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Minimal linearly dependent sets of columns

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Given two matrices

{1 2 3 3 9
1 0 1 0 2
0 2 2 3 7
2 4 6 6 18}

{9 3 9 3
9 1 3 3
9 0 2 5
6 0 0 2
3 1 3 1}

for each one:
a) list all minimal linearly dependent sets of columns;
b) list all maximal linearly independent sets of columns;
c) list all minimal sets of columns which span all columns; a',b',c') The same for rows;
d) compute the rank.

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The expert examines minimal linearly dependent sets of columns.

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Given two matrices

{1 2 3 3 9
1 0 1 0 2
0 2 2 3 7
2 4 6 6 18}

{9 3 9 3
9 1 3 3
9 0 2 5
6 0 0 2
3 1 3 1}

for each one:
a) list all minimal linearly dependent sets of columns;
b) list all maximal linearly independent sets of columns;
c) list all minimal sets of columns which span all columns;
a',b',c') The same for rows;
d) compute the rank.

Part 1. Let A=[├ █(1 2 3 3 9@1 0 1 0 2@0 2 2 3 7@2 4 6 6 18)]┤

Using elementary row operations, we obtain the reduced row echelon form of A

"rref"(A)=[├ █(1 0 1 0 2@0 1 1 □(3/2) □(7/2)@0 0 0 0 0@0 0 0 0 0)]┤

a)
Let's denote the column vectors of A by c_1,c_2,c_3,c_4,c_5 and the row vectors of A by r_1,r_2,r_3,r_4. Since "rref"(A) has leading 1's in the first and second column, it follows that {c_1,c_2} is a basis for the column space of A and {[1 0 1 0 2],[0 1 1 □(3/2) □(7/2)]} is a basis for the row space of A.

Since the column space of A has dimension 2, it follows that any set of 3 column vectors of A is linearly dependent. We see that no column of A is a scalar multiple of another column, so the minimal linearly dependent sets of A must all contain 3 column vectors of A. We see that there are (■(5@3))=10 possible combinations of 3 columns. So all minimal linearly dependent sets of columns of A are {c_1,c_2,c_3}, {c_1,c_2,c_4}, {c_1,c_2,c_5}, ...

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