Solving Homogeneous System of n Linear Equations with Positive Integer k

[soul5]

Homework Statement

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.

The Attempt at a Solution

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

 

[Mark44]

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

 

[Mark44]

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

 

[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.

 

[Mark44]

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

 

[soul5]

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

Which is?

 

[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 A^kx= 0 has only the trivial solution. Then A^k+1x= A(A^kx)= 0. What can A^kx be? And what does that tell you?