You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
Because one row cannot be composed of the others. Since it's in RREF all first instances of 1 within each row are the only non-zero values in their respective column. For example: how do you get 1 from adding 0 to a 0. Answer: you can't
I'm not sure I understand your comment. The row space of a matrix A, is the subspace spanned by its rows. Row operations do not change the row space. So, performing row reduction on the original matrix and examining the final non-zero rows provides a basis for the row space. That's what was done in the video. I'm not sure where you think a transpose should go?
@@AdamPanagos I thought the row space is the column space transposed. So after putting the matrix in rRef you transpose the matrix. So instead of it being: Basis of Row(A) = {[1 0 1 1 1 ], [0 1 1 -1 2]} it is Basis of Row(A) = {[1, 0, 1, 1, 1 ], [0, 1, 1, -1, 2]} going down.
this was exactly what i needed to get through my homework, thanks so much
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
why do you take rows out of the rref of A for the row space but the columns of A for the column space?
SIMPLE AND SWEET
Why in your Column basis video, you take the basis of the original matrix A but in this one you take the row basis from the equvi. of A?
Do you already know the reason?
thanks bud, very clear. all the best.
Thanks!
Thanks!
How do we know that the non zero rows are linearly independent?
Because one row cannot be composed of the others. Since it's in RREF all first instances of 1 within each row are the only non-zero values in their respective column.
For example: how do you get 1 from adding 0 to a 0. Answer: you can't
You forgot to transpose the matrix.
I'm not sure I understand your comment. The row space of a matrix A, is the subspace spanned by its rows. Row operations do not change the row space. So, performing row reduction on the original matrix and examining the final non-zero rows provides a basis for the row space. That's what was done in the video. I'm not sure where you think a transpose should go?
@@AdamPanagos I thought the row space is the column space transposed. So after putting the matrix in rRef you transpose the matrix. So instead of it being: Basis of Row(A) = {[1 0 1 1 1 ], [0 1 1 -1 2]} it is Basis of Row(A) = {[1, 0, 1, 1, 1 ], [0, 1, 1, -1, 2]} going down.