Prove Why $B^2 = A$ exists?

Question

Define $$A = \begin{pmatrix} 8 & −4 & 3/2 & 2 & −11/4 & −4 & −4 & 1 \\ 2 & 2 & 1 & 0 & 1 & 0 & 0 & 0 \\ −9 & 8 & 1/2 & −4 & 31/4 & 8 & 8 & −2 \\ 4 & −6 & 2 & 5 & −7 & −6 & −6 & 0 \\ −2 & 0 & −1 & 0 & 1/2 & 0 & 0 & 0 \\ −1 & 0 & −1/2 & 0 & −3/4 & 3 & 1 & 0 \\ 1 & 0 & 1/2 & 0 & 3/4 & −1 & 1 & 0 \\ 2 & 0 & 1 & 0 & 0 & 0 & 0 & 5 \\ \end{pmatrix} \in M_8(\mathbb R)$$

Justify why there is a matrix $B$ such that $B^2 = A$.

I think it has something to do with the Jordan normal form so I found it.

\begin{pmatrix} 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 5 \\ \end{pmatrix}

(I know the ones are under the diagonal, but I think it doesn't make a difference.)

I have no idea what should I do next, a help will be appreciated!

Thanks!

Answer 1

By putting $A$ into Jordan normal form you have found a non-singular matrix $P$ such that $$ A = P^{-1}CP $$ where $C$ is your matrix shown above. It is easy to find a matrix sqare root of $C$, for example, in the upper left start with $\pmatrix{\sqrt{2} &0\\\frac14\sqrt{2}& \sqrt{2}}$

Then $$(P^{-1}DP)^2 = P^{-1}D(PP^{-1})DP = P^{-1}D^2P=P^{-1}CP=A $$

Answer 2

The theory reasons why are given above. To prove this however one simply needs to find a matrix such that $B^2=A$ in this case

$B=$ \begin{array}{cccccccc} \frac{1}{90} \left(-\frac{75}{\sqrt{2}}-70 \sqrt{2}+181 \sqrt{5}\right) & -\frac{2}{\sqrt{5}} & \frac{1}{180} \left(-\frac{75}{\sqrt{2}}-70 \sqrt{2}+91 \sqrt{5}\right) & \frac{1}{\sqrt{5}} & \frac{1}{120} \left(-\frac{75}{\sqrt{2}}-130 \sqrt{2}+73 \sqrt{5}\right) & \frac{4}{9} \left(\frac{3}{\sqrt{2}}+10 \sqrt{2}-\frac{17 \sqrt{5}}{2}\right) & \frac{4}{45} \left(\frac{15}{\sqrt{2}}+20 \sqrt{2}-\frac{43 \sqrt{5}}{2}\right) & \frac{1}{2 \sqrt{5}} \\ \frac{15}{16 \sqrt{2}} & \sqrt{2} & \frac{15}{32 \sqrt{2}} & 0 & \frac{29}{64 \sqrt{2}} & 0 & 0 & 0 \\ \frac{1}{90} \left(\frac{285}{\sqrt{2}}+320 \sqrt{2}-362 \sqrt{5}\right) & \frac{4}{\sqrt{5}} & \frac{1}{180} \left(\frac{285}{\sqrt{2}}+320 \sqrt{2}-182 \sqrt{5}\right) & -\frac{2}{\sqrt{5}} & \frac{1}{120} \left(\frac{285}{\sqrt{2}}+260 \sqrt{2}-146 \sqrt{5}\right) & \frac{1}{9} (-8) \left(\frac{3}{\sqrt{2}}+10 \sqrt{2}-\frac{17 \sqrt{5}}{2}\right) & \frac{1}{45} (-8) \left(\frac{15}{\sqrt{2}}+20 \sqrt{2}-\frac{43 \sqrt{5}}{2}\right) & -\frac{1}{\sqrt{5}} \\ \frac{1}{12} \left(\frac{21}{2 \sqrt{2}}-8 \sqrt{2}+8 \sqrt{5}\right) & 2 \left(\sqrt{2}-\sqrt{5}\right) & \frac{1}{24} \left(\frac{21}{2 \sqrt{2}}-8 \sqrt{2}+8 \sqrt{5}\right) & \sqrt{5} & \frac{1}{16} \left(\frac{5}{2 \sqrt{2}}+40 \sqrt{2}-40 \sqrt{5}\right) & \frac{1}{3} (-2) \left(\frac{3}{\sqrt{2}}+5 \sqrt{2}-5 \sqrt{5}\right) & \frac{1}{3} (-2) \left(\frac{3}{\sqrt{2}}-\sqrt{2}+\sqrt{5}\right) & 0 \\ -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{2 \sqrt{2}} & 0 & \frac{5}{4 \sqrt{2}} & 0 & 0 & 0 \\ -\frac{1}{2 \sqrt{2}} & 0 & -\frac{1}{4 \sqrt{2}} & 0 & -\frac{3}{8 \sqrt{2}} & \frac{5}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & 0 \\ \frac{1}{2 \sqrt{2}} & 0 & \frac{1}{4 \sqrt{2}} & 0 & \frac{3}{8 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{3}{2 \sqrt{2}} & 0 \\ \frac{1}{2} \left(-\frac{1}{\sqrt{2}}-2 \sqrt{2}+2 \sqrt{5}\right) & 0 & \frac{1}{4} \left(-\frac{1}{\sqrt{2}}-2 \sqrt{2}+2 \sqrt{5}\right) & 0 & \frac{1}{8} \left(-\frac{3}{\sqrt{2}}-2 \sqrt{2}+2 \sqrt{5}\right) & 0 & 0 & \sqrt{5} \\ \end{array}

Answer 3

I borrow (and slightly modify) an explanation that can be found in the excellent thread https://mathoverflow.net/questions/14106/finding-the-square-root-of-a-non-diagonalizable-positive-matrix

$f$ being any function, sufficient derivable, one can write, for example for a $4 \times 4$ Jordan block:

$$f\left( \left[ \begin{array} [c]{cccc}% \lambda & 1 & & \\ & \lambda & 1 & \\ & & \lambda & 1\\ & & & \lambda \end{array} \right] \right) =\left[ \begin{array} [c]{cccc}% f\left( \lambda\right) & f^{\prime}\left( \lambda\right) & \frac{1} {2!}f^{\prime\prime}\left( \lambda\right) & \frac{1}{3!}f^{\prime \prime\prime}\left( \lambda\right) \\ & f\left( \lambda\right) & f^{\prime}\left( \lambda\right) & \frac {1}{2!}f^{\prime\prime}\left( \lambda\right) \\ & & f\left( \lambda\right) & f^{\prime}\left( \lambda\right) \\ & & & f\left( \lambda\right) \end{array} \right]$$

(it is based on a certain Taylor expansion of $I+N$ where $N$ is a nilpotent matrix).

It suffices now to take $f(x)=\sqrt{x}$ to get for the square root of a Jordan block of similar form as above but $3 \times 3$:

$$\left[ \begin{array} [c]{ccc}% \sqrt{\lambda} & \dfrac{1}{2\sqrt{\lambda}} & -\dfrac{1}{8 (\lambda)^{3/2}} \\ & \sqrt{\lambda} & \dfrac{1}{2\sqrt{\lambda}} \\ & & \sqrt{\lambda} \end{array} \right]$$

Answer 4

The question does not specify the coefficient field. If it is the complex numbers, then it would have sufficed to check that the matrix is nonsingular, since every nonsingular complex matrix has a square root. If the fields is the rational numbers, then no square root exists for this matrix. If it is the real numbers, then it suffices that all (complex) eigenvalues are real and positive.

In general for a nonsingular matrix over a field $K$ of characteristic$~0$, having a split characteristic polynomial all of whose roots are squares in$~K$ (so positive numbers in case $K=\Bbb R$) is a sufficient condition for having a matrix square root defined over$~K$. The proof of the latter is the same as for the complex case: split each restriction $R_\lambda$ to a generalised eigenspace for $\lambda\in K$ as $R_\lambda=\lambda I+N$ with $N$ nilpotent, then $R_\lambda$ has square root $\sqrt\lambda\sum_{k=0}^{d-1}\binom{1/2}kN^k$ where $\sqrt\lambda$ is some square root of$~\lambda$ in$~K$, and $d$ is the degree of nilpotency of $N$. (Characteristic$~0$ is used for the binomial coefficients. Note that one does not need to find a Jordan normal form, just the generalised eigenspaces.)

For references to other matrix square root questions see this answer.

Added: In fact the condition of characteristic$~0$ can be weakened to characteristic not$~2$, since the binomial coefficients are well defined for that case after cancelling all odd factors from the denominator in the definition. Indeed, since only the existence of appropriate coefficients matters here, it suffices to remark that $1+X$ has an $n$-th root in the formal power series ring $K[[X]]$ whenever $n$ is invertible in$~K$ (here for $n=2$), which is easy to show through a process of Hensel lifting.