If $A^2=I$ and $\lambda \not =-1$, then $A=I$

Question

Given an $n\times n$ matrix $A$ with $A^2=I$, assume that $-1$ is not an eigenvalue of $A$. Prove $A=I$.

Proof attempt: Since $A^2=I,$ we have $A^{-1}=A$. Using that fact that $\det(A)=\prod_{i=1}^n\lambda_i$, we see that each of the eigenvalues is nonzero. For any eigenvector $v$ of $A$, we see $Av=\lambda v$, so $\frac{1}{\lambda} v = A^{-1}v=Av$. This implies $\lambda^2=1$. By assumption, we conclude that each eigenvalue is $1$. Hence, $\det A=1$.

(1) How do we know that $A$ actually has any eigenvectors? E.g., the map which rotates a vector in the plane by $\pi/2$ clockwise is invertible but has no eigenvectors. (Perhaps there should be some given information about the underlying field.)

(2) How do I use $\lambda=1$ to conclude $A=I$?

I tried using $A=A^{-1} = \frac{C^T}{\det A}=C^T$, where $C$ is the matrix of cofactors of $A$. From this, I conclude that each entry of $A$ equals its corresponding co-factor -- i.e., $A_{i,j} = \det(\overline{A_{i,j}})$, where $\overline{A_{i,j}}$ is the sub-matrix of $A$ obtained by deleting row $i$ and column $j$. Using the co-factor expansion of the determinant along row $i$, we also conclude $$1=\det A=\sum_{j=1}^n A_{i,j}(-1)^{i+j}\det(\overline{A_{i,j}})=\sum_{j=1}^n(A_{i,j})^2(-1)^{i+j}.$$

Answer 1

Hint: Compute $(A+I)(A-I)$ and then think about what the fact that $-1$ is not an eigenvalue tells you about $A+I$.

Answer 2

Remember that a square matrix is diagonalizable iff its minimal polynomial is the product of different linear factors. We're given $\;A^2=I\iff (A-I)(A+I)=0\implies\;$ the minimal polynomial of $\;A\;$ is a product of different linear factors and thus it is diagonalizable.

Now use that $\;-1\;$ is not an eigenvalue...

Answer 3

$(A+I)^2=A^2+2A+I=I+2A+I=2(A+I)$. Since $A+I$ is invertible, we get $A+I=2I$, hence $A=I.$

Answer 4

$A+I=A+A^2=A^2+A=A(A+I)$. Since $A+I$ is invertible , we derive $I=A.$