Jonathan L row space is preserved through row operations. in other words, row(A) = row(rref(A)). as a result, we need only find where the pivots are and select those rows since the pivot locations will not change through row operations if you reduce all the way to rref you may get a different basis than you would using just the REF but both are still bases!
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?
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 🙏
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I thought you need a reduced row echelon form(rref)
Jonathan L row space is preserved through row operations. in other words, row(A) = row(rref(A)). as a result, we need only find where the pivots are and select those rows since the pivot locations will not change through row operations
if you reduce all the way to rref you may get a different basis than you would using just the REF but both are still bases!
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?
Cool vids bro.
Is the dimension of the row space the same as the dimension of the column space of a matrix?
Yes. We define it as the "Rank" of a matrix. See the videos where i talk about rank, that may help too
yes math.mit.edu/~gs/linearalgebra/ila5/linearalgebra5_3-5.pdf
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 ?!
in 3:20 right bottom shouldnt it have a non trivial solution to have the vectors dependent?
Can you find the basis row reducing A^T and finding the pivots on that matrix?
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There's four numbers in each vector in the basis which has 2 dimensions so how o you represent that in r 2 graph ..
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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 🙏
yes, it is a subspace math.mit.edu/~gs/linearalgebra/ila5/linearalgebra5_3-5.pdf
nice
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).
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in 3:20 right bottom shouldnt it have a non trivial solution to have the vectors dependent?