I'm having trouble with doing this matrix proof
The question is "Given some matrix A has the property A*2=A^-1, show that determinant A = 1, i.e |A| = 1"
I've tried for ages, but I can't seem to do it, this is what I got to
A^2= A^-1
|A^2| = |A^-1|
|A|^2 = 1/|A|
Can someone please help?
The statement isn't true, you have to make additional assumptions. E.g. one could add the condition that all the matrix elements are real. Then |A|^3 =1 and you know that |A| must be a real number, so |A| must be 1.
The condition that the matrix elements are real is sufficient, but not necessary. E.g., take the 2x2 diagonal matrix with exp(2 pi i/3) and
exp(-2 pi i/3) on the diagonal.
Sure, I'd be happy to help you with the matrix proof!
To prove that determinant A is equal to 1, we need to start from the given equation A^2 = A^(-1).
First, let's start by taking the determinant of both sides of the equation.
|A^2| = |A^(-1)|
According to the determinant property, the determinant of a product of matrices is equal to the product of their determinants. Therefore, we can simplify the equation as:
|A|^2 = |A|^(-1)
Next, we can multiply both sides of the equation by |A| to get rid of the exponent on the right-hand side:
|A|^2 * |A| = |A|^(-1) * |A|
Simplifying further:
|A|^3 = 1
Now, since we know that the determinant of A cubed is equal to 1, we can conclude that the determinant of A must be the cube root of 1:
|A| = 1^(1/3)
Finally, simplifying:
|A| = 1
Therefore, we have shown that the determinant of matrix A is equal to 1.
I hope this explanation helps! If you have any further questions, please let me know.