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I am doing a course on systems of differential equations. I have the following introductionary problem, to find repeated eigenvalues, before continuing on complex eigenvalues:

Solve: $$X' = \color{#C80}{\begin{pmatrix} \hfill 1 & -2 & \hfill 2\\\ -2 & \hfill 1 & -2 \\ \hfill2 & -2 & \hfill1\end{pmatrix}}X$$

Taking $det(\lambda I-A)$ I get: $$\color{#C80}{\begin{vmatrix} \lambda -1 & 2 & -2\\\ 2 & \lambda -1 & 2 \\ -2 & 2 & \lambda -1\end{vmatrix}}$$

Which after some computation gives me a long polynomial, that is shortened to $\lambda^3 -3\lambda^2-9\lambda -5$. One can guess easily that one solution is -1, so the next step is polynomial division by $\lambda+1$. In the end I get to the correct answer which is $(\lambda+1)^2(\lambda-5)$. The whole process, taking the determinant, polynomial division, without including verification took a whooping 15 minutes.

**EDIT: Besides, it is wrong. I just noticed that the correct solution is $-(\lambda+1)^2(\lambda-5)$. This does not change the eigenvalues, but is incorrect!!

Since this forum is home of many smart and resourceful people, and I enjoy discussions,are there any shortcuts I could use to skip some of the steps? All those steps are long, error prone, and I don't have this kind of time at the exam, especially since I did not even begin the complex eigenvalues part! (No matlab allowed :)!)

Rodrigo de Azevedo
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Dovendyr
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    I would suggest that your time would improve significantly with practice. Also if you know that a cubic has a repeated root, it will share that root with its derivative (the derivative of $x^3+ax^2+bx+c$ ia $3x^2+2ax+b$) so you can test the two roots of the quadratic. Here you could also simplify the computations somewhat by setting $x=\lambda-1$ before expanding the determinant. – Mark Bennet Apr 19 '21 at 08:27
  • I didn't know that! Great tips! Thank you. – Dovendyr Apr 19 '21 at 08:28
  • @Dovendyr If you are allowed to use a programmable calculator, maybe you can load up programs for 1) computing characteristic polynomial, 2) solving cubic equations. I've done 2) before, but I've never tried 1). – 光復香港 時代革命 Free Hong Kong Apr 19 '21 at 11:21
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    @Dovendyr Your answer wasn't really "wrong". It is true that $\det(\lambda I - A) = (\lambda+1)^2(\lambda - 5)$. Presumably, the other answer was found by computing $\det(A - \lambda I)$, which is indeed equal to $-(\lambda + 1)(\lambda - 5)$. As you point out, both lead to the correct eigenvalues. – Ben Grossmann Apr 19 '21 at 13:45
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    The fastest way to compute the eigenvalues in this case is to recognize that this matrix is a rank 1 update of a multiple of the identity matrix. – Ben Grossmann Apr 19 '21 at 13:47
  • Oh thanks for pointing this out @BenGrossmann! – Dovendyr Apr 19 '21 at 16:16

2 Answers2

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Well, to give a few points, though you might know most of them already:

  • For calculating the determinant (or the characteristic polynomial) of a 3x3 matrix is use the Rule of Sarrus (it should be fast enough that you don't need to use any other tricks).
  • If you have a cubic polynomial and no calculator at hand in an exam that can solve it, for usual at least one of its roots is rational (guessable), so you can effectively find it using the rational root theorem
  • To check whether your found eigenvalues are correct, simply compare it to the trace of the matrix (as the sum of the eigenvalues equals the trace).

Besides these pointers, the method you used was pretty certainly already the fastest there is.

Other methods exist, e.g. we know that, given that we have a 3x3 matrix with a repeated eigenvalue, the following equation system holds:

$$ \left|\matrix{\text{tr}(A)= 2\lambda_1 + \lambda_2 \\ \det(A)= \lambda_1^2 \lambda_2 } \right| $$

You could solve it and obtain up to three possible sets of eigenvalues, of which one is correct. But without calculator, the equation system isn't the easiest to solve, and you'd have to check whichof the sets is correct.

More generally, you can obtain the three eigenvalues of a 3x3 matrix as seen here, but just as the other method, the equation system isn't really easy to solve.

Sudix
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For a $3\times3$ matrix, the coefficients of the characteristic polynomial are

$$1,-\text{tr}(X),\frac{\text{tr}^2(X)-\text{tr}(X^2)}2,-\text{det}(X)$$ which could be easier to compute. In many exercises, a solution can be found by means of the rational root theorem.

In the case of three equal values on the main diagonal, you might as well have solved for $\lambda-1$.