Suppose A is a square matrix satisfying the equation A^3 - 2I = 0. Prove that A and (A - I) respectively are invertible. (the hint is to find an explicit equation for A^-1.

To prove A is invertible, this is what I did although I don't that it's right:
A^3 - 2I = 0
A(A^2 - 2A^-1) = 0 because A * A^-1 = I
so (A^2 - 2A^-1) must = A^-1
(A^2 - 2A^-1) = A^-1
so (1/3)A^2 = A^-1 thus proving A is invertible. However I don't know that this is correct, and I don't know how to prove that (A - I) is also invertible. I know I need to factor the equation, but the 2I makes it difficult.

Well I guess no one can help. :/

Oh well, I'll figure it out

To prove that matrix A is invertible, we can indeed start with the given equation A^3 - 2I = 0, where I represents the identity matrix.

First, let's rearrange the equation: A^3 = 2I.

We know that for a matrix to be invertible, it must have a non-zero determinant. Therefore, if we can show that det(A) ≠ 0, then A is invertible.

Now, let's take the determinant of both sides of the equation: det(A^3) = det(2I).

Using the property det(AB) = det(A) * det(B), we can simplify the equation to (det(A))^3 = 2^n * det(I).

Since det(I) = 1, we have (det(A))^3 = 2^n.

From this equation, we can see that det(A) ≠ 0 because otherwise, (det(A))^3 would be zero.

Therefore, we have proven that A is invertible.

Now, let's prove that (A - I) is also invertible.

We know that (A - B)^(-1) = (B^(-1))(A^(-1)). Using this property, we can find the inverse of (A - I) by finding the inverse of A - I.

In order to find the explicit equation for A^(-1), we need to solve the equation A(A^2 - 2A^(-1)) = 0 that you mentioned in your attempt. Let's go through it step by step:

A(A^2 - 2A^(-1)) = 0

Distribute A into the parentheses:

A^3 - 2A^(-1)A = 0

Since A^3 = 2I, substitute it into the equation:

2I - 2A^(-1)A = 0

Now, we can rearrange the equation to solve for A^(-1):

2A^(-1)A = 2I

A^(-1)A = I

Therefore, we have found that A^(-1) exists, and its equation is A^(-1) = A^2.

To prove that (A - I) is invertible, we can use similar reasoning. We want to show that (A - I) has a non-zero determinant.

Assume that (A - I) is not invertible, which means the determinant of (A - I) is zero.

det(A - I) = 0

Expanding this determinant using the cofactor expansion, we have:

det(A) - det(I) = 0

Since det(I) = 1, we get:

det(A) - 1 = 0

det(A) = 1

But we already proved earlier that det(A) ≠ 0, so det(A) can't simultaneously be equal to 1.

This contradiction means that our assumption was incorrect, and (A - I) must be invertible.

Therefore, we have proven that both A and (A - I) are invertible matrices.