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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!

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  • $\begingroup$ How did you find the Jordan normal form? $\endgroup$ – Marc van Leeuwen May 18 '16 at 4:39
  • $\begingroup$ @MarcvanLeeuwen You know, found the characteristic polynomial, found the geometric multiplicity and found the possible jordan blocks according to the length. $\endgroup$ – AriNubar May 18 '16 at 11:13
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    $\begingroup$ Seems to me like computing the characteristic polynomial requires either a computer algebra system or some judicious ad hoc techniques; certainly evaluating a general $8\times8$ determinant of $XI_8-A$ seems humanly impossible! $\endgroup$ – Marc van Leeuwen May 18 '16 at 11:20
  • $\begingroup$ @MarcvanLeeuwen You are right, I found it via computer. $\endgroup$ – AriNubar May 18 '16 at 11:24
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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 $$(P^{-1}DP)^2 = P^{-1}D(PP^{-1})DP = P^{-1}D^2P=P^{-1}CP=A $$

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  • $\begingroup$ Thanks, I had written a matrix which was the inverse of the element rather than the square root. I I am making the correction. $\endgroup$ – Mark Fischler May 18 '16 at 20:51
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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}

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    $\begingroup$ This is hilarious $\endgroup$ – ÍgjøgnumMeg May 17 '16 at 23:04
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    $\begingroup$ Thanks took work. (was bored wanted a technically correct different answer for no other reason then for the lols) $\endgroup$ – shai horowitz May 17 '16 at 23:06
  • $\begingroup$ Amazing thank you very much! $\endgroup$ – AriNubar May 17 '16 at 23:30
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    $\begingroup$ You broke math.stackexchange.com! I shudder to think what this looks like on a phone. $\endgroup$ – candied_orange May 18 '16 at 6:31
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    $\begingroup$ @StevenStadnicki "The hardest part of this answer by far would be the TeX conversion" this is what I was talking about. The time it took to type this up carefully would have been work in itself, and we shouldn't assume that the answerer just dumped it into alpha - they may very well have come up with the solution themselves. Regardless, the other answers are much easier to come up with (including thinking and typing) than this one. I understand why some may consider this to not be a "real" answer however - you don't really learn anything from this answer. $\endgroup$ – Irregular User May 18 '16 at 21:20
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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 $3 \times 3$:

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

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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.

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  • $\begingroup$ Thank you very much for your answer and the reference to your past treatment of the subject "this answer". $\endgroup$ – Jean Marie May 18 '16 at 6:02
  • $\begingroup$ Thank you for reminding, I added the field. $\endgroup$ – AriNubar May 18 '16 at 11:07

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