Linear Algebra - Lecture 15 - Linear Independence

[@sankoktas420]

Dude is x10 better than my college prof

[@mikesgarage18]

I took my favorite prof and advisor for Linear Algebra... and... sadly he let me down. Almost failed. Had to modify my major for fear of failing out of college!!

[@snehatimilsena3908]

Your explanations are great. I learned in so much short time. Thank you and god bless you for saving our time and lives!!!

[@micah2936]

I’ve been teaching myself with your lectures, I was caught off guard when you said many people have trouble with this, it seems very straight forward

[@sana181019811]

Your course is awesome and i would say that it really changed my visualization about linear algebra!!!!!

[@saulorocha3755]

Always direct to the point and clear. Thanks

[@shauntecodner9400]

I'm mind-blown how easily you explained this! Thank you so much! I have my finals tomorrow and I just found your page. I'm basically learning everything from your videos

[@christophertech7462]

Sir God bless you, the way you explain mind-blowing

[@eshuuu052]

Hello, your videos are super helpful! Thank you for putting in so much work to create them. By any chance, do you have a video on LU decomposition?

[@atodaz0826]

How would you explain this graphically? If you have linearly dependent vectors, is one not needed to span?

[@Jin-ec1vc]

thank you for the awesome lecture!

[@maxpercer7119]

If I understood this correctly, given a set of m vectors { a_1, ... a_m} where a_i ∈ R^n, call this set of vectors B, if we apply 'row reduced echelon form' on B, rref(B) for short, then by looking at the number of row pivots and column pivots of rref(B) we can answer if there are any redundant vectors (linear dependence) in B and whether B spans R^n. 1. Does rref(B) have a pivot in every column (i.e. no free variables) ? Yes - B is linearly independent, there are no redundant vectors. No - B is linearly dependent, there is at least one redundant vector. 2. Does rref(B) have a pivot in every row (i.e. no row of zeros)? Yes - B spans R^n. No - B does not span R^n. Also, note that If you answered yes to both questions then B is a basis for R^n, which in that case, m = n.

[@HamblinMath]

This is essentially correct. Your first statement is what I call the "Spanning Columns Theorem" and the second statement is the "Linearly Independent Columns Theorem." The only correction I would make is that we don't say "A spans R^m" but rather "the columns of A span R^m." Similarly, we don't say "B is linearly independent" but rather "the columns of B are linearly independent."

[@maxpercer7119]

@@HamblinMath I see. I should have been clearer. Let me make a second attempt. Given a set of vectors B = {b_1, b2, ... , b_m} with b_i ∈ R^n, where the distinction between row vector and column vector isn't emphasized (unless you define elements of R^n as column vectors or n x 1 matrices, though that seems like a representational choice since you could also define them as row vectors or 1 x n matrices). Then we construct the n x m matrix A whose columns are the vectors in B represented as column vectors or n x 1 matrices (here the choice of vectors being represented by column versus row does matter, and we could call A the matrix of column vectors from B, a subtle yet important distinction between A and B). Then at the risk of being pedantic or overly concerned about details, let me correct... 1. rref(A) has a pivot in every column (no free variables) if and only if the vectors in B are linearly independent (B has no redundant vectors). 2. rref(A) has a pivot in every row (i.e. no row of zeros) if and only if the vectors in B span R^n. Also i don't seem to see much of a practical difference between a 'column vector' and a 'n x 1 matrix'. That is, we can work entirely in terms of matrices, and consider any n dimensional vector to be an 'n x 1 column matrix' . see here math.stackexchange.com/a/112854/266200

[@LLai-zh6bk]

thx for making these awesome videos !!!

[@ajwaabid-ng4dx]

Thanks for videos and please more lectures provided

[@thantzinmaung-yq6cu]

you are a godsent.

[@dktchr3332]

it may be in lectures to come but what is the meaning of these terms, linear independence/dependence and dependence relation. Its clear what they desccribe but its not clear why those terms.

[@medardoramirez4610]

What does 0 = 0 tell us? ; 4:50 Do we just say C1 = 0; C2 = 0; C3 = 0.

[@HamblinMath]

0=0 doesn't tell us anything

[@mikesgarage18]

Wish I had this in college. My lectures were only proofs and theory. Almost no examples given. Needless to say, I almost failed the course because I failed to understand critical concepts fully. I've been studying Linear Algebra on and off since then... for almost 12 years now. If I knew then what I know now...

[@tecrahmutungulu1873]

please be slow on elementary matrices so i can follow carefully

[@xoppa09]

dont choose x3 = 0 for linear dependence relation. :P