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I've been having trouble finding the characteristic polynomial for this matrix: $$ \begin{pmatrix} 1 & -1 & -1\\ 1 & 3 & 1\\ -3 & 1 &-1 \end{pmatrix} $$

I know there are online calculators, but they don't show how to get to a "factored" (I think that's the word for it?) polynomial.

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Hint

Simply set the determinant of $A-\lambda I$ equal to zero, that is $$ \begin{vmatrix} 1-\lambda & -1 & -1\\ 1 & 3-\lambda & 1\\ -3 & 1 &-1 -\lambda \end{vmatrix}=0 $$

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  • $\begingroup$ obviously i know that, my trouble is with calculating it $\endgroup$ Apr 18 '20 at 18:15
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Compute $$ \det(t\cdot I-A)=t^3 - 3t^2 - 4t + 12=(t + 2)(t - 2)(t - 3). $$ For the factorisation use the Rational root theorem.

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  • $\begingroup$ Thanks but we did not study that, there should be a technical way to reach the factored form $\endgroup$ Apr 18 '20 at 18:17
  • $\begingroup$ This is the easiest way to factor it. Doesn't matter that you haven't studied it. It is not difficult. $\endgroup$ Apr 18 '20 at 18:36
  • $\begingroup$ @reggiN : In Algebra they should have taught you about finding rational roots of polynomials. $\endgroup$ Apr 19 '20 at 4:10
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The conventional thing to do is to expand $\det(A-\lambda I)$ and factor it. For artificial exercises like this one, the roots are often “nice,” so you can use the rational root theorem to make some educated guesses as to what the roots might be.

Before diving into all of that, it can be fruitful to try to find some eigenvectors by playing around with simple linear combinations or columns and rows of the matrix. The eigenvalues of a matrix and its transpose are identical, so left eigenvectors—eigenvectors of the transpose—are just as good if we’re mainly interested in the eigenvalues.

Here, we can quickly find that adding the first and second rows produces $(2,2,0)$, so $(1,1,0)$ is a left eigenvector with eigenvalue $2$. Similarly, adding the first and third rows produces $(-2,0,-2)$, so $(1,0,1)$ is a left eigenvector with eigenvalue $-2$. You can get the last eigenvalue “for free” by using the fact that the trace of a matrix is equal to the sum of its eigenvalues: $(1+3-1)-2+2=3$.

Therefore, the characteristic polynomial is $(\lambda+2)(\lambda-2)(\lambda-3)$.

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You can factor $t^3 - 3t^2 - 4t + 12$ by grouping: $$ t^3 - 3t^2 - 4t + 12 = t^2(t-3)-4(t-3)=(t^2-4)(t-3)=(t + 2)(t - 2)(t - 3) $$ by this is ad hoc. The rational root theorem is more systematic.

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