"The rank of a matrix is the number of linearly independent column vectors that can be used to construct all of the other column vectors." Perfect. Thank you so much!
These vidoes make linear algebra so much more clear. I am always lost trying to decipher my teachers lectures and the text, but then i watch these vids and it all makes so much sense! Thank you !
you just taught me in less than 13 minutes, what my prof failed to teach us for the past 2 weeks. Thanks alot to take time out of your daily life to help out students, especially for free.
At the time this video series came out, all people thought about was that it was great and that it explained things better than university professors, which allowed people to pass. In the future though, these videos will be considered as strong evidence for why in person university education is obsolete. These kinds of series will be the new basis vectors for education ;)
I have a linear algebra exam tomorrow. It's hard and I don't have a good prof and my text book is written in such a way that you have to unscramble what they are trying to get at... The guy I was paying to tutor me was terrible. I don't know why I bothered with all of that when you make something that seems SO FOREIGN so like, normal and not that hard! Your videos make me like linear algebra! (kind of... that might be an exaggeration) but THANK YOU SO MUCH FOR SAVING STUDENTS EVERYWHERE!
I passed this class in college, and idk how did I passed it, but I came back for engineering grad school, and you have saved me!!!! This is supposed to be material I should've already known to understand more advanced topics :/
I hope that you know that you are the reason i graduated!!!! haha at U of I to say the least you come in clutch and i can actually understand what your saying
K I watched some videos, and I think it's what I said, except you do not go back to the original un-reduced matrix to find the linearly independent rows that comprise the basis for the row space of A: you just use those rows you found in the reduced form. Nicely, once you find the row/column space, it is an easy task to find the column/row space, since the reduction exposes both in the matrix.
if you, patrick jmt, and thenewboston, and other like that were to open a college....your education would be way more valuable than any other college because the kids coming out of your college would actually know stuff
I really love your videos, they are awesome. You explain things much better than most of the professors in my university. Thank you very much. Although, you made a small mistake at 4:45. You said minus one plus minus one is zero and minus two plus minus two is also zero. Well, the second minus one and the second minus two were actually plus one and plus two. And minus one plus minus one would be minus two, not zero. :)
When you rref the column space (or span), don't you have to like turn the columns into rows first and THEN go ahead and do the substractions? (And then turn the rows back to columns)
To calculate Col A, do u use echolon form or reduced echolon form, watching some videos on it and every video has different method. Some are using reduced and some echolon only. So im getting confused which one is supposed be used?
THANKKKKKKKKKKKK YOU SO MUCH . PLEASE PUT MORE AND MORE VIDEOS. EXPLAIN A SLOWER AND AT THE END OF THE VIDEO SAY MISPLAY WHAT YOU DID . JUST LIKE WHAT YOU ASIDE AT THE END OF THIS ONE. MORE AND MORE VIDEOS PLEASE
Dimension (Number of linearly independent vectors) of Column Space: Rank Dimension (Number of linearly independent vectors) of Null Space: Nullity And Rank + Nullity = Number of Columns in the matrix.
in this lecture you said dimension =number of pivot columns in c(A) however in the previous video called dimension of null space and nullity you said dimension= number of non pivot columns in N(B) so which one is it? # of pivot or # non pivot columns? or does that change based on if its a null space V.S a column space? and what is the difference between the two?Thanks.
Yes, the set of pivot columns will be linearly independent. Think about it this way: what makes a set of vectors linearly independent? A set of vectors, A, is LI iff the only solution to Ax = 0 is the trivial solution (i.e. x is the zero vector). So, no linear combinations of vectors in A will equal 0 except for the case 0x1 + 0x2 + ... + 0xn = 0. By definition, a pivot column has a leading value and all entries underneath it are zero. So in the matrix: 1 0 0 1 No matter what nonzero scalar c1, c2 we multiply each of the vectors by, we will never be able to get a nonzero value when we add the vector components together.
@g1tfisted My thoughts exactly. My linear algebra teacher gave us all a take-home test that we have to turn in in two days, and no one really understood his explanation of column space and nullspace. Thank you so much for your help, this will surely help me answer some questions on my test. :P
Hello Khan academy, you have left me extremely confused since by the theorem of linear independence: If a set contains more vectors than there are entries in each vector than the set is linearly dependent. I must have misunderstood you so in other words: are you saying that the subset of A(a1, a2 , a4) is independent or that A (a1,a2,a3,a4,a5) is independent?
I really think you're on to something...if our fucked up govt. ever decides to relieve the poor of their education (koch brothers have tried) a system like the one you have started would definitely even the playing field.
"The rank of a matrix is the number of linearly independent column vectors that can be used to construct all of the other column vectors."
Perfect. Thank you so much!
These vidoes make linear algebra so much more clear. I am always lost trying to decipher my teachers lectures and the text, but then i watch these vids and it all makes so much sense! Thank you !
you just taught me in less than 13 minutes, what my prof failed to teach us for the past 2 weeks.
Thanks alot to take time out of your daily life to help out students, especially for free.
thank you so much Sal Khan, though you released this more than a decade ago, these videos are still as useful and relevant. a BIG THANK YOU!
Facts
Math will never be irrelevant.
Thank you sooooo soooooo much, you are much better than a loooooot of profs in my university
At the time this video series came out, all people thought about was that it was great and that it explained things better than university professors, which allowed people to pass. In the future though, these videos will be considered as strong evidence for why in person university education is obsolete. These kinds of series will be the new basis vectors for education ;)
I have a linear algebra exam tomorrow. It's hard and I don't have a good prof and my text book is written in such a way that you have to unscramble what they are trying to get at... The guy I was paying to tutor me was terrible. I don't know why I bothered with all of that when you make something that seems SO FOREIGN so like, normal and not that hard! Your videos make me like linear algebra! (kind of... that might be an exaggeration) but THANK YOU SO MUCH FOR SAVING STUDENTS EVERYWHERE!
I passed this class in college, and idk how did I passed it, but I came back for engineering grad school, and you have saved me!!!! This is supposed to be material I should've already known to understand more advanced topics :/
You explain in 10 minutes what my teacher could not in two weeks. Thank you for saving my grade :)
that video will definitely help me in my final exam tomorrow. thank you!
dim(null space)=number of non pivoted columns where as dim(column space)= number of pivoted columns.
very nicely explained thank you
thanks for making linear algebra easy, i wish my lecturer could explain things as simply as you do
must have been a great experience learning online in those days.
I hope that you know that you are the reason i graduated!!!! haha at U of I to say the least you come in clutch and i can actually understand what your saying
Best math teacher of all time!!!!!
i love you, man...
this is the first time i really understood everything someone told me about this stuff
I donate some to khan .. So that this incredible work will not stop in future..
amazing... my linear algebra final is next tuesday. I only wish I knew about this throughout the whole semester. I bet I ace it with this help.....
Juliusblue Did you pass the exam?
K I watched some videos, and I think it's what I said, except you do not go back to the original un-reduced matrix to find the linearly independent rows that comprise the basis for the row space of A: you just use those rows you found in the reduced form. Nicely, once you find the row/column space, it is an easy task to find the column/row space, since the reduction exposes both in the matrix.
this video lecture help to learn to basis , rank and dimension for the subspace of column vector which i did not find a good one anywhere. thank you
thanks a lot , that helped me a lot before my final !
thanks for making "ranking" easy for me to understand,
if you, patrick jmt, and thenewboston, and other like that were to open a college....your education would be way more valuable than any other college because the kids coming out of your college would actually know stuff
thx..i've been reading about the rank and i don't understand at all... thx 2 u i understand it now
Best video on youtube for this topic
THANKS SAL YOU ARE LIFE SAVER
I really love your videos, they are awesome. You explain things much better than most of the professors in my university. Thank you very much.
Although, you made a small mistake at 4:45. You said minus one plus minus one is zero and minus two plus minus two is also zero. Well, the second minus one and the second minus two were actually plus one and plus two. And minus one plus minus one would be minus two, not zero. :)
This is so crucial. I’m thankful
You are simply charming.saved my quiz day. Thank you
Thank you for your videos! Very helpful.
Also, the dimC(A) = Number of columns - caracteristic of (A) = 3
:)
Hey, can u reply me
u are da bossss! im gonna ace my test tmrw!
Nathaniel Sarkissian did you ace it? Lmaooooo
Thank God for Khan Academy.
I start to realize that Khan is much better than my prof.....
I can truly relate ✌
11:30 for dimension, your welcome
do you mind putting better tracability on your videos so we can know which one is the next one
Great Video! Thank you!
thank you mr Kahn. you are doing Gods work my good sir
You save my life , thank you so much .
Thank you.
Thank you so much! :D This was so helpful!
Who else is watching this 10 mins before their exam?!?!!?
Welcome to the club😂
Still you had time to type here.. 😁😁
Thank you much! That was really helpful! :)
thank you so much, your lecture is sooooooo great!
When you rref the column space (or span), don't you have to like turn the columns into rows first and THEN go ahead and do the substractions? (And then turn the rows back to columns)
Perfect!! Thanks so much.
thanks man
To calculate Col A, do u use echolon form or reduced echolon form, watching some videos on it and every video has different method. Some are using reduced and some echolon only. So im getting confused which one is supposed be used?
Good job !
Thanx I find it helpful :-)
Sir, you are awesome!
very helpful!
THANKKKKKKKKKKKK YOU SO MUCH . PLEASE PUT MORE AND MORE VIDEOS. EXPLAIN A SLOWER AND AT THE END OF THE VIDEO SAY MISPLAY WHAT YOU DID . JUST LIKE WHAT YOU ASIDE AT THE END OF THIS ONE. MORE AND MORE VIDEOS PLEASE
Thank you so so much!! xx
what is the difference between the dim. of subspace and col. space ????
Does span of column space define the space of vector space?
thank you !!!!!
Dimension (Number of linearly independent vectors) of Column Space: Rank
Dimension (Number of linearly independent vectors) of Null Space: Nullity
And Rank + Nullity = Number of Columns in the matrix.
So can we directly calculate the rank of the matrix to find the dimension and the linear independent vectors ?
yup
THANK YOUU!!!!!!!
so the dimension of the null space = # of pivot variables of the original matrix and the dimension of the column space = # of non pivot variables?
in this lecture you said dimension =number of pivot columns in c(A) however in the previous video called dimension of null space and nullity you said dimension= number of non pivot columns in N(B) so which one is it? # of pivot or # non pivot columns? or does that change based on if its a null space V.S a column space? and what is the difference between the two?Thanks.
thx
6:50 dont you mean column 4?
no, the pivot column has to have only zeros and one 1
yeah he meant 4, you know its true
Hey Sal, thank you so much for all your videos. So, pivot columns are always linearly independent right?
Yes, the set of pivot columns will be linearly independent.
Think about it this way: what makes a set of vectors linearly independent? A set of vectors, A, is LI iff the only solution to Ax = 0 is the trivial solution (i.e. x is the zero vector). So, no linear combinations of vectors in A will equal 0 except for the case 0x1 + 0x2 + ... + 0xn = 0.
By definition, a pivot column has a leading value and all entries underneath it are zero.
So in the matrix:
1 0
0 1
No matter what nonzero scalar c1, c2 we multiply each of the vectors by, we will never be able to get a nonzero value when we add the vector components together.
ur perfect
thanx u r just amazing
@g1tfisted My thoughts exactly. My linear algebra teacher gave us all a take-home test that we have to turn in in two days, and no one really understood his explanation of column space and nullspace. Thank you so much for your help, this will surely help me answer some questions on my test. :P
This guy sounds like WoodysGamertag. Great vid.
WTF i was literally thinking about that yesterday and I haven't watched woodysgamertag in years
How to figure out which the pivot column
THAAANKK YOUUUU !!!!
are dimension and rank basically the same thing?
Hello Khan academy, you have left me extremely confused since by the theorem of linear independence: If a set contains more vectors than there are entries in each vector than the set is linearly dependent. I must have misunderstood you so in other words: are you saying that the subset of A(a1, a2 , a4) is independent or that A (a1,a2,a3,a4,a5) is independent?
ya I too have this doubt
what if two pivot columns are the same in reduced row echelon form? Is the rank then 2 if you have 3 pivot columns as two are linearly dependent?
you are awesome
i love you khan
Omg i love you!
No, it doesn't matter. There is no "one right way" to rref a matrix, you can do whatever you want.
so i can say, dim(C(A)) is equal with rank(A) ?
may I know which text book u follow please
why is the 3rd column not a pivot entry
two people teach linear algebra. Sal and the guy who disliked this video.lol
so .... rank is nullity?
No! nullity is the dimension of the null space while rank is the dimension of the column space
Column space is column rank?
thanks! and I wished my bald , boring professor take some notes :p
Hm. Well this probably isn't helpful to you now, but I -think- you literally just take the ROW that the pivot point is in, rather than the column.
I still don't understand what a column span is
ah I should've skipped right to the end but the rest of the video wasn't too bad to watch either.
I feel your pain!
1 person needs to work on their clicking accuracy
@hilaryeeee I Pay $4500 as an international student........ But it's not better than this............
Why does he sound like Tom Hanks, lol?
lol same apart from I have to pay even more :(
@lebplayer004 khanacademy [dot] org
I really think you're on to something...if our fucked up govt. ever decides to relieve the poor of their education (koch brothers have tried) a system like the one you have started would definitely even the playing field.
i pay $4500 and my prof still sucks like never before!
i love u
*no homo*
:)
Thank you so much for this!
Thank you so much!
thank you!
lol same apart from I have to pay even more :(
thanku so much!!!!