### Properties

##### 4. Trace and determinant

The eigenvalues are the roots of the characteristic polynomial `det`(**A** - `λ`**I**). To see why `det`(**A** - `λ`**I**) is a polynomial of `λ`, we consider a general 3 by 3 matrix **A** = [**a**_{1} **a**_{2} **a**_{3}]. Then

`det`(**A** - `λ`**I**) = `det`(**a**_{1} - `λ`**e**_{1}, **a**_{2} - `λ`**e**_{2}, **a**_{3} - `λ`**e**_{3}).

By multilinearity, we have

`det`(**A** - `λ`**I**) = `det`(**a**_{1}, **a**_{2}, **a**_{3})

+ [`det`(**e**_{1}, **a**_{2}, **a**_{3}) + `det`(**a**_{1}, **e**_{2}, **a**_{3}) + `det`(**a**_{1}, **a**_{2}, **e**_{3})] (-`λ`)

+ [`det`(**e**_{1}, **e**_{2}, **a**_{3}) + `det`(**e**_{1}, **a**_{2}, **e**_{3}) + `det`(**a**_{1}, **e**_{2}, **e**_{3})] (-`λ`)^{2}

+ `det`(**e**_{1}, **e**_{2}, **e**_{3}) (-`λ`)^{3}.

Noting that the terms are obtained by making choices among **a**_{i} and - `λ`**e**_{i} in each variable, and the power of -`λ` is the number of choices of - `λ`**e**_{i}. The observation holds in general, and we have the following conclusion.

For an `n` by `n` matrix **A**, `det`(**A** - `λ`**I**) is a polynomial of degree `n`.

Looking at `det`(**A** - `λ`**I**) in more detail, we see the constant term `det`(**a**_{1}, **a**_{2}, **a**_{3}) = `det`**A**, and the top degree term `det`(**e**_{1}, **e**_{2}, **e**_{3}) (-`λ`)^{3} = (-1)^{3}`λ`^{3}. The observation also holds in general.

Let us consider the second highest degree term. By the column operations (-`a`_{31})[col 1] + [col 3], (-`a`_{32})[col 2] + [col 3], we have

`det`(**e**_{1}, **e**_{2}, **a**_{3}) = `det`(**e**_{1}, **e**_{2}, `a`_{31}**e**_{1} + `a`_{32}**e**_{2} + `a`_{33}**e**_{3}) = `det`(**e**_{1}, **e**_{2}, `a`_{33}**e**_{3}) = `a`_{33}.

Similarly, we have `det`(**e**_{1}, **a**_{2}, **e**_{3}) = `a`_{22}, `det`(**a**_{1}, **e**_{2}, **e**_{3}) = `a`_{11}. Thus the second order term is

(`a`_{11} + `a`_{22} + `a`_{33}) (-`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)

`tr`**A** = `a`_{11} + `a`_{22} + ... + `a`_{nn}

and the characteristic polynomial is of the form

`det`(**A** - `λ`**I**) = `det`**A** + ... + (-1)^{n-1}`tr`**A** `λ`^{n-1} + (-1)^{n} λ^{n}.

Let `λ`_{1}, `λ`_{2}, ..., `λ`_{n} be all the roots of `det`(**A** - `λ`**I**). Since the top degree term of `det`(**A** - `λ`**I**) is (-1)^{n} λ^{n}, we have

`det`(**A** - `λ`**I**) = (`λ`_{1} - `λ`)(`λ`_{2} - `λ`)...(`λ`_{n} - `λ`)

= `λ`_{1}`λ`_{2}...`λ`_{n} + ... + (-1)^{n-1}(`λ`_{1} + `λ`_{2} + ... + `λ`_{n}) `λ`^{n-1} + (-1)^{n} λ^{n}.

Comparing the two formulae for `det`(**A** - `λ`**I**), we conclude the following.

Let `λ`_{1}, `λ`_{2}, ..., `λ`_{n} be all the eigenvalues of an `n` by `n` matrix **A**. Then

`λ`_{1}`λ`_{2}...`λ`_{n} = `det`**A**

`λ`_{1} + `λ`_{2} + ... + `λ`_{n} = `tr`**A**.

We note that decomposing a polynomial into a product of `λ`_{i} - `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 |