do not like maths and hence i didn't pay attention in classes or try the work at home and i had no idea what basis or dimensions are but now i do know a bit so i think i'll survive the paper tomorrow.
i'm a bit confused here, isn't the reduced form implies that it is linearly dependent, and therefore it is not a basis? (since the one condition to be a basis is the vectors must be linearly independent). cmiiw
to prove linear independence your essentially saying that x1*u1+x2*u2..xn*un=0(x is a scalar and u being a vector). the matrix in the video is a homogeneous system(every row =0) and if you reduce the matrix to echelon form you will see that the only solution to the system is x1=x2.....=xn=0. Witch proves linear independence
when I use TI 83 to do Ref for column space form, the calculator could not do that. How come ? it is weird This girl could reduce 5 by 4 , but calculator could not do
The TI calculators are a little weird about putting mxn matrices into rref. If you simply add an extra column of zeroes to your matrix to make it a 5x5 matrix, the TI83 should have no issue doing that. And it doesn't change the result either. You will still find that your first 3 columns have pivots, the 4th does not, and the 5th is inconsequential since you added it on anyway (and it's still just zeroes).
If your set is linearly dependent, it's possible to take some vectors away so that your new set is LI. You can only take one out by finding out which vector is is in terms of the others by finding a dependency relationship!
going to sound dumb but why do you solve the matrix listing the vectors as rows and not columns? In graphing they're listed as column vectors. Why not here?
In this case it would be really difficult to do. These vectors are spanning a 3-dimensional subspace in 5-dimensional space (R5). There's just no way to accurately graph that. That's the hard part of linear algebra and the reason so many people walk away from it scratching their heads even when they understand how to do the computations.
This is just perfect. So clear and so crisp. I used the second method myself so was kinda grateful that she explained it.
thank you so much for showing both ways, rows/columns... I was so confused.
That was the loudest "OOOOOOH" I've ever released. I finally understand
GarraOfTheFunk14 even me dude!
me too in 2020!
My prof couldn't explain this at all. This video is gonna help me pass my midterm!
Hopefully my final 😂
Which uni is this?
So what happened ?
@@Honest-King Dude, it has been for 5 years past
😂😂
I think if someone is actually putting chalk to board, working through an example is something very integral to the learning process. Most (not all) of my professors rely heavily on slideshows and it drives me up the wall.
she finally cleared all my doubts.great job and thanks a lot
Professor Ana Rita Pires, thank you for an outstanding video/lecture on Basis and Dimension in Introductory Linear Algebra. The example in this video is an excellent building block for understanding Bases and Dimensions. This is an error free video/lecture on TH-cam TV with Ana Rita Pires.
Great explanation; I hope I get to see more such sessions of Prof. Ana Rita Pires.
Just awesome !!!
Please add all videos on Linear Algebra.
Omg,thanks so much, I am having exams tomorrow and after doin' nothing for the hole semester I can still pass. Gpod luck on exams everyone
me too having exams, this video is really helpful and good luck to your exam
you're a piece of shit for doin' nothing you piece of filth.
6 years.. I don't think you guys still even remember how did you do in these exams
@@mostafaatif0 I did well. Was a good exam for me!
@@justkravchis8966 hope you all the best dude wherever you are
Very simple and accurate explanation..... No fancy terms used 👌👌👌
I love this girl.
Katy Lee sbian ?? ;p
@@user-em9mw9ch3y Ingenious
Are you still in love with her? it's been 6 years.
@@Astra20284 lol
please can i talk with you please
This is really an clear explanation. Thank you !
But the set of vectors is linearly depandant and doesn't form basis, because while checking depandancy or indepandancy of the set of vectors we have to take 4 unknowns, and the rank of the matrix formed by these vactors is 3, so here rank of the matrix < number of unknowns, means that system has non zero solution. therefore these vectors are linearly depandant and doesn't form basis.
ik thats whats confusing me , there are infinte many solutions so a basis cant be formed
exactly!
I saw plenty of videos on these topics but this is a good one. The teacher teaching on a blackboard with chalk gives a different feel...
TYSM!! I couldn't get this just from the explanations I got from my lecturer... only abstract theory with no practice. This helped me so much tysm :)
amazing to the point explanation
She put the vectors into rows, not columns.
From 6:19 where she shows how to do it writing the vectors as columns. There the columns with pivot variables ended up in the basis and the columns of the free variables didn't.
So it's wrong, right?
This video is old, but gold.
Thanks, much more clearer than the lecture itself.
4:48 In which other cases, besides this case study, matrix can have a useless equation in its rows?
How you can transpose a matrix and still get a basis which is also happened to be lighter? Is it because they are in the same field?
Does it matter how each vector is written in the matrix form?
Cause I notice you write the vectors as rows but others write it as a column.
Simple and elegant, could not have any more simpler
you're an amazing teacher!~
now i understand how to work with basis and span.
THANKS!!~
:)
Finally understand what the basis is, thank you so muchXD
i am greatful to you peoples work here on u tube
thank you. you're an angel
OMG This has helpped me a lot !! THANK YOU!
After one hour my paper will be start this video really help me thanks Mam
ultimate... this is just ultimate... thanks so much... u made it so easy to understand.
Okay so the dimension is 3 yet doesn't that mean it can be graphed on R 3 graph. How do you graph a vector with 5 components in 3d space? Someone tell me why I'm wrong with all this.
If you have figured it out please explain it to me, I'm having a hard time understanding this😅
@@andybeyond9546 Ok so the only conclusion I've come to is that the dimension of the basis is simply defined as the cardinality of the number of vectors that are in that basis, in which here it is simply 3. However, we're confusing dimension of say R^2 , R ^3 as we know them with graphing just because their basis dimension is 3 in R^3, and 2 in R^2 doesn't mean that other vectors that have more than 3 components have to exist in R^3, to have a basis dimension of 3. Simply put 3 dimension space is different than thinking about dimension of a basis. Please correct me if anyone thinks otherwise.
whata easy explaination......... love and respect ma'am..
Very clear. Thank you so much!
Amazing video teaching so much to grasp easily in the big subject.
This is not the method of getting basis vectors that I'm seeing everywhere else online and now I'm more confused. Everyone else seems to use parameters for the non pivot columns and turns the pivot columns into a set of basis vectors based on their parameter values or something. I'm more confused now.... :(
did you find out why?
She's really bright
She makes it looks easy! Nice
Thanks for solving by Column also for a while I was afraid that my answer gone wrong that I done by column way when you explained in last then I take a long breadth.
akka nuvvu super eh
Finally got what it is! Thank you
Dear Ana Rita Pires
I am Ghani, from Indonesia. This tutorial video is very interesting to understand more for linear independence and basis. I just want to know the reason, why we need to prove that the vector space "linear independent" first before we call that vector space a basis? Or maybe is it related to "the uniqueness of solution"?
Thank you.
Well I think it's just to avoid redundancy, the BASIS are the root vectors together making the set and including any dependent vector will just mean including a linear combination of the root vector which has no importance at all.
Hope it helped.
There's a mistake at 3:35
-4 + (-2) = -6, not -2
thanks for clear explanation
Great work mam!!
Thank you! This is really helpful!
Wonderful explanation !
Why do you write the vectors horizontally like that?
Graet! Is there all lessons of linear algbera? Here
So would you then find the transpose of those vectors in the column space to find the basis for the original matrix?
correct
why cant you use the columns of the eliminated matrix? didnt understand that part
I decided to just listen since I was given only 4 seconds to try it myself
Awesome video, thanks!
shouldn't we interchange the rows in the matrix formed by the vectors ? then we get an other basis
I should have found this video earlier.. it's great
2:55 can someone explain how she obtained that new 4th row?
R4 => R4 - 2xR1
Thank you. It is very helpful and clear
WOW!! Thank you so much, this was very useful :)
Thank you 💛💛
I think u should write the vectors as v1= of just big blanket the vectors
Superb explanation
Thank you😋😋
Thnx ma'am. You helped me alot. Thanks alot
simple and awesome!
Writing vectors in column of matrix is a natural form..If you make system of linear equations from vectors, you will get this matrix.
Thank you very much.
wow thank you very much your way of teaching is awesome.
i have linear paper on Thursday i.e tomorrow and i so do not like maths and hence i didn't pay attention in classes or try the work at home and i had no idea what basis or dimensions are but now i do know a bit so i think i'll survive the paper tomorrow.
Excellent explanation
When you do it vertically all vectors form a basis
i'm a bit confused here, isn't the reduced form implies that it is linearly dependent, and therefore it is not a basis? (since the one condition to be a basis is the vectors must be linearly independent). cmiiw
to prove linear independence your essentially saying that x1*u1+x2*u2..xn*un=0(x is a scalar and u being a vector). the matrix in the video is a homogeneous system(every row =0) and if you reduce the matrix to echelon form you will see that the only solution to the system is x1=x2.....=xn=0. Witch proves linear independence
Thank you😊
If you put the initial vectors as column vectors instead are the results are the same... Hmm...
how does -2-0=2 how can someone explayn 2:42
take the 2nd row and subtract with the 1st. So 0 - (-2) = 0 + 2 = 2
one of a hell expansive chalkboard
when I use TI 83 to do Ref for column space form, the calculator could not do that. How come ? it is weird This girl could reduce 5 by 4 , but calculator could not do
The TI calculators are a little weird about putting mxn matrices into rref. If you simply add an extra column of zeroes to your matrix to make it a 5x5 matrix, the TI83 should have no issue doing that. And it doesn't change the result either. You will still find that your first 3 columns have pivots, the 4th does not, and the 5th is inconsequential since you added it on anyway (and it's still just zeroes).
What is pivot?
Are those vectors column vectors in default? If yes, why the matrix does not contain four columns in this video?
wait you said that the elements must be linearly independent. If your basis has free variables that implies it is linearly dependent... I'm confused
If your set is linearly dependent, it's possible to take some vectors away so that your new set is LI. You can only take one out by finding out which vector is is in terms of the others by finding a dependency relationship!
The Matlab reduced that matrix were different from her. I don't know why ?
You nailed it 👌
Do you know why dimension is 3 yet 5 components?
going to sound dumb but why do you solve the matrix listing the vectors as rows and not columns? In graphing they're listed as column vectors. Why not here?
blessed video
Very nice teacher
Can anyone explain what she did with the elimination?
After taking these vectors as columns
We get different basis. why?
great video! thank you so much! :D
AWESOME thank you teacher
THANK YOU SO MUCH !!!!
Ahhh, we all feel smart watching a video from MIT Eh?
Why are we writing them as linear combinations ? Does it come per definition of span If yes why do we show it so ?
Thanks
thanks for your videos, its was nice explanation.
elegant and clear ==> thanks + :)
Amazing speed sis
in reduced echeloen form why is collumn 4 not a basis? (0,-4,1,0)??
more understandable than my discussion's TA and I go to another top US math school
Oh boy this is so understandable! TU!
Thank you so much mam
Amazing and thanks
There's a mistake in the 2nd step.
You're a mistake
good explanation, thank you!
Im a visual learner, ummm, how would you represent the basis of a matrix graphically?
In this case it would be really difficult to do. These vectors are spanning a 3-dimensional subspace in 5-dimensional space (R5). There's just no way to accurately graph that. That's the hard part of linear algebra and the reason so many people walk away from it scratching their heads even when they understand how to do the computations.
TH-cam 3b1b
Thanks for make this video