Traces and Determinants

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 13 August 2013

Determinants

Let $a$ be an $n \times n$ matrix with entries $a_{i j}$ .

The trace of $a$ is
$tr (a) = \sum_{i = 1}^{n} a_{i i}$ .
The determinant of $a$ is
$\det (a) = \sum_{w \in S_{n}} \det (w) a_{1 w (1)} a_{2 w (2)} \dots a_{n w (n)}$

Let

𝔽

be a field and let

n \in ℤ_{> 0}

(a) Up to constant multiples,

tr : M_{n} (𝔽) \to 𝔽

is the unique function such that

(a1) If

c \in 𝔽

and

a, b \in M_{n} (𝔽)

then

tr (a + b) = tr (a) + tr (b) and tr (c a) = c tr (a) .

(a2) If

a, b \in M_{n} (𝔽) then

tr (a b) = tr (b a)

(b) Identify

M_{n} (𝔽)

with the

𝔽

-module

\underset{\underset{n times}{⏟}}{𝔽^{n} \times \dots \times 𝔽^{n}}

\begin{matrix} M_{n} (𝔽) & \tilde{⟶} & \underset{\underset{n times}{⏟}}{𝔽^{n} \times \dots \times 𝔽^{n}} \\ a & ⟼ & (a_{1} | a_{2} | \dots | a_{n}) \end{matrix}

, where

a_{i}

are the columns of

a

The function

\det : M_{n} (𝔽) \to 𝔽

is the unique function such that

(b1) (columnwise linear) If

i \in {1, 2, \dots, n}

and

c \in 𝔽

then

\det (a_{1} | \dots | a_{i} + b_{i} | \dots | a_{n}) = \det (a_{1} | \dots | a_{i} | \dots | a_{n}) + \det (a_{1} | \dots | b_{i} | \dots | a_{n})

and

\det (a_{1} | \dots | c a_{i} | \dots | a_{n}) = c \det (a_{1} | \dots | a_{i} | \dots | a_{n})

(b2) If

i \in {1, 2, \dots, n - 1}

then

\det (a_{1} | \dots | a_{i} | a_{i + 1} | \dots | a_{n}) = - \det (a_{1} | \dots | a_{i + 1} | a_{i} | \dots | a_{n})

(b3) if

a, b \in M_{n} (𝔽) then

\det (a b) = \det (a) \det (b)

(Laplace expansion) $det (\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{matrix}) = \sum_{i} p_{i} det (\begin{matrix} a_{i_{1} j_{1}} & \dots & a_{i_{1} j_{p}} \\ ⋮ & ⋱ & ⋮ \\ a_{i_{p} j_{1}} & \dots & a_{i_{p} j_{p}} \end{matrix}) (\begin{matrix} a_{i_{p + 1} j_{p + 1}} & \dots & a_{i_{p + 1} j_{n}} \\ ⋮ & ⋱ & ⋮ \\ a_{i_{n} j_{p + 1}} & \dots & a_{i_{n} j_{n}} \end{matrix})$ where $j_{1}, \dots, j_{n}$ is a fixed permutation of $1, 2, \dots, n$ and the sum is over all possible divisions of $1, 2, \dots, n$ into two sets $i_{1} < \dots < i_{p} and i_{p + 1} < \dots < i_{n},$ and $p_{i} = + 1$ or $- 1$ according as $i_{1}, .., i_{n}$ and $j_{1}, \dots, j_{n}$ are like or unlike derangements of $1, 2, \dots, m .$

If $A$ and $B$ are $n \times n$ matrices then $det (A B) = det (A) det (B) .$

Proof.

To show:

det (A B) = det (A) det (B)

\begin{matrix} det (A B) & = & \sum_{w \in S_{n}} det (w) {(A B)}_{1, w (1)} \dots {(A B)}_{n, w (n)} \\ = & \sum_{w \in S_{n}} det (w) \sum_{k_{1}, \dots, k_{n}} A_{1, k_{1}} B_{k_{1}, w (1)} \dots A_{n, k_{n}} B_{k_{n}, w (n)} \\ = & \sum_{k_{1}, \dots, k_{n}} A_{1, k_{1}} \dots A_{n, k_{n}} \sum_{w \in S_{n}} det (w) B_{k_{1}, w (1)} \dots B_{k_{n}, w (n)} \\ = & \sum_{k = (k_{1}, \dots, k_{n}) \in S_{n}} det (k) A_{1, k_{1}} \dots A_{n, k_{n}} \sum_{w \in S_{n}} det (k^{- 1}) det (w) B_{k_{1}, w (1)} \dots B_{k_{n}, w (n)} \\ = & \sum_{k = (k_{1}, \dots, k_{n}) \in S_{n}} det (k) A_{1, k_{1}} \dots A_{n, k_{n}} \sum_{w \in S_{n}} det (w k^{- 1}) B_{k_{1}, w k^{- 1} (k_{1})} \dots B_{k_{n}, w k^{- 1} (k_{n})} \\ = & \sum_{k \in S_{n}} det (k) A_{1, k_{1}} \dots A_{n, k_{n}} \sum_{w k^{- 1} \in S_{n}} det (w k^{- 1}) B_{1, w k^{- 1} (1)} \dots B_{n, w k^{- 1} (n)} \\ = & det (A) det (B) \end{matrix}

$□$

a)	Let $B$ be the matrix obtained by switching two rows of $A .$ Then $det (B) = - det (A) .$
b)	Let $B$ be the matrix obtained by adding a multiple of a row of $A$ to another row of $A .$ Then $det (B) = det (A) .$
c)	Let $B$ be the matrix obtained by multiplying a row of $A$ by a constant $c \in R .$ Then $det (B) = c det (A) .$

HW: Show that if two rows of $A$ are the same then $\det (A) = 0$ .

Inverses and Cramer's rule"

The ${(i, j)}^{th}$ signed minor or cofactor, $A_{i j},$ of $A$ is $det (\hat{A})$ where $\hat{A}$ is the matrix $A$ with the $i^{th}$ row and the $j^{th}$ column removed.

$\sum_{i = 1}^{n} a_{i k} A_{h i} = δ_{h k} det A .$

If $det (a)$ is a unit in $R$ then $a^{- 1} = det {(a)}^{- 1} (A_{i j})$ . SHOULD THERE BE A TRANSPOSE HERE?

Cramer’s rule for $A X = B .$

Put Thms VII-X of Hodge and Padoe as exercises.

$λ -matrices$

If $A$ is a $p \times q λ -matrix$ of rank $r$ and $E_{1} (λ), \dots, E_{r} (λ)$ are its invariant factors then there exist $M \in {G L}_{p} (K [λ])$ and $N \in {G L}_{q} (K [λ])$ such that $M A N = (\begin{matrix} E_{1} (λ) \\ E_{2} (λ) & 0 \\ ⋱ \\ E_{r} (λ) \\ 0 & 0 \\ ⋱ \\ 0 \end{matrix}) .$

Two $p \times q λ -matrices$ $A$ and $B$ are equivalent if and only if they have the same invariant factors and if and only if they have the same elementary divisors.

Note that these proofs work for any Euclidean domain.

The characteristic polynomial of $A$ is the polynomial $det (A - t I_{n}) .$

Cayley-Hamilton Theorem.

Note that the proof of Theorem II §10 Hodge and Padoe!
Theorem III (Hodge and Padoe) gives minimal polynomial.

page history