Currently checking the homework of an engineering freshman (this job gets assigned to math students at my university, I'm an undergrad math student myself) and the students were supposed to calculate the eigenvalues of a (symmetric) matrix. However, instead of using the usual means (Laplace expansion or Sarrus' rule) the student used a formula, which I've never encountered and which he, unfortunately, didn't write down explicitly, but the answer is correct. Maybe one of you can explain the formula that was used in this example, I'd like to understand why this works.

The matrix given was

$A=\begin{pmatrix} 3&-1&1\\-1&3&1\\1&1&3 \end{pmatrix}$

and the first expression on his homework sheet reads

$\chi_A(\lambda)=\lambda^3-9\lambda^2+(8+8+8)\lambda-\det A=\lambda^3-9\lambda^2+24\lambda-16$.

The answer is correct, but I don't see where he got that coefficient of $8+8+8$ from. I tried to do some Google research on that topic, but I couldn't find anything.

I do know that for an $n\times n$ matrix $M$, requiring that the characteristic polynomial be monic, the $(n-1)$-th coefficient always equals $-\mathrm{tr}(M)$ and the zeroth coefficient always equals $-\det M$. But where did he get the coefficient of $\lambda$ from? I'd like to understand if there is a closed formula for (symmetric) $3\times3$ matrices that I don't know about.

  • $\begingroup$ Do you know how calculate the determinat of a $3\times 3$ matrix? $\endgroup$ – Emilio Novati Dec 16 '15 at 20:59
  • $\begingroup$ Here is one formula that involves the trace in the coefficient in front of $\lambda$ en.wikipedia.org/wiki/Eigenvalue_algorithm#3.C3.973_matrices I'm sure with simplifications (especially because the diagonal values are all the same), you can figure out how it works. $\endgroup$ – rajb245 Dec 16 '15 at 21:03
  • $\begingroup$ en.m.wikipedia.org/wiki/Invariants_of_tensors, see the spot on symmetric 3x3 tensors (this is just a 3x3 matrix) $\endgroup$ – Triatticus Dec 16 '15 at 21:17
  • $\begingroup$ @EmilioNovati of course I do - as I mentioned above, standard methods include Sarrus' rule or Laplace expansion, both of which I know how to apply. I was just wondering about the specific expression leading to the coefficient in front of $\lambda$. $\endgroup$ – Sora. Dec 16 '15 at 21:37

$8+8+8$ is the sum of the principal $2 \times 2$ minors of the matrix.

In general (not just for size $3 \times 3$), the top coefficient in the characteristic polynomial is just $1$, the next is minus the trace (and the trace is the sum of the diagonal elements, i.e., the sum of the principal $1 \times 1$ minors), the next one is the sum of the principal $2 \times 2$ minors, the next one is minus the sum of the principal $3 \times 3$ minors, and so on.

This is a standard fact, which is mentioned in the Wikipedia article Characteristic polynomial, for example.

  • $\begingroup$ Didn't learn about this in my linear algebra lecture, we were always told to evaluate $\det(\lambda I-A)$ using the standard methods (Laplace expansion, basic row operations, Sarrus' rule for $3\times3$ matrices etc.) in order to obtain the characteristic polynomial of a matrix $A$ - this is exactly what I've been looking for. Guess I've been googling too specifically which is why I didn't come across this. Thanks! $\endgroup$ – Sora. Dec 16 '15 at 21:45

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