Reflect And Transform

A transformation with eigenvalues $1,-1,-1$ is not necessarily a reflection about a line.

solution 2 is very brilliant if you add $A$ is symmetric, nevertheless solution 1 is not a solution. One matrix $A$ is able to have to $det(A - \lambda * I) = - (\lambda + 1)^2(\lambda - 1)$ and it has not to be a diagonalizable matrix and much less to be a reflection about a line nor an orthogonal matrix. Solution 2 is very brilliant,and I'm going to say you why. Since $A^2 - I = 0$ it implies that its minimum polynomial $p_m (x) \text{ is } (x - 1) \text{ or } (x + 1) \text { or } (x -1)(x + 1)$ . The two first cases doesn't work for $A$ , so $p_m (x) = (x - 1)(x + 1) \Rightarrow$ the matrix $A$ is diagonalizable with eigenvalues 1, -1 due to the theorem: One square matrix is diagonalizable if and only if its minimum polynomial can be descomposed as one product of linear factors not repeated and because every eigenvalue is a root of the $p_m (x)$ ,and due to $det(A) = 1$ the only possibility is a reflection about a line because the eigenvalues will be 1, -1 and -1 and because $A$ is symmetric and therefore in this case $A$ is an orthogonal matrix.