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Infinite number of square roots of a complex matrix
I asked given $A \in \mathcal{M}_n(\mathbb{C})$ diagonalizable that has a non-zero eigenvalue with a multiplicity greater than 2 whether we could find an infinite number of diagonalizable matrices $M$ such as $M^2=A$.
The answer I had was :
Which is really helpful. But I have no clue why the matrix $B$ is diagonalizable, for me $B$ as presented here is triangularizable.
Could anyone help me further please?
Hint: An upper triangular matrix with different diagonal entries is always diagonalizable (with the same diagonal), as it has $n$ different eigenvalues, hence an eigenbasis (or, prove by using elementary basis transformations).
In particular, $\pmatrix{1&z\\0&-1}$ is similar to $\pmatrix{1&0\\0&-1}$.