Matrix manipulations/rank of a matrix

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In summary: So since the bolded parts contradict each other, I assume its false but I'm not sure how to go about proving it. Any ideas?
  • #1
mirandasatterley
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My question has 5 short parts and for each, I'm supposed to give a counterexample if the statement is false or give an argument to prove that a statement is true.

1.Q: Linear system with m equations and n variables. If m> or = to n, then the system can have at most 1 solution.
A: I think this one is false because if the # of parameters= n-rank, since rank < or = n, if we take it to be less than n, then # of parameters = n - (some # less than n), which gives a # > 0, then this means there is at least 1 parameter and infinate solutions possible.

2.Q: A and B are matrices, and the product AB is defined. Then rank (AB) = rank A
A: I would say that this is false because if A and B are different sizes, say A is 5x4 and B is 4x6, the rank of both A and B would have to be < or = 4, but AB produces a 5x6 matrix, meaning that the rank must be < or = 5. So couldn't A and B each have a rank of 4, while AB has a rank of 5? But if this is true, how can it be proved?

3.Q: A and B are nxn matricies and AB=0. Then at least one of A,B must have a determinant 0.
4.Q: An nxn matrix A satisfies AB=BA for every nxn matrix B, then A must be the identity matrix.

I'm not sure about 3 or 4, any hints/points in the right direection would be appreciated.

5.Q: If the system Ax=b has no solution, then the system Ax=0 has only the trivial solution.
A: I think this is false because the statements: The system Ax=0 has the only trivial solution x=0 and Ax=b has a unique solution for any vector b, according to the invertible matrix theorem. So since the bolded parts contradict each other, I assume its false but I'm not sure how to go about proving it. Any ideas?
 
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  • #2
mirandasatterley said:
My question has 5 short parts and for each, I'm supposed to give a counterexample if the statement is false or give an argument to prove that a statement is true.

1.Q: Linear system with m equations and n variables. If m> or = to n, then the system can have at most 1 solution.
A: I think this one is false because if the # of parameters= n-rank, since rank < or = n, if we take it to be less than n, then # of parameters = n - (some # less than n), which gives a # > 0, then this means there is at least 1 parameter and infinate solutions possible.
If all the equations were independent (i.e. rank m) (which is not given [and could only happen if m=n]) then there would bo no solutions so this is correct.

2.Q: A and B are matrices, and the product AB is defined. Then rank (AB) = rank A
A: I would say that this is false because if A and B are different sizes, say A is 5x4 and B is 4x6, the rank of both A and B would have to be < or = 4, but AB produces a 5x6 matrix, meaning that the rank must be < or = 5. So couldn't A and B each have a rank of 4, while AB has a rank of 5? But if this is true, how can it be proved?
Even if A and B were of the same size, it might happen that AB= 0 which has rank 0!

3.Q: A and B are nxn matricies and AB=0. Then at least one of A,B must have a determinant 0.
4.Q: An nxn matrix A satisfies AB=BA for every nxn matrix B, then A must be the identity matrix.

I'm not sure about 3 or 4, any hints/points in the right direection would be appreciated.
3. Do you know that det(AB)= det(A)det(B)? That makes 4 trivial. Even if you don't know that, you should know that a matrix has an inverse if and only if its determinant is not 0. If A does not have determinant 0, multiply both sides of AB= 0 by A-1.

4. Look at AB and BA for A any diagonal matrix.

5.Q: If the system Ax=b has no solution, then the system Ax=0 has only the trivial solution.
A: I think this is false because the statements: The system Ax=0 has the only trivial solution x=0 and Ax=b has a unique solution for any vector b, according to the invertible matrix theorem. So since the bolded parts contradict each other, I assume its false but I'm not sure how to go about proving it. Any ideas?

In fact, it is exactly the other way around. If Ax= b has no solution, then A is not invertible. Ax= 0 has an infinite number of solutions
 
  • #3
HallsofIvy said:
4. Look at AB and BA for A any diagonal matrix.

I have tried quite a few different things and it seems that If both A and B are diagonals this works, but since diagonal matrices can be row reduced to the Identity matrix, does this make number 3 true? When only one of A or B are diagonals, it doesn't seen to work out unless one is the identity matrix. Does this mean that it must be true, or are there special cases that i am missing here?
 

1. What is a matrix and why is it important in science?

A matrix is a rectangular array of numbers or symbols, arranged in rows and columns. It is important in science because it allows for the representation and manipulation of large amounts of data, making it a useful tool in fields such as statistics, physics, and computer science.

2. What is the rank of a matrix?

The rank of a matrix is the maximum number of linearly independent rows or columns in the matrix. In other words, it is the number of rows or columns that cannot be expressed as a linear combination of the other rows or columns.

3. How is the rank of a matrix calculated?

The rank of a matrix can be calculated by performing row operations on the matrix until it is in its reduced row echelon form. The number of non-zero rows in the reduced matrix is equal to the rank of the original matrix.

4. Why is the rank of a matrix important?

The rank of a matrix is important because it provides information about the linear independence of the rows or columns of the matrix. It can also be used to determine the solutions to systems of linear equations and the invertibility of a matrix.

5. Can the rank of a matrix change?

Yes, the rank of a matrix can change if row operations are performed on it. However, the rank will not change if only column operations are performed. Additionally, the rank of a matrix is always less than or equal to the number of rows or columns in the matrix.

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