Let $A\in M_{n*n}(\mathbb C)$ then there exists $B$ such that $B^2=A.$

Question

Let $A\in M_{n*n}(\mathbb C)$, then check the following statements with valid reasons:

1.$A$ is diagonaligable.

2.$A$ has a eigen value.

3.$A$ is upper traingularigable.

4.There exists $B$ such that $B^2=A.$

I think 1 and 2 is correct as the characteristic polynomial over $\mathbb C$ has roots over $ \mathbb C$. But I have no idea about 3 and 4.

Answer 1

Let $$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ then $A$ has $0$ the only eigenvalue $0$ so if $A$ is diagonalizable it would be similar to the zero matrix which's impossible. Why?

Since by the fundamental theorem of algebra every polynomial has a complex root then $2.$ is true. (Just we consider the characteristic polynomial).

$3.$ is true

Notice that $A$ is nilpotent with index $2$ so if there's $B$ such that $B^2=A$ then $B^4=0$ and then $B$ is also nilpotent and then $B^2=A=0$ which's a contradiction so $4.$ isn't true.