I need to find the characteristic polynomial of $B$ such that $B^2=\left[\begin{array}{cc}5&-3\\-3&5\end{array}\right]$. One can find the characteristic polynomial of the given matrix and show that the roots are $2$ and $8$, so that we know the eigenvalues of $B^2$ are $2$ and $8$. But that only implies that eigenvalues of $B$ can be any two of the 4 values $\pm\sqrt{2},\pm\sqrt{8}$. So how can one determine the exact characteristic polynomial of $B$?
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1Such $B$ is not unique. For example, $B$ and $-B$. Thus any $B$ whose characteristic polynomial has roots $\pm \sqrt{2}$ and $\pm \sqrt{8}$ is suffice. Thus overall there will be 4 choices. – induction601 Nov 28 '17 at 06:13
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Four matrices satisfy
$${B}^{2} = \left[\begin{array}{cc}5&{-3}\\ {-3}&5 \end{array}\right]$$
they are
$$B = \pm \frac{\sqrt{2}}{2} \left[\begin{array}{cc}{-3}&1\\ 1&{-3} \end{array}\right] \quad \text{or} \quad B = \pm \frac{\sqrt{2}}{2} \left[\begin{array}{cc}1&{-3}\\ {-3}&1 \end{array}\right]$$
There are also four characteristic polynomials
$$\det \left(t I-B\right) = {t}^{2} \pm 3 \sqrt{2} t+4 \quad \text{or} \quad \det \left(t I-B\right) = {t}^{2} \pm \sqrt{2} t-4$$
This can be obtained by solving directly
$$\left[\begin{array}{cc}5&{-3}\\ {-3}&5 \end{array}\right] = {\left[\begin{array}{cc}a&b\\ c&d \end{array}\right]}^{2} = \left[\begin{array}{cc}{a}^{2}+b c&b \left(a+d\right)\\ c \left(a+d\right)&{d}^{2}+b c \end{array}\right]$$
This implies ${a}^{2} = {d}^{2}$ and $a+d \neq 0$, hence $a = d$ and $b = c$. It remains ${a}^{2}+{b}^{2} = 5$ and $2 a b =-3$. It follows that ${\left(a+b\right)}^{2} = 2$ and ${\left(a-b\right)}^{2} = 8$, hence $a+b = {\varepsilon} \sqrt{2}$ and $a-b = 2{{\varepsilon}'} \sqrt{2}$ with ${\varepsilon} = \pm 1$ and ${{\varepsilon}'} = \pm 1$. Hence
$$a = d = \left({\varepsilon}+2 {{\varepsilon}'}\right) \frac{\sqrt{2}}{2} \quad \text{and} \quad b = c = \left({\varepsilon}-2 {{\varepsilon}'}\right) \frac{\sqrt{2}}{2}$$
Gribouillis
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