rank(a) = rank(transpose of a) | Matrix transformations | Linear Algebra | Khan Academy
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- Опубліковано 10 лют 2025
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Rank(A) = Rank(transpose of A)
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Summary -
Actually no of Lineraly Independent rows[dimension of rowsp(A)] = no of Lineraly Independent colums[dimension of columnsp(A)]
Only in row reduced echelon form of A ie rref(A) we get both linearly independent Rows and columns
Row reduced echelon form is aslo known as canonical form/row canonical form
But in simple echelon form we get linearly independent rows only
We don't get linearly independent Columns in simple echelon form
So row rank of A = column rank of A = Rank of A
In some books authors define rank as no of Lineraly independent rows and some define it as dim of columnsp (A) ie no of Lineraly Independent colums
Both are same
This thing can also be seen from the result rank(Aᵀ) = rank of A
As then no of linearly independent rows and columns of Aᵀ = no of Lineraly Independent colums and rows of A RESPECTIVELY
Xavier here i would have never thought😅
its remarkable, you can say more in 10 minutes than professors at universities(i.e. MIT) can say in an hour!
ex: A is rectangular 100-by-40. 100 row-vectors can span 40-dimensional space at max. (maxrank(C(A^T))=40). AT LEAST 60 row-vectors are linearly dependent and do NOT have pivots in rref(A) (minrank(Null(A^T))=60).
40 column-vectors in 100 dimensions span at MAX 40-dimensional subspace as well (maxrank(C(A))=40,minrank(Null(A))=0). suppose it turns out rank(C(A))=38 < maxrank(C(A))=40, implies rank(Null(A))=2. implies 2 columns with no pivots in rref(A), which means 2 extra rows with no pivots.
The crux of the proof is that rref(A) has the same rank as A, and I probably missed the part where you proved it, but without that, the proof is trivial (since it can be trivially proven that the rank is invariant to elementary column operations)
It baffles me how a 10yr comment on linear algebra can still be educational and insightful. Math truly transcends time.
thank you Sol
Is it correct to say of basis of column space of A instead of basis for column space of A
Which one is correct
There is no name for the dimension of the subspace spanned by row vectors because of this theorem; because it equals the rank of a matrix.
apologies, i meant
minrank(Null(A))=60, minrank(Null(A^T))=0 and rank(Null(A^T))=2.