# Relation between matrix power and Jordan normal form

(a) Assume $$A\in\mathbb{C}^{n\times n}$$ has $$n$$ distinct eigenvalues. Prove that there are exactly $$2^n$$ distinct matrices $$B$$ such that $$B^2 = A$$ (i.e., in particular, there are no more than $$2^n$$ matrices with this property).

(b) How many such matrices $$B\in\mathbb{C}^{3\times3}$$ exist if $$A=\begin{pmatrix}2&0&0\\0&2&0\\0&0&1\end{pmatrix}$$? Why?

(a) It is clear that if $$\lambda_i$$ is an eigenvalue of $$A$$, then $$\pm\sqrt\lambda_i$$ is an eigenvalue of $$B$$, therefore $$B = \begin{pmatrix}\pm\sqrt\lambda_1&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&\pm\sqrt\lambda_n\end{pmatrix}$$ satisfies $$B^2=A$$ and there are $$2^n$$ such matrices $$B$$ because of $$2^n$$ possibilities of rearranging different numbers of plus and minus signs on the diagonal.

(b) We can construct $$2^3=8$$ such matrices $$B$$ using approach of (a), but $$B$$ is not necessarily diagonal. Consider $$B=\begin{pmatrix}\sqrt2&-1&0\\0&-\sqrt2&0\\0&0&1\end{pmatrix}$$ which satisfies $$B^2=A$$ as well. So, there exist more than $$2^3$$ such matrices.

I can't give an explanation why there are no more than $$2^n$$ possibilities in the first case and more than $$2^3$$ in the second. It definitely follows from the fact that Jordan normal form is unique in (a) and isn't unique in (b) because of an eigenvalue with algebraic multiplicity $$2$$, however, I can't formulate it into a self-contained statement.

• Did you mean $$B=\begin{pmatrix}\sqrt2&-1&0\\0&-\sqrt2&0\\0&0&1\end{pmatrix}?$$ – Angina Seng Jun 29 '19 at 16:50
• @LordSharktheUnknown, yes, it was a typo and I edited the question. Thank you. – Hasek Jun 29 '19 at 17:06

Can we reduce finding matrix roots to finding roots of Jordan blocks?

In (b), your matrix $$A$$ has $$2$$ Jordan blocks associated to the same eigenvalue $$2$$. Then $$A$$ admits an infinity of square roots.

Here the infinity of matrices $$B\in M_2(\mathbb{C})$$ s.t. $$tr(B)=0,\det(B)=-2$$ are among the square roots of $$2I_2$$.

EDIT. I forgot to write that, in case (a), there are no more than $$2^n$$ square roots because, necessarily $$AB=BA$$.

You showed that $$A$$ has at least $$2^n$$ square roots. Assume $$B^2 = A$$. Since $$\sigma(B)^2 =\sigma(B^2) = \sigma(A)$$, it follows that $$\sigma(B) = \{\varepsilon_1\sqrt{\lambda_1}, \ldots, \varepsilon_n\sqrt{\lambda_n}\}$$ for some $$\varepsilon_1, \ldots, \varepsilon_n \in \{-1,1\}$$ where $$\sigma(A) = \{\lambda_1, \ldots \lambda_n\}$$. Set $$D = \operatorname{diag}(\varepsilon_1\sqrt{\lambda_1}, \ldots, \varepsilon_n\sqrt{\lambda_n})$$. $$A$$ has $$n$$ distinct eigenvalues so it is diagonalizable, i.e. there exists an invertible matrix $$P$$ such that $$P^{-1}AP = \operatorname{diag}(\lambda_1, \ldots, \lambda_n) = D^2$$.

Assume that $$C^2 = A$$ and that $$\sigma(C) = \sigma(B)$$. $$B$$ and $$C$$ both have $$n$$ distinct eingenvalues so there exist invertible matrices $$Q,R$$ such that $$Q^{-1}BQ = D = R^{-1}CR$$ It follows $$Q^{-1}AQ = Q^{-1}B^2Q = (Q^{-1}BQ)^2 = D^2 = (R^{-1}CR)^2 = R^{-1}C^2R = R^{-1}AR$$

If $$e_1, \ldots, e_n$$ is the standard basis, it follows that $$A(Pe_i) = \lambda_iPe_i, A(Qe_i) = \lambda_iQe_i, A(Re_i) = \lambda_iRe_i$$.

Since $$\lambda_1, \ldots, \lambda_n$$ are all distinct, the eigenvectors are unique up to scalar multiplication so it follows that $$Qe_i = \sigma_i Pe_i, Re_i = \pi_i Pe_i$$ for some nonzero scalars $$\sigma_i, \pi_i$$, or $$Q = P\Sigma, R = P\Pi$$ for some invertible diagonal matrices $$\Sigma, \Pi$$.

Finally, $$B = QDQ^{-1} = (P\Sigma)D(P\Sigma)^{-1} = P(\Sigma D\Sigma^{-1})P^{-1} =\\ PDP^{-1} = P(\Pi D\Pi^{-1})P^{-1} = (P\Pi)D(P\Pi)^{-1} = R^{-1}DR = C$$

Hence $$B$$ is uniquely determined by its spectrum $$\sigma(B)$$, which is uniquely defined by choosing $$\varepsilon_i \in \{-1,1\}$$ which can be done it $$2^n$$ ways. We conclude that there are $$2^n$$ such matrices $$B$$.