# Thread: A^4 = I and det(A) = -1, show that A^2 = I and Tr(A) = 0

1. ## A^4 = I and det(A) = -1, show that A^2 = I and Tr(A) = 0

assume A is a 2x2 matrix with real entries and A^4 = I and det(A) = -1, show that A^2 = I and Tr(A) = 0. so i was thinking of using eigenvalues and eigenvectors for this problem but i'm not sure because it's not guaranteed that the eigenvalues will be real. if they are real then i can just find an upper triangular matrix and show A^2 = I. do the assumptions imply that the eigenvalues will be real?

2. $\displaystyle A^4-I=0$ implies $\displaystyle p(\lambda)=\lambda^4-1=(\lambda^2+1)(\lambda-1)(\lambda+1)$ is an annihilator polynomial of $\displaystyle A$. So, you can determine the possible minimal polynomials of $\displaystyle A$. Analyzing all the possible characteristic polynomials of $\displaystyle A$ with the hypothesis $\displaystyle \det(A)=-1$ and $\displaystyle A$ real, you'll easily find that $\displaystyle A$ is similar to $\displaystyle \textrm{diag}(1,-1)$.

Regards.

Fernando Revilla

3. Originally Posted by oblixps
assume A is a 2x2 matrix with real entries and A^4 = I and det(A) = -1, show that A^2 = I and Tr(A) = 0. so i was thinking of using eigenvalues and eigenvectors for this problem but i'm not sure because it's not guaranteed that the eigenvalues will be real. if they are real then i can just find an upper triangular matrix and show A^2 = I. do the assumptions imply that the eigenvalues will be real?
I'm sorry. I misread the question.

4. sorry i am not familiar with annihilator polynomials. this is what i did. Ax=λx so (A^4)x = (λ^4)x and Ix = (λ^4)x then the characteristic polynomial will be λ^4 - 1 = (λ^2+1)(λ-1)(λ+1) so λ = 1, -1. λ could also be i and -i. i am not sure how to determine the minimal polynomials of A. could you explain that a little more? thanks!

5. Originally Posted by oblixps
sorry i am not familiar with annihilator polynomials.
No problem, there are alternatives.

this is what i did. Ax=λx so (A^4)x = (λ^4)x and Ix = (λ^4)x
Right.

then the characteristic polynomial will be λ^4 - 1 = (λ^2+1)(λ-1)(λ+1)
That is not right. Take into account that characteristic polynomial has degree 2. According to your argument:

$\displaystyle \lambda^4 x=x\;(x\neq 0)\Rightarrow (\lambda^4-1)x=0 \Rightarrow \lambda^4=1 \Rightarrow \lambda \in \left\{{1,-1,i,-i}\right\}$

If $\displaystyle \lambda_1,\lambda_2$ are the eigenvalues of $\displaystyle A$ then:

$\displaystyle \lambda_1\lambda_2=\det (A)=-1$.

Now the possible couples for $\displaystyle (\lambda_1,\lambda_2)$ (up to permutations) are:

$\displaystyle (\lambda_1,\lambda_2)\in \left\{{ (1,-1),(i,i),(-i,-i) }\right\}$

But if a real matrix has a complex eigenvalue, also has its conjugate, so necessarily $\displaystyle (\lambda_1,\lambda_2)=(1,-1)$ (up to permutation).

This implies that $\displaystyle A$ is diagonalizable and

$\displaystyle A=PDP^{-1}$ for some regular matrix $\displaystyle P$ and $\displaystyle D=\textrm{diag}(1,-1)$

Conclusion:

(a) $\displaystyle A^2=PD^2P^{-1}=PIP^{-1}=I$

(b) $\displaystyle \textrm{tr}A=\textrm{tr}D=1-1=0$ (similar matrices have the same trace).

Regards.

Fernando Revilla

6. Wonderful explanation. Thanks for the useful post.