Finding Basis for the Row Space of a Matrix | Linear Algebra
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- เผยแพร่เมื่อ 10 มี.ค. 2023
- We use elementary row operations to reduce a matrix to row echelon form in order to find a basis for the row space. After completing this process, the nonzero rows in the echelon form (which are also those with leading 1s) will form a basis for the row space. Thus, the number of row vectors that are nonzero and make up the basis also define the dimension of the row space and hence the rank of the matrix. #linearalgebra #matrices
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This video is the basis...of a great introduction to finding row spaces!
This video is just about perfect, thanks
Glad to help, thanks for watching! Column space next!
Thank you so much for making this video! This helped me a lot!!
Awesome, thanks for watching! I'm working on more linear algebra lessons, let me know if you ever have any questions!
I have a test the day after tomorrow and after searching a lecture to understand row space and column space for nearly 2 days, I found yours. Blessed. Crisp explanation 😌
Best of luck!
you are doing god's work man. thanks a lot
My pleasure! Just released a similar video on the column space.
Excellent video. No wasted time and good explanations
Glad it was helpful! Lots more in my lin alg playlist:
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U saved my timeee
❤❤🎉🎉 Perfect one
Thanks for watching!
I have a question, why basis for row space are in rref form while basis for column space are in original form?
This is because elementary row operations do not change the row space of a matrix, however they do change the column space, so we cannot just use the nonzero columns of the rref to span the column space.
wow didn't realize this, thanks for the answer!@@WrathofMath
precise explanation😌🥰
Thank you!
sorry, why does it have to do another ero (replacing R3 by R3-R2)?
This video is awesome!!! You know you can actually overtake my teacher (I understood your video faster
Since the basis combined have 6 different elements, I'm guessing the dimension is 6, correct me if I'm wrong please...
for colume space of a matrix, is it the same process but after we get row echlon form, we look at non-zero columes? Pleaase correct me if im wrong!
yes thats correct but then the basis of the column space will have the columns of the original matrix .. not the ref one.
is it the same process for colume space of the matrix? So we would find the row echlon form aand then instead of looking for non-zero rows, we would look at non-zero columes and they become our "basis for colume space of matrix"? Please correct me if im wrong
Not entirely. You find the linearly independent columns by doing ref, but you have to go back to the original matrix because the columns are changed when you perform row operations.
So to find a basis for a row, you have to put the matrix in REF first right? Not RREF? As in RREF is not required during the process, only REF is fine?
Also, if the question is "find a basis". Do we list all of the rows that return as a basis or can we just choose one of them to present as the answer, generally speaking?
I think he wrong, we need to put the matrix in RREF to find the basis for the Row Space
@@GB2G we need to list all of the rows has a leading 1 (or nonzero row)
cant the basis of RS(A) be done as CS(At) [ie. col space of the transpose of matrix A]??? we would get a correct answer even then rt?
I think so, yes. I've tried it on one example and it seems to be doing the same thing. Because what you're really doing when finding a basis, is seeing which vectors you need to make all of the vectors in the matrix with some lineair combination. When you transpose a matrix you're making the rows of the original matrix into the columns of the new matrix. If you then try to find the basis for that column space, youre again, looking for the vectors that can together create all the vectors in the columns of the matrix with a linear combination. So when you find the basis of the columns of the transposed matrix, youre really finding the basis for the rows of the original matrix. I think.
hi