I came across the following question:

The characteristic polynomial of a $3 \times 3$ matrix $A$ is $|\lambda I -A| = \lambda^3 + 3 \lambda^2+4 \lambda +3$. Find $trace(A)$ and $det(A)$.

I know that I have to find the eigenvalues $x$ from the characteristic equation $x^3 + 3 x^2+4 x +3 =0$, then add them to find $trace(A)$ and multiply them to find $det(A)$. However, I am not able to find roots the cubic equation.

I tried finding its solution by del Ferro's depressed cubic form (explained in Sec-$1.3$ of Complex Analysis by Newman and Bak). I replaced $x$ by $y-1$ in the characteristic equation and got the cubic equation $y^3+y+1=0$. On solving further as per the method, I got

$x=u+v= \sqrt[3]{\frac{-1}2+ \sqrt{\frac 14 + \frac 1{27}}}+\sqrt[3]{\frac{-1}2- \sqrt{\frac 14 + \frac 1{27}}}$.

I cannot solve it any further, hence can't find trace or determinant. Any idea how to it solve further to find the root. Also any idea on how to do it by any other method (like by writing down its companion matrix - I don't know how to do it).

Thank you.


It is hard to find out the roots, but all we need is the sum and product of all the roots. In a polynomial $x^3+ax^2+bx+c$, the sum of all roots is simply $-a$, which is $trace(A)$, and the product is $(-1)^3c$ which is $det(A)$.

To sum up, $trace(A)=-3$, $det(A)=-3$.

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    $\begingroup$ Why do we always sum up, and never multiply out? $\endgroup$ – Gerry Myerson Jun 18 '15 at 9:53
  • $\begingroup$ @GerryMyerson, love your comment!! $\endgroup$ – Indominus Jun 18 '15 at 9:57
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    $\begingroup$ @Indominus Well, since $c=3$ don't you think $det(A)=(-1)^3c=-3$? ............Anyways, I missed that part completely... Thanks.... $\endgroup$ – Ritu Jun 18 '15 at 10:20
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    $\begingroup$ Yes, you are right, thanks! Let $\lambda = 0$, we can get $c=det(-A)$, so $det(A)=(-1)^3c$ $\endgroup$ – Indominus Jun 18 '15 at 10:28

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