Thank you! -Christina, 48 year old mother of 7 children finishing a Secondary Math Ed I started 30 years ago. :) Your solution was clear and organized. Awesome!
I'm just gonna say again, I don't really understand what my professor said but I'm able to understand the explanation from this video. It really helped me a lot, no matter I'm gonna fail this subject or not, thank you for making this video.
The number of pivot variables = number of independent basis vectors that make up the column space of A. Very insightful, Sal! It took me a while to process but now I get it ☺️
I have always heard good things about Khan Academy and it definitely checks out. This video explained a topic I have been struggling with clear as day.
Hello, is there not a mistake done in the first place when you were subtracting 2 times row 1 from row 2? You said so but you subtracted row 2 from 2 times row 1 and it changed all the result. I try to understand linear algebra and everything coming up with it so I may be wrong but this is opposite to what I learned from MIT open courseware and what you said in this very video. Please clarify this point for me or I ill get lost!
I think it makes a bit more sense to apply Elementary Row Operations upon the Matrix before figuring out the Column Space. You'll see already before if the system of equations collapses the vector to a line, plane or 3d hyper-plane. It also has then a nicer form to check for the results of the Rank-nullity theorem.
the last part may not necessary to find the basis u can just pick it form the reduced encholen form which have pivot in each column in this case it is column 1 and 2
are pivot variables always the linearly independent ones? can't you write the pivot variables in terms of the free variables here as well? ack it's kinda coming together for me... thx khan
@khanacademy he is probably being sarcastic or just a throll, you are doing amazing job with your amazing explanations, dont let that anonymous idiots make you lose strength to carry on. Have a nice day.
I wouldn't say it's so much over-explanation rather than thinking out loud. At least for me, this helps, not because I don't know how to subtract (subtraction being one of many things he 'over-explains'), but because I can keep track of every assumption he's making.
It's not you, it's just the human nature that can't accept the truth and the truth is majority of the teachers here don't care if the student learns or not.(not all cuz I have some great Profs at my school). But most teachers here just work for their pay check. That doesn't happen in India. People care more about each other. Now this guy explaining everything for free, that's the kind of spirit we need in teachers her. I don't want them to teach for free but just care more than they do..
1:34 "We don't know that these are linearly independent" ... yes we do, there are 4 vectors in R3, one of them will be redundant, therefore those vectors are linearly dependent. Also, after you row reduced, you just needed to see which columns had a pivot point then go back to the original matrix and take those columns and they are the basis vectors for Col(A). ... eg. there was a pivot position in the columns for x1 and x2, the basis vectors were eventually determined to be column 1 and column 2 of the original matrix. Make sure you understand what is going on in the video though, it's really important that you do.
I think you made a mistake on your second computation. -2 x Row 1 added to the remainder of the entries in Row 2 should give -1, 2 and 1, not 1, -2 and -1.
Curious, when you first proved that X3 & X4 were "free" variables, is that enough evidence to consider those vectors redundant and exclude them from the final linear independent set, or was that just coincidence?
+aaad1100 It's even more than that ,seeing that in the reduced echelon form that the non zero rows are just 2 ,and the number of columns (variables) is 4 ,then you should figure out there is two additional variables or additional redundant vectors.
What happens if you have a column consisting of only 0's, regarding the null space basis? Wouldn't that mean that the respective x-variable is neglectable?
most probalably....self study...........or.........one good teacher(lecture) who knows the subject deeply....not by just passing the exams.....by feeling maths....
Dear friend he is talking about the education standards of the US which are very very low as compared to other countries. What you are given in 12 grade her, I was given that stuff in 9th in India
An average kid here need a calculator, an equation sheet for an exam and it's provide, where as any of that stuff in Indian schools is strictly prohibited. I am not talking about the small schools in the poor villages. I am talking about the prestigious schools which we have many
So we have weird exercises to do as homework (tho we havent even done ANY exercises on this topic, all they did was throw empty definitions at us and expect us to be geniuses) where it says "Which vectors(b1,b2,b3) are in the column space of A?" A= 1 1 1 1 2 4 2 4 8 And thats all the info we have. How does one solve it?
This doesn't actually teach you what a null space is.. this basically teaches you some trick to figure out the basis of a subspace.... waste of 20 mins.
About getting RREF, you've made a mistake (that actually not criticall, but anyway), when you subtracted 2 times row 1 from row 2 you said the one thing and did another one, you didn't subtract 2xR1 from R2 but added 2xR1 to -R2
Orpheus Pericles No, because here you can see that he put 0 for x3 while proving that v4 is redundant and put 0 for x4 while proving that v3 is redundant. So, we can get rid of both v3 and v4. Also, the basis of a subspace need not span all the points in the graph because the span of the subspace can be limited. For example, here, the span is limited to a plane in R^3. What we can say is that the number of vectors in basis need not be greater than the order of dimension.
For the point you mentioned @Vishal Goel, " the basis of a subspace need not span all the points in the graph ".... I think it is not as per the definition Sal gave in a previous video that the basis is the minimum set of vectors that spans the subspace ! Also, till now I am not totally convinced how the number basis vectors of a subspace to be less than the subspace order !?
I feel like he never missteps, but this was definitely the harder way to find the column space...why not just out it in a matrix and get the leading ones? Maybe that's what you did, but it definitely seemed more consuming. I had to stop watching the video before I got confused...
There wasn't a clear goal that he was trying to get to. He wasn't doing all these steps just to get to the final goal of the linearly independent set of vectors spanning the column space of A. You need to interpret this video as being more of a exploration in the the relationships between a matrix, it's null space and it's column space, rather than an explicit problem solving exercise.
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
You explained in 25 minutes what I have been confused about for the past 200 minutes of my class. Amazing
After 100% understood
ThinkPositive00 lol, you have two separate comments. old RUclips was something else
I love how both comments have the same exact number of likes ! Math students are so precise lmao
@@kozukioden2406 wow 5 months later and its still have the same number of likes
@@hugoirwanto9905 It still has the same number of likes 223. How far will it go? I am curious...
@@shawnjames3242 Yes. It's 259 on both now
Before 0% understood
ThinkPositive00 middle 50% understood.
Well
nice
Are you a professor now?
@@neelparekh1759 Are you a professor now?
Anybody else have their linear algebra exam coming up too? haha you saved me once again khan academy, very clear and easy to follow.
Me
me
yes :)
Tomorrow 😢
Yes
Thank you! -Christina, 48 year old mother of 7 children finishing a Secondary Math Ed I started 30 years ago. :) Your solution was clear and organized. Awesome!
A faster way to find the basis for the column space is to rref and then take the column vectors with pivots
True!
dom you're right! I noticed it too and had an aha! moment. Life of a math junkie lol
check this ruclips.net/video/8o5Cmfpeo6g/видео.html
Ty! Thats what I was thinking
Yeah
I'm just gonna say again, I don't really understand what my professor said but I'm able to understand the explanation from this video. It really helped me a lot, no matter I'm gonna fail this subject or not, thank you for making this video.
The number of pivot variables = number of independent basis vectors that make up the column space of A. Very insightful, Sal! It took me a while to process but now I get it ☺️
2am in morning..."ill let you go for now"
"yes!! im free! i can go to sleep!"
thanks, i do not why i could not understand this but your video did the trick!
i've got a feeling that i'll get my bachelors in Mech Engineering with this channel
COVID-19: Oh no you won't!
same
Ohhhhhhhh thankxxxx a lot....!! Finally I understand the difference of null and column space and it works for creating basis.
I understood more in this 25-minute video than in my lecture on the topic today........ Thank you for this video
I have always heard good things about Khan Academy and it definitely checks out. This video explained a topic I have been struggling with clear as day.
It's fantastic! So straight-forward. Thank you so much!
thank you !! i know im not alone when i say that! this cleared everything up so neatly
This man has saved so many people's grades, about to take my linear algebra midterm rn 😅
thank you so much i have a final in 4 hours and this made everything simpler
how was it?
I love you Khan Academy
That's not exactly giving me the best incentive to finish
Hello, is there not a mistake done in the first place when you were subtracting 2 times row 1 from row 2? You said so but you subtracted row 2 from 2 times row 1 and it changed all the result. I try to understand linear algebra and everything coming up with it so I may be wrong but this is opposite to what I learned from MIT open courseware and what you said in this very video. Please clarify this point for me or I ill get lost!
this 25 minute lecture puts 3 weeks of lecture in class to shame, very helpful
OMG YOUR A GENIUS. I CAN'T BELIEVE I LEARNED THAT.
OMG... I was just on this studying this topic right now... and you posted this up like 10 minutes ago... WOW!!
how old are you now?
@@certified_vg2200 12
@@certified_vg2200 jk 30
@@NotmyYTchannel wow still active 8)
@@bunstie5208 yup og
You're saving my linear algebra grade, THANKS!
I assume you've graduated by now!
Very Nice explanation!
thank you so much, finally a video i can understand
this video should be of maximum 5 mins....but u are awesome in extending videos
so if there are free variables in the reduced row echelon form, does that mean that it is linearly dependent
Yup!
adamsın adam!! (trying to get it for a day long. finally you made it. thanks in advance.)
great for the review of basis, null space and column space for a matrix !
I think it makes a bit more sense to apply Elementary Row Operations upon the Matrix before figuring out the Column Space. You'll see already before if the system of equations collapses the vector to a line, plane or 3d hyper-plane. It also has then a nicer form to check for the results of the Rank-nullity theorem.
So when the determinent is zero, the system of equations collapses down to a line?
the last part may not necessary to find the basis u can just pick it form the reduced encholen form which have pivot in each column in this case it is column 1 and 2
But he just proved that columns 1 and 2 are sufficient for finding the basis
You explained it very easy thank you, god bless you
this saved my life
Great!
khan is a god
Very nicely explained
This is the single most redundant way to explain that pivot variables determine the column space but I finally got it
Thanks!
KHAN ACADEMY in HD , aaawwww yea!!
this video is pretttttyyyyyyy old yet very relevant in 2021......
When we do the echelon reduction, do we need to make sure that the pivot elements need to be 1?
Yes or else we can't use it
yes
awesome vid
Mind. Blown.
thank u very much
anaaaaaaaaaaaaaaaaaaaazing video ! Neat Clear , thanks !
I have a query: are pivot variables aka dependent variables & free variables aka independent variables?
are pivot variables always the linearly independent ones? can't you write the pivot variables in terms of the free variables here as well? ack it's kinda coming together for me... thx khan
are pivot variables always the linearly independent ones- Yes
So in order for the column space to be Liniarly independent, the rref would have to be the identity matrix, right?
@khanacademy
he is probably being sarcastic or just a throll, you are doing amazing job with your amazing explanations, dont let that anonymous idiots make you lose strength to carry on. Have a nice day.
You just saved my ass :)
Your brain*
Took me 1 day to understand span subspace basis null space column space and then remembering it
I wouldn't say it's so much over-explanation rather than thinking out loud. At least for me, this helps, not because I don't know how to subtract (subtraction being one of many things he 'over-explains'), but because I can keep track of every assumption he's making.
AWESOMENESS !!!
ITS MAGIC!!!
Very enlightening video! One question though. What software do you write on? I'd love to take notes in class using the same method
Even though i finished this video, i play it back just to hear his voice :'(
Can a vector be in both a the Null space AND the Column space of some set of vectors? Or is it one or the other...?
yeah i totally agree... but he tries to prove it more theoretically
How did I pass this subject? This is so confusing 😭
It's not you, it's just the human nature that can't accept the truth and the truth is majority of the teachers here don't care if the student learns or not.(not all cuz I have some great Profs at my school). But most teachers here just work for their pay check. That doesn't happen in India. People care more about each other.
Now this guy explaining everything for free, that's the kind of spirit we need in teachers her. I don't want them to teach for free but just care more than they do..
1:34 "We don't know that these are linearly independent" ... yes we do, there are 4 vectors in R3, one of them will be redundant, therefore those vectors are linearly dependent. Also, after you row reduced, you just needed to see which columns had a pivot point then go back to the original matrix and take those columns and they are the basis vectors for Col(A). ... eg. there was a pivot position in the columns for x1 and x2, the basis vectors were eventually determined to be column 1 and column 2 of the original matrix. Make sure you understand what is going on in the video though, it's really important that you do.
Can the basis of the column span be the columns with pivots in rref?
Yup!
But why did he referred pivots from original one but not from rref?
I think you made a mistake on your second computation. -2 x Row 1 added to the remainder of the entries in Row 2 should give -1, 2 and 1, not 1, -2 and -1.
YEA IT IS MOST IMPORTANT FPR EVERYONED , BY THIS WAY I THIK ANYBODY CAN LEARN MATH S BIN SIMPLE WAY
The basis of Nul(A) is the same spanning set of Nul (A)...
I think you forget to say that!
Its easier to say that the pivot columns of A form a basis for Col(A) :P
THANK YOU!!!!
Curious, when you first proved that X3 & X4 were "free" variables, is that enough evidence to consider those vectors redundant and exclude them from the final linear independent set, or was that just coincidence?
+aaad1100
It's even more than that ,seeing that in the reduced echelon form that the non zero rows are just 2 ,and the number of columns (variables) is 4 ,then you should figure out there is two additional variables or additional
redundant vectors.
thanks
there easier way to figure out the basis. it is the original columns that correspond to the pivot columns in its RREF.
nice
What happens if you have a column consisting of only 0's, regarding the null space basis? Wouldn't that mean that the respective x-variable is neglectable?
most probalably....self study...........or.........one good teacher(lecture) who knows the subject deeply....not by just passing the exams.....by feeling maths....
who are those ultra genius 93 people who disliked this video?
Dear friend he is talking about the education standards of the US which are very very low as compared to other countries. What you are given in 12 grade her, I was given that stuff in 9th in India
if you speed this up to 1.5 , it essentially feels like a man trying to win an argument against a whamen
Where are next videos , please tell can't find them
thanks sal sal
Thanks for the video. Hope you keep up the good work, which obviously you are =0)
Very helpful thanks, too bad I find it impossible to stay away in any sort of linear algebra lesson *yawn*
An average kid here need a calculator, an equation sheet for an exam and it's provide, where as any of that stuff in Indian schools is strictly prohibited. I am not talking about the small schools in the poor villages. I am talking about the prestigious schools which we have many
So we have weird exercises to do as homework (tho we havent even done ANY exercises on this topic, all they did was throw empty definitions at us and expect us to be geniuses) where it says
"Which vectors(b1,b2,b3) are in the column space of A?"
A= 1 1 1
1 2 4
2 4 8
And thats all the info we have. How does one solve it?
Well because you have 4 vectors in R3 so you can tell that they are linearly dependent.
thanks again ! well ,i'm gonna forget mine LA teacher but not you.
I LLOVE YOU
This doesn't actually teach you what a null space is.. this basically teaches you some trick to figure out the basis of a subspace.... waste of 20 mins.
There is a separate video for that
haha, at 0:06 ...CURL over... ..really INTEGRATE everything...
SALL KHAN is proud of MUSLIMS
I love u
About getting RREF, you've made a mistake (that actually not criticall, but anyway), when you subtracted 2 times row 1 from row 2 you said the one thing and did another one, you didn't subtract 2xR1 from R2 but added 2xR1 to -R2
shouldn't the no. of basis vectors be equal to the dimension of the subspace??
Orpheus Pericles No, because here you can see that he put 0 for x3 while proving that v4 is redundant and put 0 for x4 while proving that v3 is redundant. So, we can get rid of both v3 and v4. Also, the basis of a subspace need not span all the points in the graph because the span of the subspace can be limited. For example, here, the span is limited to a plane in R^3. What we can say is that the number of vectors in basis need not be greater than the order of dimension.
For the point you mentioned @Vishal Goel, " the basis of a subspace need not span all the points in the graph ".... I think it is not as per the definition Sal gave in a previous video that the basis is the minimum set of vectors that spans the subspace !
Also, till now I am not totally convinced how the number basis vectors of a subspace to be less than the subspace order !?
The next video explains and visualizes that point. Thanks !
just because people are in linear algebra doesnt mean they can follow simple calculations, there's some people in my class that are really dumb
At the end, didn't he mean to say column space of A "C(A)" ? Instead of column span of A?
*nullsapce*
to moeb32, he said he was doing 2r1-r2 not r2-2r1...
I feel like he never missteps, but this was definitely the harder way to find the column space...why not just out it in a matrix and get the leading ones? Maybe that's what you did, but it definitely seemed more consuming. I had to stop watching the video before I got confused...
There wasn't a clear goal that he was trying to get to. He wasn't doing all these steps just to get to the final goal of the linearly independent set of vectors spanning the column space of A. You need to interpret this video as being more of a exploration in the the relationships between a matrix, it's null space and it's column space, rather than an explicit problem solving exercise.
"Nullsapce" in the thumbnail :^)
any one could help me to find the basis of left nullspace?
a Gizzillion Times agreed!!!!!