Is it possible to have a matrix that have pivot columns in every row, in Reduced Row-Echelon Form and is linearly dependent?
Let's assume the matrix E(m,n) , with m rows and n columns, is in the reduced row echelon form.
But the meanings of the terms:
- "a matrix that has pivot columns in every row"; and,
- "a matrix is linearly dependent";
are not clear to me. I suppose that you want to say:
- "every row in the E(m,n) contains a pivot"; and,
- "the rows (or columns) of E(m,n) form a linearly dependent set".
So, I suppose that your question could be:
"Is there any matrix E(m,n), in reduced row echelon form, the rows of which form
a linearly dependent set, while every row contains a pivot?"
If that is the case (every row in the E(m,n) contains a pivot), then the number of pivots (or the rank of E) is equal to the number of rows; why? because the number of pivots is actually the rank of the matrix. So:
rank(E) = m ,
and you know that the rank of a matrix cannot be greater than the minimum amount of number of rows and number of columns, so:
rank(E) ≤ min{m,n} ⇒ m ≤ n .
It means if each row of the matrix E contains a pivot, then the number of rows must be less than, or equal to the number of columns.
For the matrix E(m,n) (and even for any matrix A(m,n) ), " the rank of the matrix is equal to the number of rows" is equivalent to saying that "the rows of the matrix form a linearly independent set," i.e.:
rank(E) = m ⇔ the rows of E form a linearly independent set.
So, it is impossible to find such a matrix (whether in reduced row echelon from or not), that the set of its rows form a linearly dependent set.
However, rank(E) = m (i.e., each row of E contains a pivotal position) does not guarantee that the columns of E form necessarily a linearly independent set and therefore you may find many matrices (either in reduced row echelon form or not), in which each row contains a pivot, but the set of its columns is linearly dependent. For example:
E(2,3) =
1 0 -1
0 1 -2
has two rows (m = 2), and two pivotal positions, and therefore
rank(E) = 2,
however, its columns form a linearly dependent set, because:
(1 , 0) + 2 (0, 1) + (-1 , -2) = 0.
Yes, exactly as u assumed:
"Is there any matrix E(m,n) , in reduced row echelon form, in which every row
contains a pivot and the set of the rows (or columns) of E form a linearly
dependent set?"
...Thanks Sir.
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