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I had a matrix algebra question. I know how to solve for $X$ if I am given $X^3$. I am stuck on this. Any help is much appreciated. Thanks!

Solve for $X$, if $$ X^3 = \begin{bmatrix} -6 & 14 \\ -7 & 15 \\ \end{bmatrix} $$

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  • $\begingroup$ Find an eigenvalue/Jordan factorization of $X^3$. Over $\mathbb{R}$, you should be able to prove that $X=\begin{bmatrix}0&2\\-1&3\end{bmatrix}$ is the only solution. Over $\mathbb{C}$, there are $9$ possible solutions. $\endgroup$ – Batominovski Dec 2 '15 at 6:40
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Hint: if $X^3$ can be diagonalized as $S D S^{-1}$, then $S D^{1/3} S^{-1}$ works, where $D^{1/3}$ is the diagonal matrix whose diagonal elements are cube roots of the diagonal elements of $D$.

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There are probably easier ways, but it can be done by brute force plus a little good guessing. First let

$$X=\begin{bmatrix}a&b\\c&d\end{bmatrix}\;.$$

Then

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}^2=\begin{bmatrix}a^2+bc&b(a+d)\\c(a+d)&bc+d^2\end{bmatrix}\;,$$

so

$$X^3=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}a^2+bc&b(a+d)\\c(a+d)&bc+d^2\end{bmatrix}=\begin{bmatrix}a^3+2abc+bcd&b(a+d)^2-b(ad-bc)\\c(a+d)^2-c(ad-bc)&abc+2bcd+d^3\end{bmatrix}\;.\tag{1}$$

Now the determinant of $X^3$ is $(-6)(15)-(14)(-7)=-90+98=8$, so the determinant of $X$ is $ad-bc=2$. Substitute that into $(1)$ to get

$$X^3=\begin{bmatrix}a^3+2abc+bcd&b\big((a+d)^2-2\big)\\c\big((a+d)^2-2\big)&abc+2bcd+d^3\end{bmatrix}\;.$$

Now

$$b\big((a+d)^2-2\big)=14$$

and

$$c\big((a+d)^2-2\big)=-7\;,$$

so

$$-2=\frac{-14}7=\frac{b\big((a+d)^2-2\big)}{c\big((a+d)^2-2\big)}=\frac{b}c\;,\tag{2}$$

and therefore $b=-2c$. Substituting that into $(2)$ gives us

$$X^3=\begin{bmatrix}a^3-4ac^2-2c^2d&-2c(a+d)^2+4c\\c(a+d)^2-2c&-2ac^2-4c^2d+d^3\end{bmatrix}\;,\tag{3}$$

so

$$c\big((a+d)^2-2\big)=-7\;.$$

Here’s where we do a little guessing: it doesn’t hurt to see whether there is a solution in which $c$ and $(a+d)^2-2$ are integers. Clearly they would have to be $-1$ and $7$ or $1$ and $-7$ in some order. If we try $c=-1$, we want $(a+d)^2-2=7$, and $a+d=3$ does the trick. There’s no guarantee that this will work, but it’s worth a try. Substituting these values into $(3)$, we get

$$X^3=\begin{bmatrix}a^3-4a-2d&14\\-7&-2a-4d+d^3\end{bmatrix}=\begin{bmatrix}a^3-2a-6&14\\-7&-6-2d+d^3\end{bmatrix}\;.$$

To finish off, see whether you can find values of $a$ and $d=3-a$ that work.

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