Determinants

The determinant of a matrix $A$, denoted $\det(A)$, is only defined when $A$ is a square matrix, i.e., the number of rows is equal to the number of columns. If $A$ is an $n\times n$ matrix then $A$ may be regarded as a function of $F^n$, sending vectors to points. If $F=\mathbb{R}$ then it turns out $A$ will send the unit hypercube in $\mathbb{R}^n$ to a parallelepiped (the $n$-dimensional analog of a parallelogram). It is known that the absolute value of the determinant of $A$ measures the volume of that parallelpiped. If $m=n=2$ the determinant is easy to define:

$$\det\left( \begin{array}{cc} a&b\\ c&d \end{array} \right)=ad-bc.$$

The easiest constructive way to define the determinant of an arbitrary $n\times n$ matrix $A=(a_{ij})$ is to use the Laplace cofactor expansion: for any $1\leq i\leq n$, we have

$$\det(A)=\sum_{j=1}^n (-1)^{i+j}\det(A_{ij}),$$ (5.2)

where $A_{ij}$ is the $(n-1)\times (n-1)$ `submatrix of $A$' obtained by omitting all the entries in the $i^{th}$ row or the $j^{th}$ column. The matrix $A_{ij}$ is called the $(i,j)$-minor of $A$ and $(-1)^{i+j}\det(A_{ij})$ is called the $(i,j)$-cofactor. A square matrix $A$ is singular if $\det(A)=0$. Otherwise, $A$ is called non-singular or invertible.

Example 5.4.1   Taking $i=2$,

$$\begin{array}{c} \det\left( \begin{array}{ccc} 1&2&3\\ ... ...\right)\\ =(-4)(18-24)+(5)(9-21)+(-6)(8-14)=0. \end{array}$$

This implies $A$ is singular. Indeed, the parallelepiped generated by $(1,\ 2,\ 3)$, $(4,5,6)$, $(7,8,9)$, must be flat ($2$-dimensional, hence have 0 volume) since $(4,5,6)=(1,2,3)+(1,1,1)$ and $(7,8,9)=(1,2,3)+2(1,1,1)$.

There is an analogous Lagrange cofactor expansion for columns as well. See [NJ] (or any other book on linear algebra), for example. An important fact about singular matrices, and one that we will use later, is the following.

Lemma 5.4.2   Suppose $A$ is an $n\times n$ matrix with entries in $F$. The following are equivalent:

• $\det(A)\not= 0$,
• there is no non-zero vector $v$ such that $Av=0$, where 0 is the zero vector in $F^n$,
• $A^{-1}$ exists.

For a proof, see any text on linear algebra. We end this section with one last key property of determinants.

Lemma 5.4.3   If $A,B$ are any two $n\times n$ matrices having entries in $F$ then $\det(AB)=\det(A)\det(B)$.

More details on all this material can be found in any text book on linear algebra, for example [NJ].