Linear Algebra - 21 - Basis for Row Space

preview_player
Показать описание
How to find a basis for the Row Space of a matrix.
Рекомендации по теме
Комментарии
Автор

This is exactly what I needed, thank you!

tree
Автор

Thank you so much. Was really sick and missed a whole bunch of stuff.

iiiblanklll
Автор

Thank you so much. You are the best engineer

Siriusthepolymath
Автор

Why tf did my textbook make this so hard 😂

andyspendlove
Автор

I fucking love you 😭. The struggle i had with these... Thank

BrotherOfHercules
Автор

I thought you need a reduced row echelon form(rref)

jonathanl
Автор

Suppose that you identify the non-zero rows of the row reduced
matrix, but then take the corresponding rows of the original matrix,
will this in general give a basis for the row space?

sultanalhammadi
Автор

in 3:20 right bottom shouldnt it have a non trivial solution to have the vectors dependent?

thanos
Автор

At last couldnt we divide 1st row to make pivot 1 and divide 2th row by 6 to make pivot 1? And basis will change what ?!

OnMyGameplay
Автор

There's four numbers in each vector in the basis which has 2 dimensions so how o you represent that in r 2 graph ..

footage
Автор

Can you find the basis row reducing A^T and finding the pivots on that matrix?

tammypham
Автор

The basis of row sapce of A contain 2 vactors that's why it's dimension is 2 but those 2 vectors are 4 dimensional Bec they have 4 components. So does it make sense a 4 dimensional vectors exist in 2 dimensional subspace. Please reply 🙏

alokraj
Автор

Is the dimension of the row space the same as the dimension of the column space of a matrix?

jacksonmckenzie
Автор

In REF shouldn't the main diagonals be 1's? Never mind, you're doing (Reduced Echelon Form), I though you were doing (Row Echelon Form).

edwinjoseph
Автор

Change your channel name to "The God" or "The ultimate Savior" sir.

thepriestofvaranasi
Автор

in 3:20 right bottom shouldnt it have a non trivial solution to have the vectors dependent?

thanos