Compute $f_A(\lambda)$ without factoring cubic polynomial?

Question

I'm given the following prompt:

"Find the points closest to the origin on the surface defined by $x_1^2+2x_2^2+3x_3^2+x_1x_2+2x_1x_3+3x_2x_3=1$."

What's the easiest way to compute the eigenvalues of $$A=\begin{pmatrix}1&\tfrac{1}{2}&1\\\tfrac{1}{2}&2&\tfrac{3}{2}\\1&\tfrac{3}{2}&3\end{pmatrix}?$$

I've messed around with $$f_A(\lambda)=\det(A-\lambda I_3)=\det\begin{pmatrix}1-\lambda&\tfrac{1}{2}&1\\\tfrac{1}{2}&2-\lambda&\tfrac{3}{2}\\1&\tfrac{3}{2}&3-\lambda\end{pmatrix}$$

to no avail. There's usually an elementary row operation hidden in plain view that greatly reduces the number of calculations needed - is there one here?

I end up getting $\det(A-\lambda I_3)=(\tfrac{1}{8})(-16\lambda^3+44\lambda^2-62\lambda+30)$, which is no fun.

Answer 1

I think you may have miscalculated your determinant. I get $$-\lambda^3 + 6\lambda^2 - \frac{15}{2} \lambda + \frac{5}{2}$$ for which you can check 1 is a root (usually a guess worth trying). I suppose you can proceed from there.

Also, a handy way to check consistency is to see if the trace $\text{Tr}(A)$ equals the $\lambda^2$ coefficient, since they both should equal the sum of eigenvalues. This does not happen in your case, so try reviewing your steps.

Answer 2

$$ \begin{pmatrix} 0 &\tfrac{1}{2}&1\\ \tfrac{1}{2}& 1 & \tfrac{3}{2}\\ 1&\tfrac{3}{2}&2 \end{pmatrix}$$ (i.e. $\lambda=1$) is singular, since the second row is double the sum of the first and third rows. Further, now that you have one eigenvalue, you don't even need the characteristic equation: you can easily find $\det{A}= 1\lambda_2 \lambda_3 =5/2$ and $\operatorname{tr}{A} = 1+ \lambda_2 + \lambda_3 = 6 $, so the remaining eigenvalues solve the quadratic $$ 0 = \lambda^2 - (\operatorname{tr}{A}-1)\lambda + 1\det{A} = \lambda^2 -5\lambda + \frac{5}{2}. $$