# Homogeneous system

1. Oct 28, 2008

### soul5

1. The problem statement, all variables and given/known data

Let A * x= 0 be a homogeneous system of n linear equations in n unknowns, that has only the trivial solution. Show that if k is any postive integer, than the system A^k * X = 0 also has only the trivial solution.

3. The attempt at a solution

I'm so lost plz help and what is trivial solution?

2. Oct 28, 2008

### Staff: Mentor

The trivial solution is the vector x = 0. All n entries in this vector are zero.

3. Oct 29, 2008

### Staff: Mentor

If A is an n x n matrix, and Ax = 0 has only the trivial solution, what things does that tell you about A?

4. Oct 29, 2008

### sutupidmath

I am just picking up on what Mark44 said. If the only sol to the matri eq is the trivial one Ax=0, that is for =0,(x-vector, A matrix (nxn), this means that A is nonsingular. Then since A is nonsingular, we know that A has an inverse, a unique one. SO:

$$A^kx=0=>A*A*A*A*A*...*Ax=0$$ Now multiplying by A^-1 , k-1 times from the left side we get to Ax=0, which as we know has only the trivial sol. so we are set.

5. Oct 29, 2008

### Staff: Mentor

But that gets you to Ax = 0. There is one more thing that needs to be done to arrive at the conclusion you want.

6. Oct 30, 2008

### soul5

Which is?

7. Oct 31, 2008

### HallsofIvy

First, any equation like Ax= 0 has the "trivial solution": it is always true that A(0)= 0. The whole question is whether that is the only solution or whether other solutions exist.

Do you know that Ax= 0 has only the trivial solution if and only if A is invertible? And so det(A)= 0? If you do and also know that det(An)= (det(A))n then the problem is trivial.

If not then proof by induction may be simplest. Since we are given that Ax= 0 has only the trivial solution, the base case, n= 1 is trivial. Suppose Akx= 0 has only the trivial solution. Then Ak+1x= A(Akx)= 0. What can Akx be? And what does that tell you?