Determinant

In this section, $S$ denotes the set $\{1, 2, \ldots, n\}$ .

DEFINITION 15.2.1  

1. A function $\sigma : S {\longrightarrow}S$ is called a permutation on $n$ elements if $\sigma$ is both one to one and onto.
2. The set of all functions $\sigma : S {\longrightarrow}S$ that are both one to one and onto will be denoted by ${\mathcal S}_n$ . That is, ${\mathcal S}_n$ is the set of all permutations of the set $\{1, 2, \ldots, n\}$ .

EXAMPLE 15.2.2  

1. In general, we represent a permutation $\sigma$ by $\sigma = \left(\begin{array}{cccc} 1 & 2 & \cdots & n \\ \sigma(1) & \sigma(2) & \cdots & \sigma(n) \end{array}\right)$ . This representation of a permutation is called a TWO ROW NOTATION for $\sigma$ .
2. For each positive integer $n$ , ${\mathcal S}_n$ has a special permutation called the identity permutation, denoted $Id_n$ , such that $Id_n(i) = i$ for $1 \le i \le n$ . That is, $Id_n = \left(\begin{array}{cccc} 1 & 2 & \cdots & n \\ 1 & 2 & \cdots & n \end{array}\right)$ .
3. Let $n = 3$ . Then
   

   $${\mathcal S}_3 = \left\{ \tau_1 = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 1 & 2 & 3 ... ...3 = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 1 & 3 \end{array}\right), \right.$$

   $$\hspace{.5in} \left. \tau_4 = \left(\begin{array}{ccc} 1 & 2 & 3 ... ..._6 = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right)\right\}$$ (15.2.5)

   
   

Remark 15.2.3  

1. Let $\sigma \in {\mathcal S}_n$ . Then $\sigma$ is determined if $\sigma(i)$ is known for $i = 1, 2, \ldots, n$ . As $\sigma$ is both one to one and onto, $\{\sigma(1), \sigma(2), \ldots, \sigma(n)\} = S$ . So, there are $n$ choices for $\sigma(1)$ (any element of $S$ ), $n-1$ choices for $\sigma(2)$ (any element of $S$ different from $\sigma(1)$ ), and so on. Hence, there are $n(n-1)(n-2) \cdots 3\cdot2\cdot1 = n!$ possible permutations. Thus, the number of elements in ${\mathcal S}_n$ is $n!$ . That is, $\vert{\mathcal S}_n\vert = n!$ .
2. Suppose that $\sigma, \tau \in {\mathcal S}_n$ . Then both $\sigma$ and $\tau$ are one to one and onto. So, their composition map $\sigma\circ \tau$ , defined by $(\sigma \circ \tau)(i) = \sigma \bigl( \tau(i) \bigr)$ , is also both one to one and onto. Hence, $\sigma\circ \tau$ is also a permutation. That is, $\sigma\circ \tau \in {\mathcal S}_n$ .
3. Suppose $\sigma \in {\mathcal S}_n$ . Then $\sigma$ is both one to one and onto. Hence, the function $\sigma^{-1}: S {\longrightarrow}S$ defined by $\sigma^{-1}(m) = \ell$ if and only if $\sigma(\ell) = m$ for $1 \le m \le n$ , is well defined and indeed $\sigma^{-1}$ is also both one to one and onto. Hence, for every element $\sigma \in {\mathcal S}_n, \; \sigma^{-1} \in {\mathcal S}_n$ and is the inverse of $\sigma$ .
4. Observe that for any $\sigma \in {\mathcal S}_n$ , the compositions $\sigma \circ \sigma^{-1} =\sigma^{-1} \circ \sigma = Id_n$ .

PROPOSITION 15.2.4   Consider the set of all permutations ${\mathcal S}_n$ . Then the following holds:

1. Fix an element $\tau \in {\mathcal S}_n$ . Then the sets $\{ \sigma \circ \tau : \sigma \in {\mathcal S}_n \}$ and $\{ \tau \circ \sigma: \sigma \in {\mathcal S}_n \}$ have exactly $n!$ elements. Or equivalently,
   $${\mathcal S}_n = \{ \tau \circ \sigma : \sigma \in {\mathcal S}_n \} = \{ \sigma \circ \tau : \sigma \in {\mathcal S}_n \}.$$

2. ${\mathcal S}_n = \{ \sigma^{-1} : \sigma \in {\mathcal S}_n \}$ .

Proof. For the first part, we need to show that given any element $\alpha \in {\mathcal S}_n$ , there exists elements $\beta, \gamma \in {\mathcal S}_n$ such that $\alpha = \tau \circ \beta = \gamma \circ \tau$ . It can easily be verified that $\beta = \tau^{-1} \circ \alpha$ and $\gamma = \alpha \circ \tau^{-1}$ .

For the second part, note that for any $\sigma \in {\mathcal S}_n, \; (\sigma^{-1})^{-1} = \sigma$ . Hence the result holds. height6pt width 6pt depth 0pt

DEFINITION 15.2.5   Let $\sigma \in {\mathcal S}_n$ . Then the number of inversions of $\sigma$ , denoted $n(\sigma)$ , equals

$$\vert\{ (i, j) : \; i < j, \; \sigma(i) > \sigma(j) \; \}\vert.$$

Note that, for any $\sigma \in {\mathcal S}_n$ , $n(\sigma)$ also equals

$$\sum\limits_{i=1}^n \vert\{ \sigma(j) < \sigma(i), \; {\mbox{ for }} \; j = i+1, i+2, \ldots, n\}\vert.$$

DEFINITION 15.2.6   A permutation $\sigma \in {\mathcal S}_n$ is called a transposition if there exists two positive integers $m, r \in \{1, 2, \ldots, n\}$ such that $\sigma(m) = r, \; \sigma(r) = m$ and $\sigma(i) = i$ for $1 \le i \ne m, r \le n$ .

For the sake of convenience, a transposition $\sigma$ for which $\sigma(m) = r, \; \sigma(r) = m$ and $\sigma(i) = i$ for $1 \le i \ne m, r \le n$ will be denoted simply by $\sigma = (m \; r)$ or $(r \; m)$ . Also, note that for any transposition $\sigma \in {\mathcal S}_n$ , $\sigma^{-1} = \sigma$ . That is, $\sigma \circ \sigma = Id_n$ .

EXAMPLE 15.2.7  

1. The permutation $\tau = \left(\begin{array}{cccc} 1 & 2 & 3 & 4 \\ 3 & 2 & 1 & 4 \end{array}\right)$ is a transposition as $\tau(1) = 3, \tau(3) = 1, \; \tau(2) = 2$ and $\tau(4) = 4$ . Here note that $\tau = (1 \; 3) = (3 \; 1)$ . Also, check that
   $$n(\tau) = \vert\{ (1,2), (1,3), (2, 3)\}\vert = 3.$$

2. Let $\tau = \left(\begin{array}{ccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 & 2 & 3 & 5 & 1 & 9 & 8 & 7 & 6 \end{array}\right)$ . Then check that
   $$n(\tau) = 3 + 1 + 1 + 1 + 0 + 3 + 2 + 1 = 12.$$

3. Let $\ell, m$ and $r$ be distinct element from $\{1, 2, \ldots, n\}$ . Suppose $\tau = (m \; r)$ and $\sigma = ( m \; \ell)$ . Then
   

   $$(\tau\circ\sigma)(\ell) = \tau\bigl(\sigma(\ell)\bigr) = \tau(m) = r, \;\;\;\; (\tau\circ\sigma)(m) = \tau\bigl(\sigma(m)\bigr) = \tau(\ell) = \ell$$

   $$(\tau\circ\sigma)(r) = \tau\bigl(\sigma(r)\bigr) = \tau(r) = m, \;\;{\mbox{ and }} \;\; ... ...i) = \tau\bigl(\sigma(i)\bigr) = \tau(i) = i \; {\mbox{ if }} i \ne \ell, m, r.$$

   
   Therefore, $\tau\circ\sigma = (m \; r) \circ (m \; \ell) = \left(\begin{array}{cccccccccc}... ...ts & \ell & \cdots & m & \cdots & n \end{array}\right) = (r \; l) \circ (r\; m)$ . Similarly check that $\sigma\circ\tau = \left(\begin{array}{cccccccccc} 1 & 2 & \cdots & \ell & \cdo... ...1 & 2 & \cdots & m & \cdots & r & \cdots & \ell & \cdots & n \end{array}\right)$ .

With the above definitions, we state and prove two important results.

THEOREM 15.2.8   For any $\sigma \in {\mathcal S}_n, \; \sigma$ can be written as composition (product) of transpositions.

Proof. We will prove the result by induction on $n(\sigma)$ , the number of inversions of $\sigma$ . If $n(\sigma) = 0$ , then $\sigma = Id_n = ( 1 \; 2) \circ (1 \; 2)$ . So, let the result be true for all $\sigma \in {\mathcal S}_n$ with $n(\sigma) \le k$ .

For the next step of the induction, suppose that $\tau \in {\mathcal S}_n$ with $n(\tau) = k+1$ . Choose the smallest positive number, say $\ell$ , such that

$$\tau(i) = i, \; {\mbox{ for }} i = 1, 2, \ldots, \ell-1 \; {\mbox{ and }} \tau(\ell) \ne \ell.$$

As $\tau$ is a permutation, there exists a positive number, say $m$ , such that $\tau(\ell) = m$ . Also, note that $m > \ell$ . Define a transposition $\sigma$ by $\sigma = ( \ell \; m)$ . Then note that

$$(\sigma \circ \tau)(i) = i, \; {\mbox{ for }} i = 1, 2, \ldots, \ell.$$

So, the definition of ``number of inversions" and $m > \ell$ implies that

$$n(\sigma \circ \tau) = \sum\limits_{i=1}^n \vert\{ (\sigma\circ \tau) (j) < (\sigma \circ \tau) (i), \; {\mbox{ for }} \; j = i+1, i+2, \ldots, n\}\vert$$

$$= \sum\limits_{i=1}^{\ell} \vert\{ (\sigma\circ \tau) (j) < (\sigma \circ \tau) (i), \; {\mbox{ for }} \; j = i+1, i+2, \ldots, n\}\vert$$

$$\hspace{1in} + \sum\limits_{i=\ell+1}^n \vert\{ (\sigma\circ \tau... ... < (\sigma \circ \tau) (i), \; {\mbox{ for }} \; j = i+1, i+2, \ldots, n\}\vert$$

$$= \sum\limits_{i=\ell+1}^n \vert\{ (\sigma\circ \tau) (j) < (\sigma \circ \tau) (i), \; {\mbox{ for }} \; j = i+1, i+2, \ldots, n\}\vert$$

$$\le \sum\limits_{i=\ell+1}^n \vert\{ \tau (j) < \tau (i), \; {\mbox{ for }} \; j = i+1, i+2, \ldots, n\}\vert \;\; {\mbox{ as }} \; m > \ell,$$

$$< (m - \ell) + \sum\limits_{i=\ell+1}^n \vert\{ \tau (j) < \tau (i), \; {\mbox{ for }} \; j = i+1, i+2, \ldots, n\}\vert$$

$$= n(\tau).$$

Thus, $n(\sigma \circ \tau) < k+1$ . Hence, by the induction hypothesis, the permutation $\sigma\circ \tau$ is a composition of transpositions. That is, there exist transpositions, say ${\alpha}_i, \; 1 \le i \le t$ such that

$$\sigma \circ \tau = {\alpha}_1 \circ {\alpha}_2 \circ \cdots \circ {\alpha}_t.$$

Hence, $\tau = \sigma \circ {\alpha}_1 \circ {\alpha}_2 \circ \cdots \circ {\alpha}_t$ as $\sigma \circ \sigma = Id_n$ for any transposition $\sigma \in {\mathcal S}_n$ . Therefore, by mathematical induction, the proof of the theorem is complete. height6pt width 6pt depth 0pt

Before coming to our next important result, we state and prove the following lemma.

LEMMA 15.2.9   Suppose there exist transpositions ${\alpha}_i, \; 1 \le i \le t$ such that

$$Id_n = {\alpha}_1 \circ {\alpha}_2 \circ \cdots \circ {\alpha}_t,$$

then $t$ is even.

Proof. Observe that $t \ne 1$ as the identity permutation is not a transposition. Hence, $t \ge 2$ . If $t = 2$ , we are done. So, let us assume that $t \ge 3$ . We will prove the result by the method of mathematical induction. The result clearly holds for $t = 2$ . Let the result be true for all expressions in which the number of transpositions $t \le k$ . Now, let $t = k+1$ .

Suppose ${\alpha}_1 = (m \; r)$ . Note that the possible choices for the composition ${\alpha}_1 \circ {\alpha}_2$ are

$$(m \; r) \circ (m \; r) = Id_n, \; (m \; r)\circ (m \; \ell) = (r... ... \; m) {\mbox{ and }} (m \; r) \circ (\ell \; s) = (\ell \; s) \circ (m \; r),$$

where $\ell$ and $s$ are distinct elements of $\{1, 2, \ldots, n\}$ and are different from $m, \; r$ . In the first case, we can remove ${\alpha}_1 \circ {\alpha}_2$ and obtain $Id_n = {\alpha}_3 \circ {\alpha}_4 \circ \cdots \circ {\alpha}_t$ . In this expression for identity, the number of transpositions is $t-2 = k -1 < k$ . So, by mathematical induction, $t-2$ is even and hence $t$ is also even.

In the other three cases, we replace the original expression for ${\alpha}_1 \circ {\alpha}_2$ by their counterparts on the right to obtain another expression for identity in terms of $t = k+1$ transpositions. But note that in the new expression for identity, the positive integer $m$ doesn't appear in the first transposition, but appears in the second transposition. We can continue the above process with the second and third transpositions. At this step, either the number of transpositions will reduce by $2$ (giving us the result by mathematical induction) or the positive number $m$ will get shifted to the third transposition. The continuation of this process will at some stage lead to an expression for identity in which the number of transpositions is $t-2=k-1$ (which will give us the desired result by mathematical induction), or else we will have an expression in which the positive number $m$ will get shifted to the right most transposition. In the later case, the positive integer $m$ appears exactly once in the expression for identity and hence this expression does not fix $m$ whereas for the identity permutation $Id_n(m) = m$ . So the later case leads us to a contradiction.

Hence, the process will surely lead to an expression in which the number of transpositions at some stage is $t-2=k-1$ . Therefore, by mathematical induction, the proof of the lemma is complete.

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THEOREM 15.2.10   Let $\alpha \in {\mathcal S}_n$ . Suppose there exist transpositions $\tau_1, \tau_2, \ldots, \tau_k$ and $\sigma_1, \sigma_2, \ldots, \sigma_{\ell}$ such that

$$\alpha = \tau_1 \circ \tau_2 \circ \cdots\circ\tau_k = \sigma_1\circ\sigma_2\circ\cdots\circ\sigma_{\ell}$$

then either $k$ and $\ell$ are both even or both odd.

Proof. Observe that the condition $\tau_1 \circ \tau_2 \circ \cdots\circ\tau_k = \sigma_1\circ\sigma_2\circ\cdots\circ\sigma_{\ell}$ and $\sigma \circ \sigma = Id_n$ for any transposition $\sigma \in {\mathcal S}_n$ , implies that

$$Id_n = \tau_1 \circ \tau_2 \circ \cdots\circ\tau_k \circ \sigma_{\ell}\circ\sigma_{\ell-1}\circ\cdots\circ\sigma_{1}.$$

Hence by Lemma 14.2.9, $k + \ell$ is even. Hence, either $k$ and $\ell$ are both even or both odd. Thus the result follows. height6pt width 6pt depth 0pt

DEFINITION 15.2.11   A permutation $\sigma \in {\mathcal S}_n$ is called an even permutation if $\sigma$ can be written as a composition (product) of an even number of transpositions. A permutation $\sigma \in {\mathcal S}_n$ is called an odd permutation if $\sigma$ can be written as a composition (product) of an odd number of transpositions.

Remark 15.2.12   Observe that if $\sigma$ and $\tau$ are both even or both odd permutations, then the permutations $\sigma\circ \tau$ and $\tau\circ \sigma$ are both even. Whereas if one of them is odd and the other even then the permutations $\sigma\circ \tau$ and $\tau\circ \sigma$ are both odd. We use this to define a function on ${\mathcal S}_n$ , called the sign of a permutation, as follows:

DEFINITION 15.2.13   Let ${\mbox{sgn}} : \; {\mathcal S}_n {\longrightarrow}\{1, -1\}$ be a function defined by

$${\mbox{sgn}}(\sigma) = \left\{\begin{array}{cc} 1 & {\mbox{ if }}... ... -1 & {\mbox{ if }} \sigma {\mbox{ is an odd permutation}} \end{array}\right..$$

EXAMPLE 15.2.14  

1. The identity permutation, $Id_n$ is an even permutation whereas every transposition is an odd permutation. Thus, ${\mbox{sgn}}(Id_n) = 1$ and for any transposition $\sigma \in {\mathcal S}_n, \; {\mbox{sgn}}(\sigma) = -1$ .
2. Using Remark 14.2.12, ${\mbox{sgn}}(\sigma \circ \tau) = {\mbox{sgn}}(\sigma) \cdot {\mbox{sgn}}(\tau)$ for any two permutations $\sigma, \tau \in {\mathcal S}_n$ .

We are now ready to define determinant of a square matrix $A$ .

DEFINITION 15.2.15   Let $A=[a_{ij}]$ be an $n \times n$ matrix with entries from ${\mathbb{F}}$ . The determinant of $A$ , denoted $\det (A)$ , is defined as

$$\det(A) = \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\s... ... \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) \prod\limits_{i=1}^n a_{i \sigma(i)}.$$

Remark 15.2.16  

1. Observe that $\det (A)$ is a scalar quantity. The expression for $\det (A)$ seems complicated at the first glance. But this expression is very helpful in proving the results related with ``properties of determinant".
2. If $A=[a_{ij}]$ is a $3 \times 3$ matrix, then using (14.2.5),
   

   $$\det(A) = \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) \prod\limits_{i=1}^3 a_{i \sigma(i)}$$

   $$= {\mbox{sgn}}(\tau_1) \prod\limits_{i=1}^3 a_{i \tau_1(i)} + {\mbo... ...3 a_{i \tau_2(i)} + {\mbox{sgn}}(\tau_3) \prod\limits_{i=1}^3 a_{i \tau_3(i)} +$$

   $$\hspace{.5in} {\mbox{sgn}}(\tau_4) \prod\limits_{i=1}^3 a_{i \tau... ...}^3 a_{i \tau_5(i)} + {\mbox{sgn}}(\tau_6) \prod\limits_{i=1}^3 a_{i \tau_6(i)}$$

   $$= a_{11} a_{22} a_{33} - a_{11} a_{23} a_{32} - a_{12} a_{21} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{13} a_{22} a_{31}.$$

   
   Observe that this expression for $\det (A)$ for a $3 \times 3$ matrix $A$ is same as that given in (2.6.1).