# Range of a linear transformation to power n

1. Aug 9, 2011

### Lostmant

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

How to prove that: the range of a square matrix A (linear transformation) to the power of n+1 is a subspace of the Range of A to the power n, for all n >= 1?

i.e. Range (A^(n+1)) is a subspace of Range (A^n)

2. Relevant equations

3. The attempt at a solution

I can prove that Kernal (A^n) is a subspace of Kernal (A^(n+1)). Not sure if this is the basis of the prove.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 9, 2011

### jbunniii

Hint:

$$A^{n+1}x = A^n(Ax)$$

3. Aug 10, 2011

### Lostmant

Thank you jbunniii. Using the hint you gave me I could prove that Kernal (A^n) is a subspace of Kernal (A^(n+1)), because A^n(A(0)) = 0. but what about range? I think if i could prove the range of A is always a subspace of domain of A, then i am done. How to prove this? Thanks.

4. Aug 10, 2011

### jbunniii

If y is in the range of $A^{n+1}$, then by definition,

$$y = A^{n+1}x$$

for some x.

But

$$y = A^n(Ax)$$

So what does that imply?

5. Aug 10, 2011

### Lostmant

Got it. Thanks so much.