Given A is a square matrix.
A is diagonalizable and has eigenvalues which are real (and bigger than zero).
Is it necessary true that $A = S^2$? I believe it is, anyone have any ideas how to solve it?
I know that if A is diagonalizable, then it has a matrix - $D$ which diagonalizes A.
I mean: $P^{-1}AP = D$
But how can I deduce from here that $A = S^2$ ?
You can take the square root of the entries in your diagonal matrix, then conjugate by $P$ to find your matrix $S$ for which $S^2=A$.
That is if $$D = diag( d_1,..., d_n)$$ then we write $$\sqrt{D} = diag(\sqrt{ d_1}, ..., \sqrt{d_n})$$
Then we have $\sqrt{D}^2 = D$. Now compute $S = P \sqrt{D} P^{-1}$, which means that $$S^2 = P \sqrt{D}^2 P^{-1} = P D P^{-1} = A$$