The eigenvalues are the roots of the characteristic polynomial det(A - λI). To see why det(A - λI) is a polynomial of λ, we consider a general 3 by 3 matrix A = [a1 a2 a3]. Then
det(A - λI) = det(a1 - λe1, a2 - λe2, a3 - λe3).
By multilinearity, we have
det(A - λI) = det(a1, a2, a3)
+ [det(e1, a2, a3) + det(a1, e2, a3) + det(a1, a2, e3)] (-λ)
+ [det(e1, e2, a3) + det(e1, a2, e3) + det(a1, e2, e3)] (-λ)2
+ det(e1, e2, e3) (-λ)3.
Noting that the terms are obtained by making choices among ai and - λei in each variable, and the power of -λ is the number of choices of - λei. The observation holds in general, and we have the following conclusion.
Looking at det(A - λI) in more detail, we see the constant term det(a1, a2, a3) = detA, and the top degree term det(e1, e2, e3) (-λ)3 = (-1)3λ3. The observation also holds in general.
Let us consider the second highest degree term. By the column operations (-a31)[col 1] + [col 3], (-a32)[col 2] + [col 3], we have
det(e1, e2, a3) = det(e1, e2, a31e1 + a32e2 + a33e3) = det(e1, e2, a33e3) = a33.
Similarly, we have det(e1, a2, e3) = a22, det(a1, e2, e3) = a11. Thus the second order term is
(a11 + a22 + a33) (-λ)2.
In general, for an n by n matrix A, the trace of A is the sum of its diagonal entries (see this exercise and this exercise for properties of trace)
trA = a11 + a22 + ... + ann
and the characteristic polynomial is of the form
det(A - λI) = detA + ... + (-1)n-1trA λn-1 + (-1)n λn.
Let λ1, λ2, ..., λn be all the roots of det(A - λI). Since the top degree term of det(A - λI) is (-1)n λn, we have
det(A - λI) = (λ1 - λ)(λ2 - λ)...(λn - λ)
= λ1λ2...λn + ... + (-1)n-1(λ1 + λ2 + ... + λn) λn-1 + (-1)n λn.
Comparing the two formulae for det(A - λI), we conclude the following.
λ1λ2...λn = detA
λ1 + λ2 + ... + λn = trA.
We note that decomposing a polynomial into a product of λi - λ is generally possible only if we allow complex numbers (or algebraically closed fields). Thus some of the eigenvalues in the statement above may be complex.
Example From this example, we have
|det[||13||-4||] = 5×15 = 75, tr[||13||-4||] = 5 + 15 = 20.|
From this example, we have
|det[||1||3||-3||] = 4×4×(-2) = -32, tr[||1||3||-3||] = 4 + 4 - 2 = 6.|