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 :)!)
- I tried to use trace as in this reply, and indeed $-1-1+5=3= tr(A)$, but I could not have guessed that either!Is there a fast way of getting the eigenvalues of this 3x3 matrix?
2 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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 $\lambda-1$.
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