Solving the characteristic equation of a $3\times 3$ matrix to find the eigenvalues.

Question

I'm working through some problems in an old linear algebra textbook, just brushing up on a few things. I came across an exercise that's been giving me some trouble. The exercise states,

Find the characteristic equation and the real eigenvalues of the matrix $A=\begin{bmatrix} 4 & 0 & -1 \\ 0 & 4 & -1 \\ 1 & 0 & 2 \\ \end{bmatrix} $.

My attempted solution:

$$ A - \lambda I= \begin{bmatrix} 4 & 0 & -1 \\ 0 & 4 & -1 \\ 1 & 0 & 2 \\ \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \\ \end{bmatrix} = \begin{bmatrix} 4 - \lambda & 0 & -1 \\ 0 & 4 - \lambda & -1 \\ 1 & 0 & 2 - \lambda \\ \end{bmatrix} \\ $$ From here we calculate $|A-\lambda I|$ $$ \begin{align} |A-\lambda I| &= (4-\lambda)^2(2-\lambda)-(4-\lambda) \\ & =(\lambda^2-8\lambda+16)(2-\lambda)-(4-\lambda) \\ & =(2\lambda^2-16\lambda+32-\lambda^3+8\lambda^2-16\lambda)-(4-\lambda) \\ & =-\lambda^3+10\lambda^2-32\lambda+32-(4-\lambda) \\ & =-\lambda^3+10\lambda^2-31\lambda+28 \\ \end{align} $$ So, unless I've made errors in my calculations, the characteristic polynomial is $-\lambda^3+10\lambda^2-31\lambda+28$ and the characteristic equation is $-\lambda^3+10\lambda^2-31\lambda+28=0$.

This is where I'm stuck. I haven't had to find the roots of a cubic polynomial in over a decade. I know the quadratic function is used to find the roots of a polynomial with form $ax^2+bx+c=0$. So I looked for a cubic type of quadratic function and found The Cubic Formula, which seems overly complicated.

I believe the roots can be found using $(4-\lambda)^2(2-\lambda)-(4-\lambda)$ but I am unsure of how to go about it. I've been searching for a method but I haven't found one, can someone pleases shed some light on my problem.

Answer 1

Your approach is fine, but not your computations. The characteristic polynomial of $A$ is $-\lambda^3+10\lambda^2-33\lambda+36$. Using the rational root theorem, you can see that its roots are $3$ (twice) and $4$.

By the way, note that the characteristic polynomial is not $(4-\lambda)^2(2-\lambda)-(4-\lambda)$. In fact, it is $(4-\lambda)^2(2-\lambda)+(4-\lambda)$, and\begin{align}(4-\lambda)^2(2-\lambda)+(4-\lambda)&=(4-\lambda)\bigl((4-\lambda)(2-\lambda)+1\bigr)\\&=(4-\lambda)(\lambda^2-6\lambda+9)\\&=(4-\lambda)(\lambda-3)^2.\end{align}

Answer 2

Even before constructing the characteristic polynomial, you can see from the matrix itself that $(0,1,0)^T$ is an eigenvector with eigenvalue $4$. However, if you didn’t happen to spot this before plunging in, computing the determinant in a different way would have also given you this partial factorization right away: The second column only has one nonzero element, so it’s a prime candidate for computing the determinant. If you expand along the second column, you get $$(4-\lambda)\left((4-\lambda)(2-\lambda)+1\right) = (4-\lambda)(9-6\lambda+\lambda^2) = (4-\lambda)(3-\lambda)^2$$ for the characteristic polynomial.