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x
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clear
i
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3131313131351515151515≈83137
How to find the rank by bordering minors
Start from a single non-zero entry (a 1×1 minor). Border it with adjacent rows/columns to form a 2×2 minor; if any 2×2 minor is non-zero, continue bordering to 3×3; and so on. The rank is the size of the largest non-zero bordered minor.
Bordering minors — worked example (4×4)
Write the initial matrix
A
:
A
=
1
0
2
1
2
1
4
0
3
2
6
1
4
1
8
3
2
STEP [0]Let's look at matrix
, there are nonzero values among its elements;
For example, there is a nonzero element at the intersection of row
1
and column
1
;
Let's refer to this element as a first order minor (
M
0
1
);
M
0
1
=
1
0
2
1
2
1
4
0
3
2
6
1
4
1
8
3
=
1
;
Since matrix
has a first order minor, rank(
) ≥ 1;
3
STEP [0]Let's try to find any nonzero
2
-order minor (
M
0
2
) bordering the
1
-order minor (
M
0
1
);
Find a
2
-order minor bordering the
1
-order minor at the intersection of row
2
and column
1, 2
;
M
0
2
=
1
0
2
1
2
1
4
0
3
2
6
1
4
1
8
3
=
1
0
2
1
=
1
;
So a nonzero
2
-order minor exists, therefore rank(
) ≥
2
;
We'll refer to this minor as
M
0
2
;
4
STEP [0]Let's try to find any nonzero
3
-order minor (
M
0
3
) bordering the
2
-order minor (
M
0
2
);
Find a
3
-order minor bordering the
2
-order minor at the intersection of row
3
and column
1, 2, 3
;
1
0
2
1
2
1
4
0
3
2
6
1
4
1
8
3
=
1
0
2
2
1
4
3
2
6
=
0
This minor equals zero;
So, if possible, we continue the search!
Find a
3
-order minor bordering the
2
-order minor at the intersection of row
3
and column
1, 2, 4
;
1
0
2
1
2
1
4
0
3
2
6
1
4
1
8
3
=
1
0
2
2
1
4
4
1
8
=
0
This minor equals zero;
So, if possible, we continue the search!
Find a
3
-order minor bordering the
2
-order minor at the intersection of row
4
and column
1, 2, 3
;
M
0
3
=
1
0
2
1
2
1
4
0
3
2
6
1
4
1
8
3
=
1
0
1
2
1
0
3
2
1
=
2
;
So a nonzero
3
-order minor exists, therefore rank(
) ≥
3
;
We'll refer to this minor as
M
0
3
;
5
STEP [0]Let's try to find any nonzero
4
-order minor (
M
0
4
) bordering the
3
-order minor (
M
0
3
);
Find a
4
-order minor bordering the
3
-order minor at the intersection of row
3
and column
1, 2, 3, 4
;
1
0
2
1
2
1
4
0
3
2
6
1
4
1
8
3
=
1
0
1
2
2
1
0
4
3
2
1
6
4
1
3
8
=
0
This minor equals zero;
So, if possible, we continue the search!
So we've checked all the
4
-order minors bordering the minor
M
0
3
, but all of them are equal to zero;
The last non zero minor was of the
3
-order, therefore rank(
) =
3
;
Answer
rank(A) = rank(
A
) =
3
;
SIZE4×4METHODBordering minors