### Properties

##### 4. Trace and determinant

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.

For an n by n matrix A, det(A - λI) is a polynomial of degree n.

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.

Let λ1, λ2, ..., λn be all the eigenvalues of an n by n matrix A. Then

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. -4 7 -4 7

From this example, we have

 det[ 1 3 -3 ] = 4×4×(-2) = -32, tr[ 1 3 -3 ] = 4 + 4 - 2 = 6. -3 7 -3 -3 7 -3 -6 6 -2 -6 6 -2