Finding roots/eigenvalues of an not so good looking polynomial

Question

So I'm given a matrix $$A=\begin{pmatrix}\frac{1}{5}& 3 & \frac{3}{5}\\ 0 &\frac{1}{2} & 0 \\ \frac{4}{5} & 1 & \frac{2}{5}\end{pmatrix}$$

whose polynomial characteristic I got was

$$P_A(\lambda)=\frac{12\lambda}{25}+\left(\frac{1}{5}-\lambda\right)\left(\frac{2}{5}-\lambda\right)\left(\frac{1}{2}-\lambda\right)-\frac{6}{25}=0$$

I've tried most of the things I know, but I still don't get the correct roots. Is there any trick here that I fail to see?

Answer 1

The second row of the matrix allows you to understand that $\lambda = \dfrac{1}{2}$ is a solution of the whole polynomial equation.

let's invent a matrix $(M - x\mathbb{1})$ in order to find the eigenvalues:

$$ M = \left( \begin{array}{ccc} 2-x & 5 & 9 \\ 0 & 12-x & 0 \\ 120 & 31 & 97-x \\ \end{array} \right) $$

The second row contains two zeroes. Hence $x = 12$ is for sure a solution of its characteristic polynomial.

Indeed, if we calculate it, we get:

$$P(x) = -x^3+111 x^2-302 x-10632$$

If $x = 12$ you can easily check that $P(12) = 0$.

The same in your matrix.

$$\lambda = \dfrac{1}{2}$$

is a solution and it is an eigenvalue.

The other two come easy by factoring the polynomial and solving it.

Answer 2

Be more judicious about how you expand the determinant and don’t be in a hurry to expand and simplify the resulting expression. The second row has only one entry that’s not identically zero, which also happens to contain $\lambda$, so if you expand along that row, the characteristic polynomial is already partially factored: $$\left(\frac12-\lambda\right)\left(\left(\frac15-\lambda\right)\left(\frac25-\lambda\right)-\frac{12}{25}\right).$$ Now you just need to work on the second factor, which at worst you’ll end up attacking with the quadratic formula. In this case, it’s pretty easy to factor that term, too.

For this matrix, the eigenvalues can instead be found by inspection pretty easily. Remember that you can sometimes find eigenvectors directly, without first somehow computing the eigenvalues of the matrix. As a general approach, look for rows or columns of the matrix for which the only nonzero entry is on the main diagonal, and try some simple linear combinations of rows and columns. (Remember that the eigenvalues of a matrix and its transpose are identical.)

Here, the second row has $\frac12$ on the main diagonal and zeros everywhere else, so one of the eigenvalues is $\frac12$, just as we found by expanding the determinant along this row. Next, those zeros in the second row suggest trying a linear combination of the first and last columns. Adding them up doesn’t get us anywhere, but subtracting the last from the first produces $\left(-\frac25,0,\frac25\right)^T$, which is a multiple of $(1,0,-1)^T$, so we know that that’s a (right) eigenvector of the matrix with eigenvalue $-\frac25$. You can then get the last eigenvalue “for free” by examining the trace—it’s equal to the sum of the eigenvalues, taking multiplicities into account. For this matrix, that gives us $$\left(\frac15+\frac12+\frac25\right)-\left(\frac12-\frac25\right)=1.$$ It’s worth trying this before plunging into the solve-the-characteristic-equation-and-compute-a-bunch-of-null-spaces method, especially for homework and test problems, in which the values are often constructed to be “nice.”