You're absolutely right. I was initially confused because I thought of an example where his definition didn't work. For example, if a=1i, b=-1i, c=1j. These three vectors are clearly independent. Also, 1a+1b+0c = 0. By the definition he gave, these vectors should be dependent. Thus, his definition doesn't work as stated. After I saw him work some problems, I realized that his definition was intended for vectors with the same number and type of components. I posted a correction stating that.
The first example (6:55)is a system of 2 equations with 2 unknowns, therefore there is one set of answers only, (0.0) The second example(11:16), however, is a system of 2 equations with 3 unknowns, the answer of c1=0, c2=0, c3=0 satisfy the equation as well. The rule like Sal said is that if all Cs are 0, then the system is independent. But the second example is concluded as linearly dependent. How do we then go about determining independence and dependence? I maybe missing something here; if anyone can clarify, or add to any holes in knowledge I may have, thank you!
the rule said that if AT LEAST one arbitrary constant is non-zero and the result of the vector combination is 0 then it's linearly dependent. You can only say that the system is independent if the ONLY solution for the arbitrary constants is 0. I hope this helps.
Thanks for the suggestion- just checked 3Blue1Brown out and it's really good. Any other suggestions- I know it's 2 year old comment...thought I would try.
This is amazing. I am doing this just now and i have a textbook that i paid £50 for, and i still couldn't understand what it means! Thank you so much. Keep making videos, i will probably be back here as my course gets harder :P
What if one of the vectors is just zeroes? c1 * [1, 2] + c2 * [2,3] + c3 * [0, 0] = [0, 0] - linearly dependent because [0, 0] can be represented by 0*[1, 2] but we can easily prove it's independence by having c1 = 0, c2 = 0, c3 = arbitrary number.
Wait now, if 3 vectors in a set and two of them are linearly dependent, co linear, then the third cannot be written as a linear combination of the other two? But you said that even if two were linearly dependent or nonlinear, then it is a linearly dependent set
You know what, I have difficulty listening to my professor's explanation. And then one day my professor actually played Khan's video in the lecture, and I was like, maybe I could just watch Khan's video instead of attending the lecture LOL
visually, the 2-dimensional vectors are linearly dependent if they are collinear and independent if not. While the 3-dimensional vectors are linearly dependent if they are coplanar and independent if not. Did I get that right?
If we represent colours as a vector integer 0-255 with r, g, b being an R3 vector then would combinations of co-linear vector colours be more pleasing to the eye than those that aren't? Hmm. I'm too colourblind to check. Would be extreme easy to do. Although reasoning this out, given that black (0, 0, 0) "goes with everything" then it follows that my hypothesis is unsound given than it can't multiply. Although this is the div/0 problem. Is it true to say that, in an abstract sense, zero fits infinitely? That's above my mathematical paygrade.
I’m stuck at why equaling a zero matrix is proof of being linearly dependent. Is this because it brings us back to an origin of 0,0 - back to where the vector started?
I don't understand why you put c3 at 15:12 as -1 and solve it. If you put c3 as 0, both c2 and c1 would be linearly independent, so I wonder if you put it as -1 and it comes out linearly dependent.
In my understanding, there is also a theorem that states If you have more vectors (i.e. columns) than you do rows, then you can already determine that the set is linearly dependent.
"Theorem 8: If a set contains more vectors than there are entries in each vector, then the set is linearly dependent. That is, any set fv1; : : : ; vpg in Rn is linearly dependent if p > n." From Linear Algebra and it's Applications
Hi! Why or how did you end up multiplying everything by 2 at min 14:40? I cannot get it :/ Thanks a ton for all your videos, I can really connect and understand from your lectures. Much respect!
why (2, 1) and (3,2) vectors are linearly independent? The vectors we get by x=2 and y=1 dot and x=3 and y=2 are on the same lane. You said that vectors are dependant if they are on the same line (collinear)
13:48 what if we put c3=0 ? If we solve these equations after putting c3=0 then c1 and c2 would also be zero, which means if would be linearly independent.
Linear independence means that this is only true if c1,c2, and c3 are all equal to 0, which means that c1=0 c2=0 c3=0 is the only solution to this equation. Since there are other non-zero solutions to this equation, it's linearly dependent. C3 is equal to minus 1 in video just to show that it has other solutions. I'm not a native English speaker, hoping you can understand.
I don't really understand why if you draw the three vectors into an ex-coordinate you can clearly see that they do not have the same slope, which should mean that they are linearly independent, right? Could someone explain that to me? By the way, thanks so much for these videos, they are fantastic!
It's more like can a combination of two vectors equal the third one. You only need two vectors to not have the same slope to get any vector in R2, which is how we know a third vector is redundant. It's difficult to find a combination for this example, but there was an example he used in the previous video.
yes, that's true, but you can't just draw them, you must actually prove the slopes are not equal. Because they might have slopes that are so close together that they are hard to distinguish, but still be not equal. Also slope is really a concept in R^2 only. You can extend it to R^n, but it's then not a number but a vector or something, so you haven't gained much if you are trying to show that the vectors are linearly independent. Although it is quite a useful concept in multivariate and vector calculus.
Not exactly. A slight modification - A set of 3+ vectors 'in the same plane' are always going to be dependent. Check this for more - thejuniverse.org/PUBLIC/LinearAlgebra/LOLA/indep/examples.html
Actually, it's your deduction that's fallacious. In your example, the set {a, b, c} is linearly dependent. But you can't deduce from that that the set composed of any pair of vectors (like {a, b}) is also dependent.
Zero would be a poor choice, because he might solve it and find that the other two cs are zero, which wouldn't show linear dependence. So you're right to point that out as an exception.
Technical question: you say that, "any one of these vectors can be represented as some combination of the other ones." I think you mean, "at least one of the vectors". Saying "any one" would imply that it's guaranteed to work for all of them, which I believe is incorrect. Take for example a set of vectors where the zero vector is in the set. You can always represent the zero vector as a combination of the other vectors, and you can always give it a non-zero weight to make your sum add up to the zero vector (with zero weights on all the other vectors). So a set of vectors with the zero vector is linearly dependent, right? However, it isn't guaranteed that including the zero vector in a linear combination will allow you to form any other vector you'd like, so the same logic won't apply to creating linear combinations for the other vectors in the set. Let me know if I'm missing something, I'm pretty new to this subject. Thanks for the great video.
Adding the zero vector to any set of vectors makes it linearly dependent, yes. And adding the zero vector to a set of vectors never increases the Span of it because adding t*Ο is "useless". You are correct
Correction: b must also be a sum of vectors. Basically, if you are given a set of vectors that are linearly dependent that are an element of R^n, you could add an additional vector that is an element of R^(n+1) that cannot be linearly dependent (since one of its components is never present in the others) and multiply it by zero. The conditions of your definition are fulfilled, but the the additional vector is independent.
I think the naming is weird. If two vectors are needed to represent a real space, then shouldn't it be called Linearly dependent instead of independent? I mean, if one is needed, then the set is dependent upon that vector. Maybe I am not understanding the concept.
how about to choose c3 as 0 "zero" ?? . Why we have to choose c3 or another c as whole number except zero ? if I choose c3 as 0 the three vectors become independent
we want to find a solution to prove if they are dependent and its an easy way to pick c3=-1 or any other value you could also solve it with Gaussian elimination and then express c1 and c2 by c3 or any of them by the other tho your comments its quite old I just felt like to explain
Your proof the definition of linear dependency is fallacious. Using the same work, I can prove that any two vectors are linearly dependent. Given vectors a and b, let vector c=-b 0a + 1b + 1c = 0 Therefore, a,b, and c are linearly dependent. Therefore, a and b are linearly dependent. This result can be generalized by making c the sum of an arbitrary number of linear combinations of other vectors.
This guy could walk into any university and explain stuff better than at least 90% of professors on staff.
Many Thanks.
Weird way to say that he is at the top 10% of professors, but alright
Sir you are really Brilliant... excellent lecturing skills... I wish you were at my University... Keep up the good work.
ĹBBBŞČ MÅÝ F W Ù ÑĒĒĐ ĒŞP ČÒÙŘŞĒŞ Ñ ĶÀHVÈÌÑ ČÒFƏĔŞHÒP Ñ MZ.ŞHÌDĐÌĒQÌ ČHÈÇĶİȚ
thank you...ive been passing exams like a mad man thanks to you.
Oh man your amazing you saved my life I got exam soon and our teacher goes so fast its hard to understand him you break down so well your heaven sent.
I never got the concept so clear as this before. Many many thanks.
I watch this in class instead of listening to the professor
The professor plays this in class instead of teaching.
I watch this instead of my online classes
I understand the concept in 30 min while my professor took like 1 hour to teach us
can't believe this is all on youtube, thanks so much
You're absolutely right. I was initially confused because I thought of an example where his definition didn't work. For example, if a=1i, b=-1i, c=1j. These three vectors are clearly independent. Also, 1a+1b+0c = 0. By the definition he gave, these vectors should be dependent. Thus, his definition doesn't work as stated. After I saw him work some problems, I realized that his definition was intended for vectors with the same number and type of components. I posted a correction stating that.
"...dealing with linearly dependent network... oh... linearly dependent set..."))) this guy, for sure, knows more than he is speaking to us)))
So much better than both my lecturer and book. Love it!
I love this video,u r much more better than my 2 teachers,i couldnt undrestánd these in 3 lection but i undrestand perfectly in 17min here,thanks alot
The first example (6:55)is a system of 2 equations with 2 unknowns, therefore there is one set of answers only, (0.0)
The second example(11:16), however, is a system of 2 equations with 3 unknowns, the answer of c1=0, c2=0, c3=0 satisfy the equation as well. The rule like Sal said is that if all Cs are 0, then the system is independent. But the second example is concluded as linearly dependent. How do we then go about determining independence and dependence? I maybe missing something here; if anyone can clarify, or add to any holes in knowledge I may have, thank you!
the rule said that if AT LEAST one arbitrary constant is non-zero and the result of the vector combination is 0 then it's linearly dependent. You can only say that the system is independent if the ONLY solution for the arbitrary constants is 0. I hope this helps.
@@patrickcordero6673 Thank you!
Thanks for great explanation! Was struggling to understand why it must equal to zero.
Amazing. These along with the 3Blue1Brown videos on Linear Algebra and you don't need any books to understand vectors and LA.
Thanks for the suggestion- just checked 3Blue1Brown out and it's really good. Any other suggestions- I know it's 2 year old comment...thought I would try.
Thank you for adding other languages subtitle
My thanks spans R^n with linearly independent vectors because my grades on my quiz were all members of zero
thanks for explain.thats nice example....this video better than previous video
These videos make me thirsty for knowledge like never before. Thank you so much !!!!!!!!!!
i think am now ready for this linear algebra exams thanks a lot.... i mean it actually seems easier when u explain
I searched the whole internet but this is the video which clear my doubt , thanks sal sir
I was 8 years when this was uploaded am watching it at 20 yrs .......thank you for sharing your knowledge
This is amazing. I am doing this just now and i have a textbook that i paid £50 for, and i still couldn't understand what it means! Thank you so much. Keep making videos, i will probably be back here as my course gets harder :P
Sir, you are so great, thank you!
You are absolutely legendary!!! Thank you so much, I would never have understood this if it wasn't for you!!!!!
Thank you for making math simpler and even intriguing;)
What if one of the vectors is just zeroes? c1 * [1, 2] + c2 * [2,3] + c3 * [0, 0] = [0, 0] - linearly dependent because [0, 0] can be represented by 0*[1, 2] but we can easily prove it's independence by having c1 = 0, c2 = 0, c3 = arbitrary number.
He doesn't explain it properly for this but if c1=0, c2=0 then c3 MUST = 0. There can't be a situation where it is some arbitrary number
Thank you so much for doing this! So helpful!!!!
As you said, we can pick any c3, in your case -1. What if we take c3 as 0?
I am also having same doubt. if c3=0 then c1,c2 becomes zero then are they independent?
brilliant explanation, so easy to follow along
Brilliant as always Sal the great.
Wait now, if 3 vectors in a set and two of them are linearly dependent, co linear, then the third cannot be written as a linear combination of the other two? But you said that even if two were linearly dependent or nonlinear, then it is a linearly dependent set
Thank you so much Sir.
btw, do you provide assignments as well?
You know what, I have difficulty listening to my professor's explanation. And then one day my professor actually played Khan's video in the lecture, and I was like, maybe I could just watch Khan's video instead of attending the lecture LOL
visually, the 2-dimensional vectors are linearly dependent if they are collinear and independent if not. While the 3-dimensional vectors are linearly dependent if they are coplanar and independent if not. Did I get that right?
If we represent colours as a vector integer 0-255 with r, g, b being an R3 vector then would combinations of co-linear vector colours be more pleasing to the eye than those that aren't? Hmm. I'm too colourblind to check. Would be extreme easy to do.
Although reasoning this out, given that black (0, 0, 0) "goes with everything" then it follows that my hypothesis is unsound given than it can't multiply.
Although this is the div/0 problem. Is it true to say that, in an abstract sense, zero fits infinitely? That's above my mathematical paygrade.
I’m stuck at why equaling a zero matrix is proof of being linearly dependent. Is this because it brings us back to an origin of 0,0 - back to where the vector started?
The only way to add two different vectors and have it equate to the zero-vector is if they are colinear.
@@Decimated_By_A_Train thanks!
@@jeffreychavey4161 I was a little late but no problem
I don't understand why you put c3 at 15:12 as -1 and solve it. If you put c3 as 0, both c2 and c1 would be linearly independent, so I wonder if you put it as -1 and it comes out linearly dependent.
Thanks for the lectures ^^
i can be better prepared now for the exam
you are just amazing
In my understanding, there is also a theorem that states If you have more vectors (i.e. columns) than you do rows, then you can already determine that the set is linearly dependent.
"Theorem 8: If a set contains more vectors than there are entries in each vector, then the set
is linearly dependent. That is, any set fv1; : : : ; vpg in Rn is linearly dependent if
p > n."
From Linear Algebra and it's Applications
A great professor! Thank you!
Hi! Why or how did you end up multiplying everything by 2 at min 14:40? I cannot get it :/
Thanks a ton for all your videos, I can really connect and understand from your lectures. Much respect!
Gaussian_elimination. So C1 can be cancelled to find C2. A way of solving linear equations.
the speed should be increased x2.5 but a good explanation :)
@dimitridandeniya If the set does not contain enough constraints then it's automatically linearly dependent. He is just proving that with an example
i don't understand why you chose c3 as negative 1? Can you just pick a random number?
why (2, 1) and (3,2) vectors are linearly independent? The vectors we get by x=2 and y=1 dot and x=3 and y=2 are on the same lane. You said that vectors are dependant if they are on the same line (collinear)
They are not on the same line, just drawn that, try again : )
thnx alot this video is very helpful
Thank you sir
can someone explain me why I can not pick c3=0 at 13:44? Thank you!!
wait why would the span containing the zero vector mean its linearly dependent? Wouldn't all vectors spans contain the 0 vector?
thank you SO much
you saved my Linear algebra mark from reaching below a 50! :D
At 15:48 What made C1 + 4 = 0 equal C1 = -4? What made it negative?
Also again done at 15:20 where he changes positive 3 to negative 3 and back to positive 3
13:48 what if we put c3=0 ? If we solve these equations after putting c3=0 then c1 and c2 would also be zero, which means if would be linearly independent.
Linear independence means that this is only true if c1,c2, and c3 are all equal to 0, which means that c1=0 c2=0 c3=0 is the only solution to this equation. Since there are other non-zero solutions to this equation, it's linearly dependent. C3 is equal to minus 1 in video just to show that it has other solutions. I'm not a native English speaker, hoping you can understand.
Thank you so much!!
I don't really understand why if you draw the three vectors into an ex-coordinate you can clearly see that they do not have the same slope, which should mean that they are linearly independent, right? Could someone explain that to me? By the way, thanks so much for these videos, they are fantastic!
It's more like can a combination of two vectors equal the third one. You only need two vectors to not have the same slope to get any vector in R2, which is how we know a third vector is redundant. It's difficult to find a combination for this example, but there was an example he used in the previous video.
+NITEMARE1CHIMERA Thank you so much! I think I understand now. A set of three vectors is always going to be dependent. Thanks again.
yes, that's true, but you can't just draw them, you must actually prove the slopes are not equal. Because they might have slopes that are so close together that they are hard to distinguish, but still be not equal. Also slope is really a concept in R^2 only. You can extend it to R^n, but it's then not a number but a vector or something, so you haven't gained much if you are trying to show that the vectors are linearly independent. Although it is quite a useful concept in multivariate and vector calculus.
Not exactly. A slight modification - A set of 3+ vectors 'in the same plane' are always going to be dependent. Check this for more - thejuniverse.org/PUBLIC/LinearAlgebra/LOLA/indep/examples.html
Actually, it's your deduction that's fallacious. In your example, the set {a, b, c} is linearly dependent. But you can't deduce from that that the set composed of any pair of vectors (like {a, b}) is also dependent.
Thank you
ربنا يكرمك :)
Don't go too far.. It could be counted as shirk... He's just a human being with exceptional talent given by God. That's it
Man you're my Saviour Khan!!
I tried choosing c3 as 5 and 3. Did not work in both cases.
Isn't that 3eqns and 2 vars, not the other way round (13:39)
for the last example why did he pick c3 =-1? Could he pick either c1 or c2 to equal to positive number and still show its linearly dependent?
+twilights A , you can pick literally any real number for any of the three constants.
even a zero ?
Zero would be a poor choice, because he might solve it and find that the other two cs are zero, which wouldn't show linear dependence. So you're right to point that out as an exception.
Thank You Khan (y)
Technical question: you say that, "any one of these vectors can be represented as some combination of the other ones." I think you mean, "at least one of the vectors". Saying "any one" would imply that it's guaranteed to work for all of them, which I believe is incorrect. Take for example a set of vectors where the zero vector is in the set. You can always represent the zero vector as a combination of the other vectors, and you can always give it a non-zero weight to make your sum add up to the zero vector (with zero weights on all the other vectors). So a set of vectors with the zero vector is linearly dependent, right? However, it isn't guaranteed that including the zero vector in a linear combination will allow you to form any other vector you'd like, so the same logic won't apply to creating linear combinations for the other vectors in the set. Let me know if I'm missing something, I'm pretty new to this subject. Thanks for the great video.
Adding the zero vector to any set of vectors makes it linearly dependent, yes. And adding the zero vector to a set of vectors never increases the Span of it because adding t*Ο is "useless". You are correct
Just check at 2:12 , and pray to pass.
Correction: b must also be a sum of vectors. Basically, if you are given a set of vectors that are linearly dependent that are an element of R^n, you could add an additional vector that is an element of R^(n+1) that cannot be linearly dependent (since one of its components is never present in the others) and multiply it by zero. The conditions of your definition are fulfilled, but the the additional vector is independent.
I think the naming is weird. If two vectors are needed to represent a real space, then shouldn't it be called Linearly dependent instead of independent? I mean, if one is needed, then the set is dependent upon that vector. Maybe I am not understanding the concept.
Thank you :)
i freakin love this academy :D saved my ass for linear algebraaa test :D:D:D: i personally appreciate n tell'em theyre the best at lin algebraa :D
the equation 09:03 is always giving c1 and c2=0
thanks bud :D
Why does Sal subtract one equation from the other rather than adding? 14:41
If u will add, u will get another equation in terms of C1 and C2 rather than the value of C2.
ty a looot :C
this video makes me feel like a einstein
lol
how about to choose c3 as 0 "zero" ?? . Why we have to choose c3 or another c as whole number except zero ? if I choose c3 as 0 the three vectors become independent
we want to find a solution to prove if they are dependent and its an easy way to pick c3=-1 or any other value you could also solve it with Gaussian elimination and then express c1 and c2 by c3 or any of them by the other tho your comments its quite old I just felt like to explain
why would you multiply it by 1 half though -9:04
@3:50 what if V1 was actually 0 vector ?
Yeah it can be
Think geometrically !
I don't like how you did the last one though, still not convinced of your method
🙏
yahahaha this saved my life
Does someone know what software he's using?
+Lexus ISF MS Paint
@Brianaust138 in that case it is nothing......i don't think u'll be presented with such a problem
Where did he get that C3=-1????????
but if c3=0 it becomes independent, why did you choose c3= -1????
It is because we want one redundant vector of others.
Because that's the definition of linear dependance. There exists some c_i != 0 such that c_1 * v_1 + ... c_n * v_n = 0
what is R2?
Loop loop not sure if you’ve had this answered already, but it is 2 dimensions of the real numbers
what is span
Check out his previous videos
Lol I just realised Sal is Salman Khan, the founder of Khan academy. I feel special that he is talking to me xD
Watching speed *2, 10 minutes to exam😂😂
do yall not know what an augmented matrix is?
Khan 2012
i wait and wait and nothing is said
Look, c 1 and c2 always gives 0 in example 1 . 😂😂😂 why? Is it true?
they should play your video during lecture instead of the prof teaching =p
Kamogelo Mohlabeng
im come here instead of going to my lectures LMFAO
@youareme888
do you have FA? thats ridiculous!!!!
Just found out this video is actually 10 years old 😮😮
I feel old n dumb
Amazing job on the explanations, however this is really really boring...
Most of the time I love Sal but sometimes all the mumbo jumbo gets on my nerves.
Your proof the definition of linear dependency is fallacious. Using the same work, I can prove that any two vectors are linearly dependent.
Given vectors a and b, let vector c=-b
0a + 1b + 1c = 0
Therefore, a,b, and c are linearly dependent. Therefore, a and b are linearly dependent. This result can be generalized by making c the sum of an arbitrary number of linear combinations of other vectors.