Find eigenvalues given the characteristic polynomial without finding the roots

Question

I am having trouble developing an approach to a problem about eigenvalues and the characteristic polynomial of a $3 \times 3$ matrix. The problem considers a $3 \times 3$ matrix with the characteristic polynomial $x^3 - 3x^2 + 5x +9$ with eigenvalues $\lambda_1, \lambda_2, \lambda_3$. It asks to give the value $\frac{1}{\lambda_1}+\frac{1}{\lambda_2}+\frac{1}{\lambda_3}$ but without computing the roots of the characteristic equation.

My first idea was to provide a formula for the characteristic polynomial of a $3 \times 3$ matrix by just deriving it out using the determinant, then algebraically working out each eigenvalue from that formula, but that seems like a very long and unneccessarily complicated approach. I'm stuck on what else I could try.

Answer 1

Given $\lambda^3 - 3\lambda^2 + 5\lambda +9=0$ has roots $\lambda_1,\lambda_2,\lambda_3$, use the Vieta's formulas: $$\frac{1}{\lambda_1}+\frac{1}{\lambda_2}+\frac{1}{\lambda_3}=\frac{\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3}{\lambda_1\lambda_2\lambda_3}=\frac{5}{-9}=-\frac59.$$

Answer 2

Hint:

$$(\lambda_1\lambda_2\lambda_3) \left( \frac1{\lambda_1} + \frac1{\lambda_2}+\frac1{\lambda_3}\right)=\lambda_2\lambda_3+\lambda_1\lambda_3+\lambda_1\lambda_2$$

Some of expression aboves can be read of from the coefficients of the charactheristic polynomial.

Answer 3

Alt. hint:   if the polynomial $\,P(x)=x^3 - 3x^2 + 5x +9\,$ has roots $\,\lambda_{1,2,3}\,$ then the polynomial $\,x^3 P\left(\dfrac{1}{x}\right)=1 - 3x+5x^2+9x^3\,$ has roots $\,\dfrac{1}{\lambda_{1,2,3}}\,$ and their sum is given by Vieta's relations.

Answer 4

$$ p(\lambda)=( \lambda - \lambda _1)( \lambda - \lambda _2 )( \lambda - \lambda _3)=\\ \lambda^ 3 -(\lambda_1 +\lambda_2 +\lambda_3) \lambda ^2 + (\lambda_1\lambda_3+\lambda_1\lambda_2 +\lambda _2 \lambda _3 ) \lambda -\lambda_1\lambda_2 \lambda_3 = \lambda^3 - 3\lambda^2 + 5 \lambda +9$$

Answer 5

If $\lambda$ is eigenvalue of $A$, then $1/\lambda$ is eigenvalue of $A^{-1}$. The problem asks for the trace of $A^{-1}$, hence we first find the characteristic polynomial of $A^{-1}$ by replacing $\lambda$ with $1/\lambda$ (and then multiplying with $\lambda^3$, i.e. the reciprocal polynomial. This is $1-3\lambda+5\lambda^2+9\lambda^3$.

The sum of its roots is (by Vieta formulas) 5/9.