Revision as of 00:31, 9 May 2020 by Mag1c (talk | contribs) (determinants in terms of simpler determinants)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

The determinant is an important notion in linear algebra.

For an $n\times n$ matrix $a = (a_{ij})$, the determinant is defined by the sum \begin{align*} \det a &= \sum_{\sigma \in S_n} \text{sign} (\sigma) a_{1\sigma(1)} a_{2\sigma(2)} \dotsc a_{n\sigma(n)} \\ &= \sum_{\sigma \in S_n} \text{sign}(\sigma) \prod_{i=1}^n a_{i\sigma(i)} , \end{align*} where $S_n$ is the set of all permutations on the set $\{ 1, \dotsc, n\}$, and $\text{sign}(\sigma)$ is the parity of the permutation $\sigma$.

For example, the determinant of a $2\times 2$ matrix $\begin{pmatrix} a&b \\c&d \end{pmatrix}$ is $ad - bc$.

This quantity may seem unwieldy, but surprisingly, it is multiplicative. That is, for any $n \times n$ matrices $a,b$ (over the same commutative field), \[\det (ab) = \det (a) \cdot \det (b) .\]

More generally, if $F$ is a commutative field and $a$ is an element of a (strictly power associative) $n$-dimensional $F$-algebra $A$, then the determinant of $a$ is $(-1)^n$ times the constant term of the characteristic polynomial of $a$.

Our generalized determinants also satisfy the multiplicative property when $A$ is associative.

Determinants in Terms of Simpler Determinants

An $n\times n$ determinant can be written in terms of $(n-1)\times(n-1)$ determinants: \[\det(a_{n\times n})=\sum_{i=1}^n(-1)^{i+1}\det(m_i)\] where $m_i$ is the $(n-1)\times(n-1)$ matrix formed by removing the $1$st row and $i$th column from $a$: \[m_i=\begin{pmatrix} \blacksquare & \blacksquare & ... & \blacksquare & \blacksquare & \blacksquare & ... & \blacksquare & \blacksquare\\ a_{21} & a_{22} & ... & a_{2,i-1} & \blacksquare & a_{2,i+1} & ... & a_{2,n-1} & a_{2n}\\ a_{31} & a_{32} & ... & a_{3,i-1} & \blacksquare & a_{3,i+1} & ... & a_{3,n-1} & a_{3n}\\ &&&&\vdots&&&&\\ a_{n1} & a_{n2} & ... & a_{n,i-1} & \blacksquare & a_{n,i+1} & ... & a_{n,n-1} & a_{nn} \end{pmatrix}=\begin{pmatrix} a_{21} & ... & a_{2,i-1} & a_{2,i+1} & ... & a_{2n}\\ a_{31} & ... & a_{3,i-1} & a_{3,i+1} & ... & a_{3n}\\ &&\vdots&\vdots&&&\\ a_{n1} & ... & a_{n,i-1} & a_{n,i+1} & ... & a_{nn} \end{pmatrix}\]

This makes it easy to see why the determinant of an $n\times n$ matrix $a$ is the sum of the diagonals labeled $+$, minus the sum of the diagonals labeled $-$, where "diagonal" means the product of the terms along it:

[asy] size(300); draw(arc((3.3,-1.5), 4, 155, 205)); draw(arc((-0.3,-1.5), 4, -25, 25)); label("$a_{11}$",(0,0)); label("$a_{12}$",(1,0)); label("$...$",(2,0)); label("$a_{1n}$",(3,0));  label("$a_{21}$",(0,-1)); label("$a_{22}$",(1,-1)); label("$...$",(2,-1)); label("$a_{2n}$",(3,-1));  label("$\vdots$",(0,-2)); label("$\ddots$",(2,-2));  label("$a_{n1}$",(0,-3)); label("$a_{n2}$",(1,-3)); label("$...$",(2,-3)); label("$a_{nn}$",(3,-3)); label("$a_{11}$",(0,0)); label("$a_{12}$",(1,0)); label("$...$",(2,0)); label("$a_{1n}$",(3,0));   label("$a_{11}$",(4,0)); label("$a_{12}$",(5,0)); label("$...$",(6,0)); label("$a_{1n}$",(7,0));  label("$a_{21}$",(4,-1)); label("$a_{22}$",(5,-1)); label("$...$",(6,-1)); label("$a_{2n}$",(7,-1));  label("$\vdots$",(4,-2)); label("$\ddots$",(6,-2));  label("$a_{n1}$",(4,-3)); label("$a_{n2}$",(5,-3)); label("$...$",(6,-3)); label("$a_{nn}$",(7,-3));  for (int i=0; i<4; ++i) { draw((i-0.2,0.2)--(i+3-0.2,-3+0.2)); label("$+$",(i-0.4,0.4)); } for (int i=0; i<4; ++i) { draw((i+0.2+4,0.2)--(i-3+4+0.2,-3+0.2)); label("$-$",(i+0.4+4,0.4)); } [/asy]

Matrix Determinants are Multiplicative

In this section we prove that the determinant as defined for $n\times n$ matrices is multiplicative.

We first note that \begin{align*} \det a \det b &= \sum_{\sigma \in S_n} (-1)^{\sigma} 	\prod_{i=1}^n a_{i\sigma(i)} 	\sum_{\tau \in S_n} (-1)^{\tau} \prod_{j=1}^n b_{j\tau(j)} \\ &= \sum_{\sigma, \tau \in S_n} (-1)^{\tau \sigma} 	\prod_{i=1}^n a_{i\sigma(i)} \cdot \prod_{j=1}^n b_{\sigma(j) 	\tau (\sigma(j))} \\ &= \sum_{\sigma, \tau \in S_n} (-1)^{\tau \sigma} 	\prod_{i=1}^n a_{i \sigma(i)} b_{\sigma(i) \tau\circ \sigma(i)} , \end{align*} from rearrangements of terms. If we let $\upsilon = \tau \circ \sigma$, we then have \begin{align} \nonumber \det a \det b &= \sum_{\upsilon \in S_n} (-1)^\upsilon 	\sum_{\sigma \in S_n} \prod_{i=1}^n a_{i\sigma(i)} b_{\sigma(i) 	\upsilon(i)} \\ &= \sum_{\sigma \in S_n} \sum_{\upsilon \in S_n} (-1)^\upsilon 	\prod_{i=1}^n a_{i\sigma(i)} b_{\sigma(i) \upsilon(i)} . \end{align}

On the other hand, \begin{align*} \det ab &= \sum_{\upsilon \in S_n} (-1)^\upsilon \prod_{i=1}^n (ab)_{i 	\upsilon(i)} \\ &= \sum_{\upsilon \in S_n} (-1)^\upsilon \prod_{i=1}^n \sum_{j=1}^n 	a_{ij} b_{j \upsilon(i)} \\ &= \sum_{\upsilon \in S_n} (-1)^\upsilon \sum_{\phi \in [n]^{[n]}} 	\prod_{i=1}^n a_{i \phi(i)} b_{\phi(i) \upsilon(i)} \\ &= \sum_{\phi \in [n]^{[n]}} \sum_{\upsilon \in S_n} (-1)^\upsilon 	\prod_{i=1}^n a_{i \phi(i)} b_{\phi(i) \upsilon(i)} , \end{align*} where $[n]$ is the set $\{1, \dotsc, n\}$, and $[n]^{[n]}$ is the set of functions mapping $[n]$ into itself.

From equation (1), it thus suffices to show that if $\phi \in [n]^{[n]}$ is not a permutation on $[n]$, then \[\sum_{\upsilon \in S_n} (-1)^\upsilon \prod_{i=1}^n 	a_{i\phi(i)} b_{\phi(i) \upsilon(i)} = 0 . \qquad (2)\]

To this end, suppose that $\phi$ is not a permutation. Then there exist distinct integers $a,b \in [n]$ such that $\phi(a) = \phi(b)$. Let $\psi$ be the permutation on $[n]$ that transposes $a$ and $b$ while fixing everything else. Then \[\prod_{i=1}^n a_{i \phi(i)} b_{\phi(i) \upsilon(i)} 	= \prod_{i=1}^n a_{i\phi(i)} b_{\phi(i) \upsilon(\psi(i))} ,\] as the latter product is the same as the former, with two terms switched. On the other hand $\psi$ is an odd permutation, so \[(-1)^\upsilon \prod_{i=1}^n a_{i \phi(i)} b_{\phi(i) \upsilon(i)} + (-1)^{\upsilon \psi} \prod_{i=1}^n a_{i \phi(i)} b_{\phi(i) \upsilon(\psi(i))} =0 .\] Since $\psi = \psi^{-1}$, we can partition the elements of $S_n$ into pairs $\{ \sigma, \sigma \circ \tau\}$ for which the equation above holds. Equation 2 then follows, and we are done. $\blacksquare$

Equivalence of Definitions

We now prove that our two definitions are equivalent. We first note that the definitions coincide in the case of upper-triangular matrices, as each entry in the diagonal of an upper-triangular matrix corresponds to a (generalized) eigenvalue of the matrix.

We now use the fact that every element $a$ of $M_n(F)$ is similar to an upper triangular matrix; that is, there exists an upper triangular matrix $b$ and an invertible matrix $c$ such that \[a = c b c^{-1} .\] Writing $\det\nolimits'$ for our specialized determinant for matrices and $\det$ for our generalized definition with the characteristic polynomial, we have \begin{align*} \det\nolimits'(a) = \det\nolimits'(cbc^{-1}) &= \det\nolimits'(c) \cdot \det\nolimits'(bc^{-1}) \\ &= \det\nolimits'(bc^{-1}) \det\nolimits'(c) \\ &= \det\nolimits'(b) = \det b = \det a , \end{align*} as the characteristic polynomial does not change under automorphisms of $A$ that fix $F$. Our two definitions are therefore equivalent. $\blacksquare$


  • Garibaldi, Skip, "The Characteristic Polynomial and Determinant Are Not Ad Hoc Constructions". American Mathematical Monthly 111 (2004), no. 9, p. 761, Nov. 2004. Preprint

See also