I need help with the following.
If $1$ is an eigenvalue of $A^2$ , $A$ is a $2 \times 2$ matrix, then $1$ is an eigenvalue of $A$.
If $A$ is a reflective matrix, then $A^2 = I$.
Eigenvalues of matrice.
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eigenvalues-eigenvectors
3 Answers
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If $1$ is an eigenvalue of $A^2$, then $1$ and(!) $-1$ are possible eigenvalues of $A$.
Example: $A=-I_2$. $-1$ is an eigenvalue of $A$, $1$ is not an eigenvalue of $A$.
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If $A^2$ has the Eigenvalue $1$, with the corresponding Eigenvector $v$ we have
$$(A^2-I)v=(A+I)(A-I)v=(A-I)(A+I)v=0.$$
This can arise with both $1$ or $-1$ for the Eigenvalue of $A$.
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You can power matrix by this identity: $A^n = BD^nB^{-1}$, where $D$ is the diagonal matrix with the eigenvalues, you got from solving the equation $det(A-\lambda E) = 0$, on its diagonal. $B$ is the matrix that consists of the eigenvectors. So even $A^n$ have the eigenvalue $\lambda = 1$.