Math and science::Algebra

Product of eigenvalues and sum of eigenvalues

Let $A$ be an $N \times N$ matrix. Then the following two statements are true.

The product of the N eigenvalues equals the determinant of $A$ .

The product includes repeated eigenvalues.

The sum of the N eigenvalues equals the trace of $A$ .

Product of eigenvalues equals the determinant.

Intuition followed by the angle of attack for a more formal justification.

Intuition

Consider the 2D case. The column space of $A$ can be spanned by the two independent eigenvectors. Consider the quadrilateral formed by these two vectors. Applying $A$ to the space will a stretch in volume of the space by a factor of $det (A)$ ; the quadrilateral will grow in volume by a factor $det (A)$ . The two sides defining this quadrilateral don't rotate or skew under the $A$ transformation, as they are eigenvectors—they only scale. And so the factor by which the quadrilateral grows must be decomposable into the two scaling factors by which each eigenvector grows.

Proof idea

A more formal justification is to express $det (A - λ I)$ in terms of the N polynomial roots (always possible), then setting $λ = 0$ we are left with $det (A) = λ_{1} λ_{2} . . . λ_{N}$ .

\begin{array}{rcl} det (A - λ I) = p (λ) & = & (- 1)^{n} (λ - λ_{1}) (λ - λ_{2}) \dots (λ - λ_{n}) \\ = & (- 1) (λ - λ_{1}) (- 1) (λ - λ_{2}) \dots (- 1) (λ - λ_{n}) \\ = & (λ_{1} - λ) (λ_{2} - λ) \dots (λ_{n} - λ) \end{array}

The trace equals the sum of eigenvalues.

For the 2D case, this can be shown easily through the quadratic formula.

Let:

A = [\begin{matrix} a & b \\ c & d \end{matrix}]

and then observer that:

\begin{aligned} det (A - λ I) = 0 \\ ⟹ (a - λ) (b - λ) - b c = 0 \\ ⟹ λ^{2} - λ (a + d) + a d - b c = 0 \\ ⟹ (λ - \frac{a + d}{2})^{2} + a d - b c - (\frac{a + d}{2})^{2} = 0 \\ ⟹ λ = \frac{a + d}{2} \mp \sqrt{b c - a d + (\frac{a + d}{2})^{2}} . \end{aligned}

The sum of these two eigenvalues will be the trace, $a + d$ .

Some easy examples of the trace = sum of eigenvalues proposition.

A few forms of 2D matrices make it easy to see how their trace is the sum of eigenvalues. Some examples.

General observation

An observation is that adding $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ only affects the diagonal $[\begin{matrix} x & . \\ . & y \end{matrix}]$ .

A zero entry

When there is a zero entry, there are two ways to make a matrix singular by additions/subtractions of $I$ :

the "axis" column vector (only 1 non-zero value) falls to the origin, $[0, 0]$ .
the other column vector falls onto the axis of the "axis" column vector.

\begin{aligned} A & = [\begin{array}{c} 1 & 7 \\ 0 & 2 \end{array}] \\ = [\begin{array}{c} 0 & 7 \\ 0 & 1 \end{array}] + [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] & (first eigenvalue) \\ = [\begin{array}{c} - 1 & 7 \\ 0 & 0 \end{array}] + [\begin{array}{c} 2 & 0 \\ 0 & 2 \end{array}] & (second eigenvalue) \end{aligned}

The same deal with

B

B = [\begin{matrix} 3 & 0 \\ 1 & 1 \end{matrix}]

Symmetric matrices with equal diagonal

Symmetric matrices with the additional property that all diagonal entries are the same have simple eigenvalues, as the addition/subtraction of $I$ acts the same on all columns.

\begin{aligned} C & = [\begin{array}{c} 4 & 1 \\ 1 & 4 \end{array}] \\ = [\begin{array}{c} 1 & 1 \\ 1 & 1 \end{array}] + [\begin{array}{c} 3 & 0 \\ 0 & 3 \end{array}] & (first eigenvalue) \\ = [\begin{array}{c} - 1 & 1 \\ 1 & - 1 \end{array}] + [\begin{array}{c} 4 & 0 \\ 0 & 4 \end{array}] & (second eigenvalue) \end{aligned}