I dub this lecture as "Everything your parents didn't tell you about Linear Independence and you were afraid to ask" Thanks Professor Strang, you are a wonderful human being
I feel so lucky to be able to pause, rewind, and re-watch. I might have struggled trying to take notes while letting the incredible insights sink in. Are we getting a better education online than even the students in this very course?!
I could not help smiling when he just so smoothly introduced the concept of Basis Vectors. I was never able to understand it and now it seems that it was always so easy. The pauses he takes during explanations really builds you up and makes you think. Do not fast forward. If he is speaking slow. There is a reason. I am sure he is experienced enough to technically garble all the information and leave. But he does not. Please respect that and you will actually start learning.
There are nonzero solutions to Ax = 0 if m < n 1:17 Definition of linear independence of vectors 4:36 Examples of linear independence and dependence of vectors 7:05 Linear independence of columns of a matrix 13:18 Vectors that span a (vector) space 18:05 Definition of basis of a vector space 21:26 Examples of basis of vector spaces 24:12 (Error referenced in next lecture 27:43) Definition of dimension of a vector space 33:17 Examples of dimension of a vector space 37:44
@joaopp3 Yes, some lecturers become so used to their courses that they start to forget what it is that is considered difficult by new students, and stupidly assume that "everyone" "just knows" what all math symbols and proofs mean. In my opinion, the best way to teach a new concept is to give a quick summary of what it is going to be about and then give a lot of examples, and encourage the students to help solving them. That's exactly what professor Strang always does, and that's one of the reasons why he is so amazing. I also think that it is very important that a lecturer shows a genuine fascination for the subject, and is always willing to help the students and respects that some of them may have a harder time than others with certain parts; and it is also a huge plus if the lecturer has a sense of humour and presents the lecturers in a relaxed way. And Strang seems to have all of these qualities as well.
This lecture is amazingly put together. It feels like I am watching a well written movie, where all the pieces seem to fall into place. All this is thanks to the professor.
This professor has saved me a bunch of struggle. I really appreciate it. I don't normally write on UA-cam videos but this is an exception. Great stuff, great professor. Can't wait to watch more.
His teaching style seems casual and intuitive. I go to a small public college and the course is much more formal and proof driven. These lectures are a great addition to (as well as a nice break from) formal proofs. Thanks MIT!
It is so satisfying when you struggle with linear algebra for a whole semester and then you watch these lectures and it all makes click. I bet it would be a bit harder to understand if this was the first time I heard about all these concepts because it's a lot to remember, but Gilbert Strang can explain why the rules we read in the textbooks are true very intuitively. Before I was able to solve exercises, now I am able to understand them.
Thanks for pointing this out. Also, when vectors (1, 1, 2), (2, 2, 5) and (3, 3, 8) are written as column vectors and placed in a 3x3 matrix A, A’s first two rows equal (1, 2, 3). Hence the rank of A is 2 and its nullity is 3 - 2 = 1 > 0. Thus A’s columns are linearly dependent.
This is just a perfect lecture, pure perfection! Can't thank you enough for explaining these concepts so clearly Professor Strang! I can listen to Prof Strang's lecture all day everyday!
Due to the pandemic, I have to take LA at a local community college online. And taking a course online meaning you are on your own, learning by only reading notes and textbook. Before Prof. Strang's lectures, I was struggling with the class. Now I understand the subject much better. Thank you, professor!
This is so good. Watching these in order and following it closely makes linear algebra seem like the easiest thing ever AND like magic at the same time.
Thanks so much for the lecture. Professor Strang is a teacher that devote his heart to educate people , thanks to him I have been progressing in the field of Linear Algebra.
37:19 “Independence: that looks at combinations not being zero, Spanning: that looks at all the combinations, Basis: the one that combines independence and spanning, Dimension of a space: number of vectors in any basis ( cause all bases have the same number)“
@28:35, the set of vectors given are not a basis. They don't span R^3 since the nullspace corresponding to those vectors have a common form: [c,-2c,c] which is not only the zero vector. But however, the intended idea is clear from whatever he has explained :)
On 45:55 he says that any two linearly independent vectors would span the C(A). But these two linearly independent vectors should also be in C(A) he forgot to mention that i think. Because (1,0,0) and (0,1,0) are linearly independent but they do not span C(A). I listened carefully that part 3 times before writing this comment but maybe I missed it.
"Any space with those vector in it must have all the combination of those vector in it. And if I stop there then I have got the smallest space and that is the space they span". Got a feel. ☺️ Amazing Person.
Thank you MIT and Professor Strang for those fascinating lectures. In every lessons we reuse the things we have learned before and this makes a super connection between the topics. This lecture was a great example of that. Thank you again
this guy is awesome. i'm watching his classes now because i'm taking linear algebra 2, and my professor is chinese(i live in brazil) and is using strang's book to give a quit revision about linear algebra 1, so i came here to watch his playlist, and man how i wish i had watched this during the pandemic in 2020, just like i did with david jerison's calculus playlist
Basis is spanning birthdays and surnames. For first letter and last four of social. Basis for span is the Incubation period of quail, baby, or donkey . Half raw vegetable rice, twice yearly crop years.
my thanks to MIT for providing not only a proffesor whose interested in teaching.... but good at it.. and .. providing it for free. Some day.. i hope to attend thanks
My favorite example of two bases for the same space: in the old days your showers had two knobs that let you adjust cold and warm water (per time). Modern showers let you adjust volume per time (~= warm + cold) and temperature (~= warm - cold). For every configuration of temperature and volume per time, there is some setting of cold and warm that achieves that configuration, but it might be tricky to find. [Probably in practice this system is not linear. But it feels about right.]
independence: a bunch of vectors are independent if there is no such combination among them that gives zero (other than all-zero combination). reminder: pivots represent independent columns while free columns mean there is dependence. a bunch of vectors C in a matrix A are independent when the nullspace of Ac =0 happens only when C is all zeros, a matrix of independent columns will have a rank = n (number of columns). span: vectors v1...vl span a space when that space consists of combinations of those vectors. basis of a space is a sequence of vectors v1...vd that are 1. independent 2. span that space basis test: n vectors of subspace Rn give basis if the matrix made from those vectors is invertible. given a space (such as Rn) then every basis for that space has same number of vectors (n vectors), that is what we call "Dimension". so for a space(n) we have a matrix A where rank(A) = no. of pivot columns = dimension of space n
guys when talking about matrices as LINEAR COMBINATION of some vectors, discussing linear independence dependence etc., NULL SPACE is a misused term. null space is only defined in the context of LINEAR TRANSFORMATION. both contexts use matrices in the same way for computation so there won't be a technical problem, rather, a conceptual one.
A lot equivalent propositions here, prof strang beautifully demonstrated that they are all connected with the basic eliminations process, it helps a lot.
To come at this from the other direction, it is easy to see that the vectors in the second "basis" also fail to _span_ R^3 (the other necessary condition to form a basis). Any linear combination of the three vectors will be of the form (x,x,y), so the vector (1,2,3), for example, is not in the span of the set of vectors.
Note for myself : By basis for column space, we mean LI vectors which span the whole "column" space. Basis need not always be those [1 0 0], [ 0 1 0] type vectors. Basis is defined wrt a space as follows : It is the set of LI vectors which span that space.
Intellectually stimulating and after just one drink you really hear the humor of this guy ... small example 44.10 ... "I guess I am giving you infinitely many possibilities so I can't expect a unanimous answer" ... LOL
Column space is span of pivot columns and dim(Colspace) (rank)= number of pivot cols BUT ! Even tough dim(nullspace) = number of free variables, nullspace is not span of free variable columns. Have I just get it right?
It is great that we could press the pause button to think, those "fortunate" students didn't have this luxury, hehehe... Yeah, they could still do that after lecture, of course... Internet is great and MIT is so generous to put all these lectures up for free viewing... Thank you very much
Love his lectures but I think there's a little mistake in there, correct me if I'm wrong. Professor Strang said {(1,1,2),(2,2,5),(3,3,8)} makes a basis for R^3, but can't that set of vectors only produce vectors where x1 and x2 are equal? So you cant get (1,0,0) from the vectors and certainly not the entirety of R^3, right? The first and second row would even be the exact same written as a Matrix, so they should be linearly dependent for that reason as well. Again, maybe I'm overlooking something but it doesn't add up for me. Edit: Should have sorted comments by new first, it seems a lot of people saw this too but it didn't make it to the top comments
same bug in the exmple as in pres 7, column 3 and 4 are combinations of 1 and 2, therefore, you have 2 independent 3d vectors, ie. they do not span the space. class in coma.
Professor Strang said that the span of any vectors is the smallest subspace, I didnt quite get that. Would appreciate if anyone helps me on that regard, but nonetheless, this is a gem of a lecture, Thanks professor Strang.
43:41 Columns 1 and 3 don't form a basis, though? They span the subspace but are not independent? Column 1 can be found in column 3? Wait, but no combination of 1 and 3 will ever cancel each other out, so their nullspace is only the zero vector. I would have made a major mistake if I didn't catch that, damn.
But the 3rd vector 3,3,8 wouldn't make it a basis, because all vectors in the span would have the property that the first and second coordinate are equal and thus they can't span R^3
basis of R^n spans the space of R^n, but at 43:46, the column space is R^3 and we had only 2 linearly independent vectors so how can it create a basis? because with only 2 vectors we cant span R^3.
Here we have 2 independent columns which are enough to span the 'column space' and they are not enough to span whole R^3. So rank of A is 2.and dimensionality of null space is 2 which is n - rank(A)
at 45:25 prof is talking about the other basis vectors for the column space he says finally that since the dim of the column space is 2, he says any 2 vectors that are independent can form the column space but is that possible? i mean the column space is a plane in 4 dimensional space right? so if i pull out some two random independent vectors that are both out of the original column space, then their linear combinations will span an entirely new plane right? math genius plz help this poor soul...
To begin with, that is actually a plane in 3-dimensional space because the matrix contains 3 rows, and its 4 columns means it has 4 vectors. What he indicated are not arbitrary 2 independent vectors but the ones from the original columns in the matrix, or at least the ones linearly combined by them, such as (1, 1, 1) & (2, 1, 2), (1, 1, 1) & (3, 2, 3), (2, 2, 2) & (3, 2, 3), (1, 1, 1) & (7, 5, 7), and so on.
You are doing so many people a favor by posting this. Seriously.
Mam ,Can I get your Linkedin profile .
@@hariompandey4364 why, wanna message her, creepy ahh Indian?
Just blessing for students who couldn't go in mit
I dub this lecture as "Everything your parents didn't tell you about Linear Independence and you were afraid to ask"
Thanks Professor Strang, you are a wonderful human being
😂😂😂
You had mathematicians as parents?
I feel so lucky to be able to pause, rewind, and re-watch. I might have struggled trying to take notes while letting the incredible insights sink in. Are we getting a better education online than even the students in this very course?!
yes definitely, I have problems in live lectures. I understand much better in video lectures
agree
2020 read this comment
No, they were provided the videotapes for these lectures.
maybe you are right about one side but you cannot ask to teacher whenever you want .
@ 7:40 - "Let me take an example where I have a vector and twice that vector...If the word dependent means anything, these should be dependent"
Gavino Felix ko
😝
ma boi gilbert saving my butt once again
How is you butt now? Is it safe?
I hope it is...
still stands true after six years.
@@lokidoki123 yesss
I could not help smiling when he just so smoothly introduced the concept of Basis Vectors. I was never able to understand it and now it seems that it was always so easy. The pauses he takes during explanations really builds you up and makes you think. Do not fast forward. If he is speaking slow. There is a reason. I am sure he is experienced enough to technically garble all the information and leave. But he does not. Please respect that and you will actually start learning.
Exactly
all what I am thinking is like if he is an instructor, who is mine?
There are nonzero solutions to Ax = 0 if m < n 1:17
Definition of linear independence of vectors 4:36
Examples of linear independence and dependence of vectors 7:05
Linear independence of columns of a matrix 13:18
Vectors that span a (vector) space 18:05
Definition of basis of a vector space 21:26
Examples of basis of vector spaces 24:12
(Error referenced in next lecture 27:43)
Definition of dimension of a vector space 33:17
Examples of dimension of a vector space 37:44
This is just amazing, how much you appreciate this lecture largely depends upon how much you paused during the previous ones.
29:35 The vectors are not independent. 2*[2,2,5]-[1,1,2] = [3,3,8].
Yes this was really making me second-guess myself. Thanks!
Thanks man
@joaopp3
Yes, some lecturers become so used to their courses that they start to forget what it is that is considered difficult by new students, and stupidly assume that "everyone" "just knows" what all math symbols and proofs mean.
In my opinion, the best way to teach a new concept is to give a quick summary of what it is going to be about and then give a lot of examples, and encourage the students to help solving them.
That's exactly what professor Strang always does, and that's one of the reasons why he is so amazing.
I also think that it is very important that a lecturer shows a genuine fascination for the subject, and is always willing to help the students and respects that some of them may have a harder time than others with certain parts;
and it is also a huge plus if the lecturer has a sense of humour and presents the lecturers in a relaxed way.
And Strang seems to have all of these qualities as well.
At first I thought you were a hater....
He does the lecture off the top of his head, he's great
This lecture is amazingly put together. It feels like I am watching a well written movie, where all the pieces seem to fall into place. All this is thanks to the professor.
This professor has saved me a bunch of struggle. I really appreciate it. I don't normally write on UA-cam videos but this is an exception. Great stuff, great professor. Can't wait to watch more.
His teaching style seems casual and intuitive. I go to a small public college and the course is much more formal and proof driven. These lectures are a great addition to (as well as a nice break from) formal proofs. Thanks MIT!
It is so satisfying when you struggle with linear algebra for a whole semester and then you watch these lectures and it all makes click. I bet it would be a bit harder to understand if this was the first time I heard about all these concepts because it's a lot to remember, but Gilbert Strang can explain why the rules we read in the textbooks are true very intuitively. Before I was able to solve exercises, now I am able to understand them.
This has been one my favorite lectures so far, for me really brought these concepts together beautifully for the first time.
Thanks for pointing this out.
Also, when vectors (1, 1, 2), (2, 2, 5) and (3, 3, 8) are written as column vectors and placed in a 3x3 matrix A, A’s first two rows equal (1, 2, 3).
Hence the rank of A is 2 and its nullity is 3 - 2 = 1 > 0.
Thus A’s columns are linearly dependent.
Completed 8 videos, watching 9th, 26 more to go!
This is just a perfect lecture, pure perfection!
Can't thank you enough for explaining these concepts so clearly Professor Strang!
I can listen to Prof Strang's lecture all day everyday!
It's amazing that we live in a time where we can watch such high quality educational content right in our homes.
Due to the pandemic, I have to take LA at a local community college online. And taking a course online meaning you are on your own, learning by only reading notes and textbook. Before Prof. Strang's lectures, I was struggling with the class. Now I understand the subject much better. Thank you, professor!
This is so good.
Watching these in order and following it closely makes linear algebra seem like the easiest thing ever AND like magic at the same time.
No other video about LA is better than this! respect and tks prof so much :)
+Roku Vo But he never talked about Los Angeles in the whole video? =/
LA is linear algebra not Los Angeles hahaha :v
lol
Best video I have watched so far, I was with him all the way and my concentration never dipped.
Thanks so much for the lecture. Professor Strang is a teacher that devote his heart to educate people , thanks to him I have been progressing in the field of Linear Algebra.
Gilbert Strang,
You are an awsome teacher! Congratulations for the great work done and thanks for the opportunity!
Congratulations to MIT as well.
The best linear algebra teacher I`ve ever seen (and read, his book is great)
I'm literally in awe of what I can understand thanks to Dr. Strang
37:19
“Independence: that looks at combinations not being zero,
Spanning: that looks at all the combinations,
Basis: the one that combines independence and spanning,
Dimension of a space: number of vectors in any basis ( cause all bases have the same number)“
@28:35, the set of vectors given are not a basis. They don't span R^3 since the nullspace corresponding to those vectors have a common form: [c,-2c,c] which is not only the zero vector. But however, the intended idea is clear from whatever he has explained :)
I'd just like to say thank you to Prof. Strang. The best prof I never had.
woah this guy just singlehandedly saved this semester for me. and he does it so nonchalantly. i'm stunned
Thank you so much for making these lectures public and not gatekeeping knowledge.
On 45:55 he says that any two linearly independent vectors would span the C(A). But these two linearly independent vectors should also be in C(A) he forgot to mention that i think. Because (1,0,0) and (0,1,0) are linearly independent but they do not span C(A). I listened carefully that part 3 times before writing this comment but maybe I missed it.
exactly man, i had this question in my mind... hi-fi
He says “and they span” at 46:01 but it is hard to hear
I was a bit confused after the last lecture but now it's clear why reduced row echelon form and rank are so important thanks professor
Very useful course! You can learn Linear Algebra yourself! Let knowledge spread to everyone, free and efficiently.
I cannot help but write something laudatory after watching every video.
"Any space with those vector in it must have all the combination of those vector in it. And if I stop there then I have got the smallest space and that is the space they span". Got a feel. ☺️ Amazing Person.
No words for sir strang ,he is my favorite teacher
Thank you MIT and Professor Strang for those fascinating lectures. In every lessons we reuse the things we have learned before and this makes a super connection between the topics. This lecture was a great example of that. Thank you again
one of the best teacher in the world thank you professor Gilbert Strang god bless u
this guy is awesome. i'm watching his classes now because i'm taking linear algebra 2, and my professor is chinese(i live in brazil) and is using strang's book to give a quit revision about linear algebra 1, so i came here to watch his playlist, and man how i wish i had watched this during the pandemic in 2020, just like i did with david jerison's calculus playlist
This video is amazing! AMAZING! All the principles fit reality so nicely! LONG LIVE MATRICES.
Basis is spanning birthdays and surnames. For first letter and last four of social.
Basis for span is the Incubation period of quail, baby, or donkey .
Half raw vegetable rice, twice yearly crop years.
my thanks to MIT for providing not only a proffesor whose interested in teaching.... but good at it.. and .. providing it for free. Some day.. i hope to attend
thanks
My favorite example of two bases for the same space: in the old days your showers had two knobs that let you adjust cold and warm water (per time). Modern showers let you adjust volume per time (~= warm + cold) and temperature (~= warm - cold). For every configuration of temperature and volume per time, there is some setting of cold and warm that achieves that configuration, but it might be tricky to find.
[Probably in practice this system is not linear. But it feels about right.]
It is an interesting idea.
It is an interesting idea.
It is an interesting idea.
Nicely done, prof! Some people are just good at Linear Algebra, some can make others be good at Linear Algebra!
independence: a bunch of vectors are independent if there is no such combination among them that gives zero (other than all-zero combination).
reminder: pivots represent independent columns while free columns mean there is dependence.
a bunch of vectors C in a matrix A are independent when the nullspace of Ac =0 happens only when C is all zeros, a matrix of independent columns will have a rank = n (number of columns).
span: vectors v1...vl span a space when that space consists of combinations of those vectors.
basis of a space is a sequence of vectors v1...vd that are 1. independent 2. span that space
basis test:
n vectors of subspace Rn give basis if the matrix made from those vectors is invertible.
given a space (such as Rn) then every basis for that space has same number of vectors (n vectors), that is what we call "Dimension".
so for a space(n) we have a matrix A where rank(A) = no. of pivot columns = dimension of space n
DR. Strang thank you and MIT for another great lecture on classics linear algebra topics.
guys when talking about matrices as LINEAR COMBINATION of some vectors, discussing linear independence dependence etc., NULL SPACE is a misused term. null space is only defined in the context of LINEAR TRANSFORMATION. both contexts use matrices in the same way for computation so there won't be a technical problem, rather, a conceptual one.
28:43 A small mistake here. These three vectors are not the basis.
(-v1 + 2v2 - v3=0)
True. You can also look at the rows since the rank is the same for rows and columns. [1, 2, 3] = [1, 2, 3]. Everybody goofs sometimes.
The last 30 seconds were the biggest mic drop, ever. I just stopped ... and stared in awe. Wow!
Gilbert Strang greatest mathematics professor in 21st century.... Thank you a lot.
he is a good but I disagree with being the greatest, I have seen other profs who come prepared to the lecture, you watch them like you watch Ted talk
A lot equivalent propositions here, prof strang beautifully demonstrated that they are all connected with the basic eliminations process, it helps a lot.
I actually consider Professor Gilbert Strang to be one of my heroes 🏆❤️🙏🏽🎊
To come at this from the other direction, it is easy to see that the vectors in the second "basis" also fail to _span_ R^3 (the other necessary condition to form a basis). Any linear combination of the three vectors will be of the form (x,x,y), so the vector (1,2,3), for example, is not in the span of the set of vectors.
These lectures are actually fun! The book is also great. Thanks Gilbert.
I guess the girl at 50:06 going towards the blackboard is the one who finds the error which Professor talks about in the next lecture.
Gillbert strang is a legend! He explained everything we might not ask! Or think of
Note for myself : By basis for column space, we mean LI vectors which span the whole "column" space. Basis need not always be those [1 0 0], [ 0 1 0] type vectors.
Basis is defined wrt a space as follows : It is the set of LI vectors which span that space.
I can't expect enormous answers. Like it a lot.
this guy is amazing
I NEED SOME SPACE AFTER THIS!!!
This man is a literal God at teaching linear can he please teach my lecturer how to teach.
Your videos have been saviour for me. Thanks a ton dear prof ❤️
yeah
just amazing! I feel a little bit sad because I have no chance to listen this course exactly from him :)
hadi abi hadi şov yapma, direk ondan dinliyosun zaten
You are attending his course for free, be happy
16:39 A is dependent when NA gets some vectors,independent when NA gets one solution,zero.
Lectures with my prof feel like fistfighting a grizzly bear, lessons with gilbert feel like being sung a lullaby by celine dion
Nice lecture sir.Love from India🇮🇳🇮🇳
What a brilliant man.. I wish I had a teacher like that in all my courses!
Thank you, MIT! And thank you for enabling comments! Free speech kicks ass!
Intellectually stimulating and after just one drink you really hear the humor of this guy ... small example 44.10 ... "I guess I am giving you infinitely many possibilities so I can't expect a unanimous answer" ... LOL
The best linear Algebra teacher!!
If you guys find the determinant of the matrix, it shows Det = 1*(16-15) - 2*(8-6) + 3*(5-4) = 0. So the matrix is not linearly independent.
37:37 independence span dimension; same number of
39:04 vector that is in a null space
Column space is span of pivot columns and dim(Colspace) (rank)= number of pivot cols
BUT !
Even tough dim(nullspace) = number of free variables, nullspace is not span of free variable columns. Have I just get it right?
It is great that we could press the pause button to think, those "fortunate" students didn't have this luxury, hehehe... Yeah, they could still do that after lecture, of course... Internet is great and MIT is so generous to put all these lectures up for free viewing... Thank you very much
Independence, Basis, and Dimension. Hmmm, sounds like a Robert Nozick title. Thank you Dr. Strang !
The second basis for R^3 is not a basis, since -(1,1,2)+2*(2,2,5)=(3,3,8); i.e., the vectors are not linearly independent.
Thank you
I don't why I am in Love with this Guy.
"if one of those is the zero vector, *INDEPENDENCE IS DEAD* "
not the context i expected those dramatic words in
gilbert strang makes linear algebra fucking simple
So glad you put these up
I am grateful for seeing this video
He is so slowly teaching, but is a wonderfull teacher.
i thought i've been taught linear algebra till i met this guy.
Love his lectures but I think there's a little mistake in there, correct me if I'm wrong. Professor Strang said {(1,1,2),(2,2,5),(3,3,8)} makes a basis for R^3, but can't that set of vectors only produce vectors where x1 and x2 are equal? So you cant get (1,0,0) from the vectors and certainly not the entirety of R^3, right? The first and second row would even be the exact same written as a Matrix, so they should be linearly dependent for that reason as well. Again, maybe I'm overlooking something but it doesn't add up for me.
Edit: Should have sorted comments by new first, it seems a lot of people saw this too but it didn't make it to the top comments
But is still good that you say that now so we don´t get too confused. Thank you.
And Thank you Gilbert Strang for this work!!!
18:13 to review the meaning of span
same bug in the exmple as in pres 7, column 3 and 4 are combinations of 1 and 2, therefore, you have 2 independent 3d vectors, ie. they do not span the space. class in coma.
Professor Strang said that the span of any vectors is the smallest subspace, I didnt quite get that. Would appreciate if anyone helps me on that regard, but nonetheless, this is a gem of a lecture, Thanks professor Strang.
43:41 Columns 1 and 3 don't form a basis, though? They span the subspace but are not independent? Column 1 can be found in column 3?
Wait, but no combination of 1 and 3 will ever cancel each other out, so their nullspace is only the zero vector. I would have made a major mistake if I didn't catch that, damn.
But the 3rd vector 3,3,8 wouldn't make it a basis, because all vectors in the span would have the property that the first and second coordinate are equal and thus they can't span R^3
basis of R^n spans the space of R^n, but at 43:46, the column space is R^3 and we had only 2 linearly independent vectors so how can it create a basis? because with only 2 vectors we cant span R^3.
Here we have 2 independent columns which are enough to span the 'column space' and they are not enough to span whole R^3. So rank of A is 2.and dimensionality of null space is 2 which is n - rank(A)
This man has infinite chalkboards
This professor is amazing
at 45:25 prof is talking about the other basis vectors for the column space
he says finally that since the dim of the column space is 2, he says any 2 vectors that are independent can form the column space
but is that possible? i mean the column space is a plane in 4 dimensional space right?
so if i pull out some two random independent vectors that are both out of the original column space, then their linear combinations will span an entirely new plane right?
math genius plz help this poor soul...
To begin with, that is actually a plane in 3-dimensional space because the matrix contains 3 rows, and its 4 columns means it has 4 vectors.
What he indicated are not arbitrary 2 independent vectors but the ones from the original columns in the matrix, or at least the ones linearly combined by them, such as (1, 1, 1) & (2, 1, 2), (1, 1, 1) & (3, 2, 3), (2, 2, 2) & (3, 2, 3), (1, 1, 1) & (7, 5, 7), and so on.
how he know pivot columns before doing elimination?don't we need elimination to find pivot variables 42:20
Im not sure what is more impressive. What his talking about or that there is a professor strang. Yours sincerely matt strang