Calculating the trace of a 3x3 matrix over the rationals when its determinant is 0
I encountered a problem where I had a $3\times 3$ matrix $A$ with rational numbers entries and it was given that $A^3=2A$ and $A$ could not be the zero matix. I was asked to calculate the trace of the square of $A$, namely $tr(A^2)$.
I could find that the determinant $\det(A)=0$ because otherwise it had to be $\sqrt8$ or $-\sqrt8$. This comes from $\det(A)^3=8\det(A)$. Since $\sqrt8$ is not rational the determinant has to be $0$. Is there a way to express $tr(A^2)$ in terms of $\det(A)$?
$\endgroup$ 12 Answers
$\begingroup$If $A^3=2A$, then its characteristic polynomial is $x^3-2x$.
Indeed, $x^3-2x=x(x^2-2)$ is a decomposition into irreducibles in $\mathbb Q[x]$. Now, the minimal polynomial of $A$ cannot be $x$ because $A\ne0$ or $x^2-2$ because then $\det(A)^2=8$, which cannot happen over $\mathbb Q$. Therefore, the minimal polynomial is $x^3-2x$ and so coincides with the characteristic polynomial.
The characteristic polynomial of a $3\times3$ matrix $A$ is$$ x^3 - tr(A) x^2 + \frac{tr(A)^2 - tr(A^2)}{2}x - \det(A) $$
See characteristic polynomial in terms of trace and determinant for 4x4 matrices.
Comparing coefficients, we get$tr(A)=0$ and so $tr(A^2)=4$.
$\endgroup$ 4 $\begingroup$Experiment with$$ A = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 2 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$
For example, find $A^3$ and compare that with $2A$
Also find $A^2$ and its trace
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