I want to know why there is no solution $A_{n\times n}$ for $A^2 = -I$ , when the dimension of A is odd. The question is to find a matrix $A_{n\times n}$ for $A^2 = -I$. In geometric, I can imagine that A is a rotation transformation(rotate by $\frac{\pi}{2}$), and for even number n, that can be popularized by block matrix.(Then I found the solution for even number n)
However, I can't verify why there is no solution for odd n.
To summarize, I want know why there is no solution for $A^2 = - I$ , when square matrix A is  odd demension.

Since $n$ id odd, characteristic polynomial of $A$ has a real root (i.e. any odd degree polynomial has a real root). Let $\lambda$ be eigenvalue and $x$ corresponding eigenvector (nonzero). Then, since $Ax = {\lambda}x$ we get $A^2x = {\lambda}^2x$.
At the same time, if $A^2 = -I$ then $A^2x = -x$ so, for a real number ${\lambda}^2 = -1$
