Chapter 3 Determinants

The reader probably already met determinants in calculus or algebra, at least the determinants of $2\times 2$ and $3\times 3$ matrices. For a $2\times 2$ matrix

the determinant is simply $ad-bc$; the determinant of a $3\times 3$ matrix can be found by the “Star of David” rule.

In this chapter we would like to introduce determinants for $n\times n$ matrices. I don’t want just to give a formal definition. First I want to give some motivation, and then derive some properties the determinant should have. Then if we want to have these properties, we do not have any choice, and arrive to several equivalent definitions of the determinant.

It is more convenient to start not with the determinant of a matrix, but with determinant of a system of vectors. There is no real difference here, since we always can join vectors together (say as columns) to form a matrix.

Let us have $n$ vectors $\mathbf{v}_{1},\mathbf{v}_{2},\ldots,\mathbf{v}_{n}$ in $\mathbb{R}^{n}$ (notice that the number of vectors coincides with dimension), and we want to find the $n$-dimensional volume of the parallelepiped determined by these vectors.

The parallelepiped determined by the vectors $\mathbf{v}_{1},\mathbf{v}_{2},\ldots,\mathbf{v}_{n}$ can be defined as the collection of all vectors $\mathbf{v}\in\mathbb{R}^{n}$ that can be represented as

It can be easily visualized when $n=2$ (parallelogram) and $n=3$ (parallelepiped). So, what is the $n$-dimensional volume?

If $n=2$ it is area; if $n=3$ it is indeed the volume. In dimension $1$ is it just the length.

Finally, let us introduce some notation. For a system of vectors (columns) $\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n}$ we will denote its determinant (that we are going to construct) as $D(\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n})$. If we join these vectors in a matrix $A$ (column number $k$ of $A$ is $\mathbf{a}_{k}$), then we will use the notation $\det A$,

Also, for a matrix

its determinant is often denoted by

We know, that for dimensions 2 and 3 “volume” of a parallelepiped is determined by the base times height rule: if we pick one vector, then height is the distance from this vector to the subspace spanned by the remaining vectors, and the base is the $(n-1)$-dimensional volume of the parallelepiped determined by the remaining vectors.

Now let us generalize this idea to higher dimensions. For a moment we do not care about how exactly to determine height and base. We will show, that if we assume that the base and the height satisfy some natural properties, then we do not have any choice, and the volume (determinant) is uniquely defined.

The plan of the game is now as follows: using the properties that as we decided in Section 3.2 the determinant should have, we derive other properties of the determinant, some of them highly non-trivial. We will show how to use these properties to compute the determinant using our old friend—row reduction.

Later, in Section 3.4, we will show that the determinant, i.e. a function with the desired properties exists and unique. After all we have to be sure that the object we are computing and studying exists.

While our initial geometric motivation for determinant and its properties came from considering vectors in the real vector space $\mathbb{R}^{n}$, so they relate only to matrices with real entries, all the constructions below use only algebraic operations (addition, multiplication, division) and are applicable to matrices with complex entries, and even with entries in an arbitrary field.

So in what follows we are constructing determinant not just for real matrices, but for complex matrices as well (and also for matrices with entries in an arbitrary field). The nice geometric motivation for the properties works only in the real case, but after we decided on the properties of the determinant (see properties 1–3 below) everything works in the general case.

In this section we arrive to the formal definition of the determinant. We show that a function, satisfying the basic properties 1, 2, 3 from Section 3.3 exists, and moreover, such function is unique, i.e. we do not have any choice in constructing the determinant.

Consider an $n\times n$ matrix $A=\{a_{j,k}\}_{j,k=1}^{n}$, and let $\mathbf{v}_{1},\mathbf{v}_{2},\ldots,\mathbf{v}_{n}$ be its columns, i.e.

Using linearity of the determinant we expand it in the first column $\mathbf{v}_{1}$:

Then we expand it in the second column, then in the third, and so on. We get

Notice, that we have to use a different index of summation for each column: we call them $j_{1},j_{2},\ldots,j_{n}$; the index $j_{1}$ here is the same as the index $j$ in (3.4.1).

It is a huge sum, it contains $n^{n}$ terms. Fortunately, some of the terms are zero. Namely, if any 2 of the indices $j_{1},j_{2},\ldots,j_{n}$ coincide, the determinant $D(\mathbf{e}_{j_{1}},\mathbf{e}_{j_{2}},\ldots\mathbf{e}_{j_{n}})$ is zero, because there are two equal columns here.

So, let us rewrite the sum, omitting all zero terms. The most convenient way to do that is using the notion of a permutation. Informally, a permutation of an ordered set $\{1,2,\ldots,n\}$ is a rearrangement of its elements. A convenient formal way to represent such a rearrangement is by using a function

where $\sigma(1),\sigma(2),\ldots,\sigma(n)$ gives the new order of the set $1,2,\ldots,n$. In other words, the permutation $\sigma$ rearranges the ordered set $1,2,\ldots,n$ into $\sigma(1),\sigma(2),\ldots,\sigma(n)$.

Such function $\sigma$ has to be one-to-one (different values for different arguments) and onto (assumes all possible values from the target space). The functions which are one-to-one and onto are called bijections, and they give one-to-one correspondence between the domain and the target space.¹¹1 There is another canonical way to represent permutation by a bijection $\sigma$, namely in this representation $\sigma(k)$ gives new position of the element number $k$. In this representation $\sigma$ rearranges $\sigma(1),\sigma(2),\ldots,\sigma(n)$ into $1,2,\ldots,n$. While in the first representation it is easy to write the function if you know the rearrangement of the set $1,2,\ldots,n$, the second one is more adapted to the composition of permutations: it coincides with the composition of functions. Namely if we first perform the permutation that correspond to a function $\sigma$ and then one that correspond to $\tau$, the resulting permutation will correspond to $\tau\circ\sigma$.

Although it is not directly relevant here, let us notice, that it is well-known in combinatorics, that the number of different permutations of the set $\{1,2,\ldots,n\}$ is exactly $n!$. The set of all permutations of the set $\{1,2,\ldots,n\}$ will be denoted $\operatorname{Perm}(n)$.

Using the notion of a permutation, we can rewrite the determinant as

The matrix with columns $\mathbf{e}_{\sigma(1)},\mathbf{e}_{\sigma(2)},\ldots,\mathbf{e}_{\sigma(n)}$ can be obtained from the identity matrix by finitely many column interchanges, so the determinant

is $1$ or $-1$ depending on the number of column interchanges.

To formalize that, we (informally) define the sign (denoted $\operatorname{sign}\sigma$) of a permutation $\sigma$ to be 1 if an even number of interchanges is necessary to rearrange the $n$-tuple $1,2,\ldots,n$ into $\sigma(1),\sigma(2),\ldots,\sigma(n)$, and $\operatorname{sign}(\sigma)=-1$ if the number of interchanges is odd.

It is a well-known fact from the combinatorics, that the sign of permutation is well defined, i.e. that although there are infinitely many ways to get the $n$-tuple $\sigma(1),\sigma(2),\ldots,\sigma(n)$ from $1,2,\ldots,n$, the number of interchanges is either always odd or always even.

One of the ways to show that is to introduce an alternative definition. Let $K=K(\sigma)$ be the number of disorders of $\sigma$, i.e. the number of pairs $(j,k)$, $j,k\in\{1,2,\ldots,n\}$, $j<k$ such that $\sigma(j)>\sigma(k)$, and see if the number is even or odd. We call the permutation $\sigma$ odd if $K$ is odd and even if $K$ is even. Then define $\operatorname{sign}\sigma:=(-1)^{K(\sigma)}$; note that this way $\operatorname{sign}\sigma$ is well defined.

We want to show that $\operatorname{sign}\sigma=(-1)^{K(\sigma)}$ can indeed be computed by rearranging the $n$-tuple $1,2,\ldots,n$ into $\sigma(1),\sigma(2),\ldots,\sigma(n)$ and counting the number of interchanges, as was described above.

If $\sigma(k)=k$ $\forall k$, then the number of disorders $K(\sigma)$ is $0$, so sign of such identity permutation is $1$. Note also, that any elementary transpose, which interchange two neighbors, changes the sign of a permutation, because it changes (increases or decreases) the number of disorders exactly by 1. So, to get from a permutation to another one always needs an even number of elementary transposes if the permutations have the same sign, and an odd number if the signs are different.

Finally, any interchange of two entries can be achieved by an odd number of elementary transposes. This implies that sign changes under an interchange of two entries. So, to get from $1,2,\ldots,n$ to an even permutation (positive sign) one always need even number of interchanges, and odd number of interchanges is needed to get an odd permutation (negative sign).

So, if we want determinant to satisfy basic properties 1–3 from Section 3.3, we must define it as

where the sum is taken over all permutations of the set $\{1,2,\ldots,n\}$.

If we define the determinant this way, it is easy to check that it satisfies the basic properties 1–3 from Section 3.3. Indeed, it is linear in each column, because for each column every term (product) in the sum contains exactly one entry from this column.

Interchanging two columns of $A$ just adds an extra interchange to the permutation, so right side in (3.4.2) changes sign. Finally, for the identity matrix $I$, the right side of (3.4.2) is $1$ (it has one non-zero term).

For an $n\times n$ matrix $A=\{a_{j,k}\}_{j,k=1}^{n}$ let $A_{j,k}$ denotes the $(n-1)\times(n-1)$ matrix obtained from $A$ by crossing out row number $j$ and column number $k$.