Factoring a cubic polynomial?

Question

So I have a matrix

$$A = \begin{pmatrix} -5 & -6 & 3 \\ 3 & 4 & -3 \\ 0 & 0 & -2 \end{pmatrix} $$

I'm to find the characteristic polynomial and all the eigenvalues of $A$.

I have found $\det (A - \lambda I) = -\lambda^3 - 3\lambda^2 + 4$.

But I can't understand how to factor it to find the eigenvalues. The cubic factorization methods I've found don't seem to work on this one?

Any ideas?

Answer 1

$$-λ^3 - 3λ^2 + 4=-(λ^3 -1)-( 3λ^2 -3)=$$ $$=-(λ-1)(λ^2+λ+1)-3(λ-1)(λ+1)$$ $$=(λ-1)(-λ^2-λ-1-3λ-3)=$$ $$=(λ-1)(-λ^2-4λ-4)=$$ $$=(1-λ)(λ^2+4λ+4)=$$ $$=(1-λ)(λ^2+2\cdot2λ+2^2)=$$ $$=(1-λ)(λ+2)^2$$

Answer 2

The last row of $A$ contains all zeroes except the last element, so that element $\lambda =−2$ must be an eigenvalue. Then, divide:

$$ \frac{-x^3-3x^2+4}{x+2} = -\frac{x^3+3x^2-4}{x+2} = -(x^2+x-2) $$

Answer 3

The best way to factor a cubic is in my opinion by using the rational root theorem. This states that a rational number $\frac{p}{q}$ is a root of a polynomial $\sum_{k=0}^na_kx^k,$ then $p$ is a factor of $a_0$ and q is a factor of $a_n$. Since you have a cubic, you know that the end behavior tends towards opposite infinities, and also, irrational roots and complex roots come in pairs, so you must have 1 rational root with a cubic. So for your problem, the factors of $4$ are $\pm{1}, \pm{2},$ and $\pm{4}$ and the factors of $-1$ are obviously $\pm{1}$. So your rational roots can be just $\pm{1}, \pm{2}, \pm{4}$.

Answer 4

Several good answers have been given already. Let me provide a quicker way for this particular matrix. By using a bit of theory and some observations, there's no need to explicitly calculate the characteristic polynomial at all.

1. The matrix is block upper triangular. This immediately tells us that the eigenvalues are the union of the blocks on the diagonal. In this case, the eigenvalues come from $$\left\{A'=\begin{pmatrix}-5 & -6\\3 & 4\end{pmatrix},\ (-2)\right\}$$ Therefore one of the eigenvalues is $-2$.

2. Looking at the diagonal block $A'$, notice that each column sums to $-2$. This tells us that $(1\ \ 1)^\mathrm{T}$ is an eigenvector for $(A')^\mathrm{T}$ with eigenvalue $-2$. Since transposition does not change eigenvalues, this tells us that $A'$ also has eigenvalue $-2$.

3. The last eigenvalue can be obtained from the trace of $A'$. The sum of the eigenvalues must equal the trace. The trace of $A'$ is $-1$ and we already know one of its eigenvalues to be $-2$. This tells us the last eigenvalue is $1$.

All together, we have the eigenvalues of $A$ as $\{-2,\ -2,\ 1\}$, all without calculating the characteristic polynomial.