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.