Lec 9 | MIT 18.06 Linear Algebra, Spring 2005

[AntoMelta]

Good, I like that you share this video Independence, Basis, and Dimension., I wish success always

[willamricard]

Nice Video Independence, Basis, and Dimension That You Share , So Very Nice Thanks You

[imegatrone]

I Really Like The Video Independence, Basis, and Dimension From Your

[bundawartini]

Your Video Is Very Useful Sharing Independence, Basis, and Dimension.

[TaishaLadawnck546]

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[LostAngel01234]

WAAAYYY Better than my Linear Algebra prof

[joaopp1]

This lectures are so simple to understand.

The lectures that my professor of linear algebra give at my university are so difficult to understand even at the beginning because he doesn't explain in a language that a new student understand.

He start to write the role with symbols that i doesn't seen before and then prove all the theories without speaking what's going on before.

Thanks for all this lectures.

[stran1992]

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.

[chiraal]

What a guy!

[contactit]

Thanks........

Praveen



[akhil089]

thanks professor strang

[balagopalanthampi]

great teacher .........:)

[Nolonar]

I'm not sure, if (1,1,2) (2,2,5) (3,3,8) are independent.

I mean, if we ignore the third component, (1,1) (2,2) and (3,3) are obviously NOT independent. They lie on the same line.

Now when we add the third component, the only thing that changes, is the Z-coordinate.

Those 3 vectors span a 2-Dimensional subspace in R^3 through (1,1), and are thus dependent.

[aadityakalsi]

@Nolonar They are not dependent. There is no a1, a2, a3 except a1=a2=a3=0, such that

a1*v1 + a2*v2 + a3*v3 = 0.

It becomes three linear equations in 3 variables. Try it out.

[Nolonar]

@aadityakalsi Actually, there is.

-1, +2, -1

[aadityakalsi]

@Nolonar hahah! youre absolutely right! I tend to believe him blindly.. :|

[PrinceMapleD]

how come there are people dislike these lectures?

[ghostlucian12]

@KaLNFoRc3R There are many applications where linear algebra is used, like quantum computing, cryptography, computer graphics, and many others. I started to review linear algebra because i have some courses on quantum computing, and it is based on linear algebra. Thanks for the free courses.

[fjalees]

brilliant!

[ducefalo21]

Por favor activen la opcion para traducir los subtitulos

[ducefalo21]

Por favor activen la opion para traducir los subtitulos

[pierusa123]

It is better watch some video of Math.

[THESQUATINATOR]

I was going to wait until lecture 34 before I wrote what I think about Gilbert Strang and this course but I cant wait any longer. Each lecture provides insight after insight and I feel like I want to give him a standing ovation after every single one of them. He is so amazing at teaching that I have skipped all the lectures at my university and learn from these lectures instead. Thank you Gilbert Strang and MIT.

[themole28]

mind blown

[emilfh]

Great Lecture!

[telecinecult]

Why did he change his "2"? haha. I just find that funny.

[cunucky]

man the students in this lecture are either slow or dont like to shout out answers on camera lol

[roostaj]

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!

[KaLNFoRc3R]

@roostaj i fucking hate proofs

[hvutrong]

@KaLNFoRc3R why? I think you fail how to do and read proofs in the right way. That makes you hate them.

[KaLNFoRc3R]

@hvutrong of course, i can't even begin to understand most of them and their complexities, i think they are useless unless you wana become some linear algebra lecturer.

[xyber3364]

very good teacher, very helpful video

[pmshadow]

Gilbert Strang,

You are an awsome teacher! Congratulations for the great work done and thanks for the opportunity!

Congratulations to MIT as well.

[2009Aron]

Toda raba!

[dramon231]

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

[russelljbarry15]

These lectures are so great. I am taking quantum mechanics and linear algebra is a large part of it. Looking at these helps review linear algebra and he is such a great teacher.

Thanks !!!

[fuckooo]

@russelljbarry15 I agree, he is an absolutely wonderful lecturer.

[Lemnirot]

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.

[tehinfidel]

He catches and corrects it in the next lecture.

[Mathematics215]

Its also clear because the matrix has two of the exact same rows, namely 1,2,3 appears twice.

[Mathematics215]

Alright well nevermind, thought this way was better, then went to video 10, and that was the exact reasoning he used....

[gavilanch]

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!!!

[ateen83]

@Lemnirot He corrects this in Lecture 10

[cuzimwhiterite]

@Lemnirot lol 1 + 2(5) != 8

so you're wrong

[slatz20]

@Lemnirot u are right..but what if u look @ the two first rows? its also the same..

[flapjacksmike]

@Lemnirot Not sure if you someone told you or if you already saw the next video, he did point this out. Not be mean, just wanted to let you know.

[RadekLudva]

@Lemnirot Well the main problem is that the fist two rows (1,2,3; 1,2,3; 2,5,x) are the same! He could have put any number on the last position x in vector 3 and they would be dependent nevertheless :)

[TheHackersGambit]

Great Vid!