How to show that $A$ has a positive determinant if it satisfies a certain polynomial equation? Let $A$ be a real $3\times 3$ matrix such that $A^3=tA+I_3$ for some $t\in (0, \sqrt[3]{4})$. Show that $A$ is invertible and $\det(A)\gt 0$.
 A: You can write $A (A^2-tI)=I$ and use the fact that the determinant of a product is the product of determinants for square matrices of same size (take the determinant of both sides).
You can then factor the LHS (det), which is 1, into product of three numbers and see if this helps with positivity.
A: The proof below shows that the result is true whenever $t\in (0,\frac{3}{\sqrt[3]{4}} )$
(which is slightly stronger than what the OP states).
The hypothesis says that $P=X^3-tX-1$ annihilates $A$. The eigenvalues
of $A$ must be roots of $P$. Since $P(0)=-1$, $0$ is not a root of $P$
so $A$ is invertible.
Since $P(0)<0$ and $\lim_{+\infty}P=\infty$, $P$ has a positive root
$\lambda$. Also, if we put $Q=\frac{P}{X-\lambda}$ then
$$Q=X^2+\lambda X+\lambda^2-t=\bigg(X+\frac{\lambda}{2}\bigg)^2+d$$
where
$$d=\frac{3\lambda^2}{4}-t$$
We are interested in the sign of $d$. Note that $d>0$ iff $\lambda > \sqrt{\frac{4t}{3}}$.
Putting $s=\sqrt{\frac{4t}{3}}$, we have $t=\frac{3s^2}{4}$ and
$$
P(s)=s^3-st-1=\frac{4t}{3}s-st-1=\frac{st}{3}-1=\frac{s^3}{4}-1
$$
so that
$$
P(s)<0 \Leftrightarrow s^3 < 4 \Leftrightarrow s < \sqrt[3]{4}  
\Leftrightarrow \frac{4t}{3} < 4^{\frac{2}{3}}  \Leftrightarrow 
t < \frac{3}{\sqrt[3]{4}} 
$$
Since $\sqrt[3]{4} <1.6< \frac{3}{\sqrt[3]{4}} $, this is indeed the case ; so $P(s)<0$
and we can choose $\lambda$ in $(s,\infty)$. Then $d<0$ and the roots of $P$ other than $\lambda$ (i.e. the roots of $Q$) are complex :
let us denote them by $\alpha\pm\beta i$ with $\alpha,\beta\in{\mathbb R},\beta \neq0$.
If $A$ has no complex eigenvalue, then $\lambda$ is the only eigenvalue of $A$
with multiplicity $3$, so $\det(A)=\lambda^3>0$ and we are done.
So we may assume that $A$ has at least one complex eigenvalue. Then its conjugate is also an eigenvalue since $A$ is real, so both $\alpha+\beta i$ and $\alpha-\beta i$ are eigenvalues of $A$. There is only one dimension left for the remaining eigenvalue, which must thus be real, so it must in fact be $\lambda$. So the eigenvalues of $A$ coincide
with the roots of $P$, and the determinant of $A$ is the products of the roots of $P$, which is the constant coefficient in $P$ : $\det(A)=1>0$. QED
A: The polynomial $p(x) = x^3 - tx - 1$ annihilates $A$. By Cardano's formula, the roots of $p(x)=0$ are given by
$$
\begin{cases}
a = S+T,\\
b = -\dfrac{S+T}2+\dfrac{i\sqrt{3}}2(S-T),\\
\bar{b} = -\dfrac{S+T}2-\dfrac{i\sqrt{3}}2(S-T),
\end{cases}
$$
where
\begin{align*}
S &= \sqrt[3]{\frac12 + \sqrt{\frac14 - \frac{t^3}{27}}},\\
T &= \sqrt[3]{\frac12 - \sqrt{\frac14 - \frac{t^3}{27}}}.
\end{align*}
When $0<t<\frac3{\sqrt[3]{4}}$ (in particular, when $0<t<\sqrt[3]{4}$), we have $S>T>0$. Hence $a>0$ and $b,\bar{b}$ are not real. Since non-real eigenvalues of a real matrix must occur in conjugate pairs, the spectrum of $A$ is either $\{a,a,a\}$ or $\{a,b,\bar{b}\}$. Now it is clear that $\det(A)>0$.
