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Question 1181409: Given the following matrices over Z5:
b1 b4 b7
b2 b5 b8 = B
b3 b6 b9
r(B)= 3
For easier understanding let’s say that bi is the column i in matrix B.
A is a matrix of dimensions 3x3 in which the following exists:
K = { (b1+b3) , (2b2-b3) } is a basis for the equation: Ax=0.
r(A)=1
Also:
1 2 4
3 1 2 =A*B
1 2 4
r(AB)=2.
1) find a vector basis for the vector space that is A’s columns.
2) given the following:
K = { (1,2,0) , (0,1,1) }
Find a reduced echelon form (C) for A.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to approach this problem:
**1) Finding a vector basis for the column space of A:**
* **Understanding the relationship between A and AB:** Since r(AB) = 2 and r(B) = 3, it means that the columns of AB are linearly dependent. However, since the rank is 2, there are two linearly independent columns in AB. These columns span the column space of AB.
* **Column space of AB and A:** The column space of AB is a *subspace* of the column space of A. This is a crucial point. Because r(A) = 1, the column space of A is a *line*. This means that the column space of AB (which is a plane since it's spanned by two vectors) *cannot* be a subspace of the column space of A. This is a contradiction. There must be an error in the given ranks. If r(A) = 1, then r(AB) <= 1.
* **Assuming r(A) = 2:** Let's assume that r(A) = 2 (so that r(AB) can be 2). Since the column space of AB is a subspace of the column space of A, and they both have rank 2, they must be the same. The columns of AB span the column space of A.
* **Basis for the column space of A:** Since the rank of AB is 2, any two linearly independent columns of AB will form a basis for its column space, and therefore also for the column space of A. For example, the first two columns of AB: {(1, 3, 1), (2, 1, 2)} form a basis.
**2) Finding the reduced echelon form (C) for A:**
* **Understanding the null space:** K = {(1, 2, 0), (0, 1, 1)} is a basis for the null space of A (Ax = 0). This means any linear combination of these vectors will be mapped to the zero vector by A.
* **Relating null space to matrix structure:** The vectors in K tell us about the relationships between the columns of A. Let the columns of A be a1, a2, and a3. Then:
* 1*a1 + 2*a2 + 0*a3 = 0 => a1 = -2a2
* 0*a1 + 1*a2 + 1*a3 = 0 => a3 = -a2
* **Constructing A:** We can express all columns of A in terms of a2. Let a2 = (x, y, z). Then a1 = (-2x, -2y, -2z) and a3 = (-x, -y, -z). Thus A will have the form:
```
-2x x -x
-2y y -y
-2z z -z
```
* **Reduced echelon form:** Since the rank of A is supposed to be 1, the reduced echelon form will have only one non-zero row. If we choose x=1, y=0, z=0, we get a matrix:
```
-2 1 -1
0 0 0
0 0 0
```
We can row reduce this to get:
```
1 -1/2 1/2
0 0 0
0 0 0
```
This is one possible reduced echelon form. There are infinite possible reduced echelon forms. If we choose x=0, y=1, z=0, we get a different one. The important thing is that there is only one non-zero row.
* **0 solutions:** Since K is the basis for the null space of A, the equation Ax = 0 has infinitely many solutions. You cannot have 0 solutions.
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