# Eigenvalues of arbitrary $2×2$ matrix?

I think I am supposed to use linear algebra techniques, but I remember almost nothing from the subject. If i have $$\begin{pmatrix} 1-p & p \\ q & 1-q \end{pmatrix}$$ and i know one of its eigenvalue is 1 with left eigenvector $$(q/(p+q), p/(p+q))$$ is there a technique I can use to find the other eigenvalue?

• The eigenvalues are found using $$|A - \lambda I| = 0$$ You will find $$\lambda_1 = 1, \lambda_2 = 1 - p - q$$
– Moo
Commented May 11, 2020 at 4:09
• I tried that but I ended up with a monster quadratic equation. Do you just observe and find it from $(1-p-\lambda)(1-q-\lambda) = pq$ just by thinking? Commented May 11, 2020 at 4:19
• $$|A - \lambda I| = \lambda ^2 +(-2 + p+ q)\lambda +(-p -q+1) = (\lambda -1) (\lambda +p+q-1) = 0$$ After grouping the like terms, this is just the quadratic formula to find the two roots.
– Moo
Commented May 11, 2020 at 4:21

You can always get the last eigenvalue of a matrix “for free” by using the fact that its trace is equal to the sum of the eigenvalues, taking into account their multiplicities. So, the other eigenvalue is equal to $$(1-p)+(1-q)-1 = 1-p-q.$$

You can also often find eigenvectors and eigenvalues by trying simple linear combinations of rows or columns. In this case, if you subtract the second row from the first, which corresponds to left-multiplication by $$(1,-1)$$, you get $$(1-p-q,-(1-p-q))$$. Another trick that you can use when the matrix has simple (multiplicity one) eigenvalues is that the products of left and right eigenvectors with different eigenvalues vanish. Here, we know that $$(1,1)^T$$ is a right eigenvector of $$1$$, so $$(1,-1)$$ is a left eigenvector of the other eigenvalue, which we know from the trace to be $$1-p-q$$.

If

$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \tag 1$$

is an arbitrary $$2 \times 2$$ matrix, its eigenvalues are defined to be the solutions of the equation

$$\det (A - \lambda I) = 0; \tag 2$$

that is,

$$\det \left (\begin{bmatrix} a - \lambda & b \\ c & d - \lambda \end{bmatrix} \right ) = 0; \tag 3$$

expanding this determinant yields

$$(a - \lambda)(d - \lambda) - bc = 0, \tag 4$$

or

$$\lambda^2 - (a + d) \lambda + (ad - bc) = 0; \tag 5$$

if $$\lambda_1$$ and $$\lambda_2$$ are the roots of this quadratic, we also have

$$(\lambda - \lambda_1)(\lambda - \lambda_2) = 0, \tag 6$$

or

$$\lambda^2 - (\lambda_1 + \lambda_2) \lambda + \lambda_1 \lambda_2 = 0; \tag 7$$

comparing (5) and (7) reveals that

$$\lambda_1 + \lambda_2 = a + d = \text{trace} A \tag 8$$

and

$$\lambda_1 \lambda_2 = ad - bc = \det A. \tag 9$$

Given one eigenvalue of $$A$$, we may use either of these equations to determine the other eigenvalue, viz.

$$\lambda_2 = a + d - \lambda_1, \tag{10}$$

or, assuming

$$\lambda_1 \ne 0, \tag{11}$$

$$\lambda_2 = \dfrac{ad - bc}{\lambda_1}. \tag{12}$$

In the present example,

$$\lambda_1 = 1, \tag{13}$$

$$a + d = 2 - p - q, \tag{14}$$

$$ad - bc = 1 - p - q; \tag{15}$$

it then follows from both (10) and (12) that

$$\lambda_2 = 1 - p - q. \tag{16}$$

Note we needn't use eigenvectors to address this problem.