# Faster way to compute eigenvalues of a matrix doing rows operations

I had to find the eigenvalues of this matrix:

$$\begin{bmatrix} -20 & -25 & -22 \\ 6 & 11 & 12 \\ 5 & 5 & 1 \end{bmatrix}$$

I found the characteristic polynomial:

$$P(\lambda) = -\lambda^3 -8\lambda^2 +29\lambda +180$$

Then I matched $$P(\lambda) = 0$$ and with Ruffini's rule I solved my problem.

Can I reduce the given matrix in order to ease the computations? Is there a better way? Because I spent a lot of time trying all of the divisors of 180 to find a root.

I was thinking about add the first row with 5 times the third in order to set $$a_{12} = 0$$ and then apply Laplace.

• No, doing row operations will change the characteristic polynomial and therefore the eigenvalues. Jun 20, 2019 at 1:00
• For what it's worth: the eigenvalues are $-9,-4,5$ Jun 20, 2019 at 1:11
• You could notice that if you subtract $5I$ from the original matrix, then you get a matrix in which the first and second columns are identical. That implies that $5$ is an eigenvalue. Also, if you add $9I$, then the three rows sum to zero, so $-9$ is an eigenvalue. Then the eigenvalues must sum to the trace, which is $-8$, so the third eigenvalue is $-4$. Jun 20, 2019 at 1:24
• Ok thank you. Can I apply this method instead of finding the characteristic polynomial? Jun 20, 2019 at 9:29
• Only sometimes. Jun 20, 2019 at 13:35

First off, what you write down is not the characteristic polynomial, but minus the characteristic polynomial, and then with $$\lambda$$ (which traditionally stands for an eigenvalue, i.e., a scalar) substituted for the indeterminate which I shall take to be $$X$$). So $$X^3+8X^3-19X-29X-180$$ is the actual characteristic polynomial.
But gripes apart, there is a heuristic method to find this more easily, which works only in such fabricated cases as this one where there is a simple basis of integer-coefficient eigenvalues. It consists of writing for your matrix $$A$$ the matrix of which the characteristic polynomial is the determinant, that is $$XI-A$$, and doing row and column operations on it before actually computing the determinant. And one thing I have (again heuristically) found useful is focus on reducing large coefficient (not necessarily to $$0$$) and to work by conjugations: every time you add some multiple by $$c$$ of row $$k$$ to another row $$l$$, subsequently (or simultaneously, if you can get this right in your head) subtract the multiple by $$c$$ of column $$l$$ from column $$k$$. And vice versa. This will ensure that afterwards the occurrences of $$X$$ are again only on the diagonal.
So in your example, let us first reduce that coefficient $$25$$ by subtracting column $$3$$ from column $$2$$; then for good measure add row $$2$$ to row $$3$$: $$\left|\matrix{X+20 & 25 & 22 \\ -6 & X-11 & -12 \\ -5 & -5 & X-1}\right| =\left|\matrix{X+20 & 3 & 22 \\ -6 & X+1 & -12 \\ -5 & -X-4 & X-1}\right| =\left|\matrix{X+20 & 3 & 22 \\ -6 & X+1 & -12 \\ -11 & -3 & X-13}\right|$$ That is a bit better, but not much. now we seem to be able to make more progress either by subtracting column $$1$$ from column $$3$$ or by adding row $$3$$ to row $$1$$; but these are two sides of a single conjugation, so let's do that (in the order given): $$\left|\matrix{X+20 & 3 & 22 \\ -6 & X+1 & -12 \\ -11 & -3 & X-13}\right| =\left|\matrix{X+20 & 3 & -X+2 \\ -6 & X+1 & -6 \\ -11 & -3 & X-2}\right| =\left|\matrix{X+9 & 0 & 0 \\ -6 & X+1 & -6 \\ -11 & -3 & X-2}\right|$$ Weren't we lucky, two coefficients $$0$$ created and we did not even directly try to do so. At this point we could easily compute and factor the characteristic polynomial; the factor $$X+9$$ is evident.
But with such good fortunes so far, let's see if we can continue a bit. Subtracting column $$3$$ from column $$1$$ looks promising, especically since then adding row $$1$$ to row $$3$$ creates yet another $$0$$. This time I'll do both at once $$\left|\matrix{X+9 & 0 & 0 \\ -6 & X+1 & -6 \\ -11 & -3 & X-2}\right| =\left|\matrix{X+9 & 0 & 0 \\ 0 & X+1 & -6 \\ 0 & -3 & X-2}\right|.$$ Great, although not really a big step towards computing the characteristic polynomial. After a bit of thought, now add column $$2$$ to column $$3$$ and subtract row $$3$$ from row $$2$$, giving $$\left|\matrix{X+9 & 0 & 0 \\ 0 & X+1 & -6 \\ 0 & -3 & X-2}\right| =\left|\matrix{X+9 & 0 & 0 \\ 0 & X+4 & 0 \\ 0 & -3 & X-5}\right|.$$ Now the factorisation of the characteristic polynomial is evident.
We can even get to a completely diagonal form, killing the off-diagonal $$-3$$, by subtracting $$c$$ times column $$3$$ from column $$2$$ then adding $$c$$ times row $$2$$ to row $$3$$, for $$c=\frac13$$. All in all, we have explicitly conjugated $$A$$ to a diagonal matrix with diagonal entries $$-9,-4,5$$. The matrices we used to right-multiply by (and then left-multiply by their inverse) were ones differing from identity by just one off-diagonal coefficient, namely successively $$\pmatrix{1&0&0\\0&1&0\\0&-1&1}, \pmatrix{1&0&-1\\0&1&0\\0&0&1}, \pmatrix{1&0&0\\0&1&0\\-1&0&1}, \pmatrix{1&0&0\\0&1&1\\0&0&1}, \text{and} \pmatrix{1&0&0\\0&1&0\\0&-\frac13&1}$$ Altogether we have found the diagonal matrix as $$P^{-1}AP$$ where $$P$$ is the product of those matrices, which is $$P=\pmatrix{2&\frac13&-1\\0&\frac23&1\\-1&-1&0}.$$ You can check that the columns of $$P$$ are eigenvectors of $$A$$. And we didn't even solve any linear system to find them.