Enumerating the square roots of a $2 \times 2$ matrix using its diagonalization and diagonalizability

Question

Enumerating the square roots of a $2 \times 2$ complex matrix using its eigenvalues and diagonalizability.

---

My answer:

Cases:

• If the eigenvalues are equal then infinitely many.

• If all of it's eigenvalues are distinct it has 4 unless one of them is zero and then 2.

• If it is not diagonalizable then it doesn't have any square roots.

My questions:

1. I'm very unsure about case 3, and how to prove it.

2. In the first two cases all the options are diagonalizable, so does it mean all the matrices who square to a diagonalizable matrix are diagonalizable?

Answer 1

I suspect that the first case is supposed to refer to diagonalizable matrices with equal eigenvalues. If so, then first two statements regarding diagonaliziable matrices are correct. A correct version of the third case is as follows.

Claim: Let $A$ be a non-diagonalizable $2\times 2$ complex matrix. If $A$ is has $0$ as its eigenvalue, then $A$ has no square root. If $A$ has a non-zero eigenvalue, then $A$ has two square roots.

Proof: For the case that $A$ has zero as an eigenvalue, suppose for the sake of contradiction that there exists a matrix $B$ such that $B^2 = A$. Because all eigenvalues of $A$ must be zero, all eigenvalues of $B$ must also be zero. By the Cayley-Hamilton theorem, this implies that $B^2 = 0$. Thus, we have $A = B^2 = 0$, contradicting the fact that $A$ was non-diagonalizable.

For the case that $A$ has an eigenvalue $\lambda \neq 0$: the existence of Jordan form implies that there exists a matrix $N\neq 0$ such that $$ A = \lambda I + N $$ where $I$ denotes the identity matrix. Let $B$ be a matrix such that $B^2 = A$. It follows that $$ BA = BB^2 = B^3 = B^2B = AB. $$ It can be shown that if $BA = AB$, there must exist $p,q\in \Bbb C$ such that $$ B = pI + qN. $$ With that, we have $$ B^2 = A \implies p^2 I + 2pq N = \lambda I + N \implies \begin{cases} p^2 = \lambda \\ 2pq = 1 \end{cases} \implies\\ p = \pm \sqrt{\lambda}, \quad q = \frac 1{2 p}. $$ Thus, $A$ indeed has exactly $2$ square roots.

---

Claim: Suppose that $A = \lambda I + N$ and that $AB = BA$. Then, there exist $p,q$ such that $B = pI + q N$.

Proof: Because $N$ is non-zero nilpotent, there exists a basis $\{v_1,v_2\}$ of $\Bbb C$ such that $Nv_1 = 0$ and $Nv_2 = v_1$ (the Jordan basis of $N$). From the fact that $AB = BA$, conclude that $NB = BN$.

First, write $Nv_1 = pv_1 + kv_2$ for some $p,k \in \Bbb C$. We have $$ (NB)v_1 = N(pv_1 + kv_2) = kv_1,\\ (BN)v_1 = B(0) = 0. $$ Because $NB = BN$, we have $k = 0$.

Similarly, write $N v_2 = qv_1 + (p+k) v_2$. Note that $$ (NB)v_2 = N(qv_1 + (p+k)v_2) = (p+k)v_1,\\ (BN)v_2 = Bv_1 = pv_1. $$ Thus, we must have $k = 0$. That is, there exist $p,q \in \Bbb C$ such that $Bv_1 = pv_1$ and $Bv_2 = qv_1 + p v_2$.

It follows that $B = p I + qN$.