Determinants of Matrices over Commutative Rings

Introduction

Let \(R\) be a commutative ring. Fix an integer \(n \geq 1\). Let \(\mathbf{A} = (a_{ij})\) be the \(n \times n\) square matrix

with entry \(a_{ij}\in R\) and \(1 \leq i,j \leq n\) is at the intersection of the \(i^\mathrm{th}\) row and \(j^\mathrm{th}\) column. We will define a function \(\det(\mathbf{A}) =\det (a_{ij})\) of this matrix \(\mathbf{A}\), which will also be denoted by

We will also state and prove the various familiar properties of the determinant in this setting.

The approach we follow is Emil Artin's, based on his book Galois Theory, [Artin1942]. This formulation had also appeared earlier for example in van der Waerden's Algebra, [vanderWaerden1930]. Artin bases everything over (possibly noncommutative) fields. A rewriting of these arguments to the setting of commutative rings is already available; see for example S. Lang's Algebra, [Lang2002]; but we follow Artin's original more closely. Gian Carlo Rota in [Rota1996] criticizes Artin's proofs that they are perfect but not enlightening, and so the reader is urged to make sure that he really absorbs the proof if required in repeated readings.

Axiomatic Definition

First, a note. We write a matrix \(\mathbf{A}\) also as \((\mathbf{a}_1, \mathbf{a}_2, \ldots , \mathbf{a}_n)\), where

is a column vector, in this case the \(j^\mathrm{th}\) column of \(\mathbf{A}\).

Thus the determinant \(\det(\mathbf{A})\) also written as \(\det(\mathbf{a}_1, \mathbf{a}_2, \ldots , \mathbf{a}_n)\),

For further development, we first take an axiomatic approach first and prove the existence later.

Definition. A function \(\det\) of the column vectors is called a determinant if it satisfies the following conditions.

1. The function \(\det\) is linear operating as \(D_j\) of a fixed \(j^\mathrm{th}\) column \(\mathbf{a}_j\) of \(\mathbf{A}\). That is to say,

   \(D_j(r\mathbf{a}_j + \mathbf{a}^\prime_j) = r.D_j(\mathbf{a}_j) + D_j(\mathbf{a}^\prime_j)\)

   where \(r \in R\) is a scalar.

2. If adjacent columns \(\mathbf{a}_j = \mathbf{a}_{j+1}\), then \(\det(\mathbf{A}) = 0\).

3. Let \(\mathbf{a}_j\) be given by the Kronecker delta, i. e., \(a_{ij} = 1\) if \(i =j\) and \(0\) if \(i \neq j\). Then, \(\det(\mathbf{A})=1\).

We now derive consequences of these axioms, and as part, uniqueness of the determinant(if it exists).

Properties of determinants

1. Determinant is zero if a column is zero.

   • To see this, put \(\mathbf{a}^\prime_j = 0\) and \(r = 0\) in equation eqref{eq:detlinear}.

2. Determinant does not change if a scalar multiple of one column is added to an adjacent one.

   • From linearity, \(D_j(\mathbf{a}_j + r\mathbf{a}_{j+1}) = D(\mathbf{a}_j) + rD(\mathbf{a}_{j+1})\) and from axiom 2, we have \(D_j(\mathbf{a}_{j+1}) = 0\).

3. Determinant changes sign if two adjacent columns are exchanged.

   • Let columns \(j, j+1\) be \(\mathbf{a}_j\) and \(\mathbf{a}_{j+1}\) respectively. Add column \(j\) to column \(j+1\).

   • Now subtract column \(j+1\) from column \(j\). Then add column \(j\) to column \(j+1\).

   • The \(j^\mathrm{th}\) and \((j+1)^\mathrm{th}\) columns are now \(-\mathbf{a}_{j+1}\) and \(\mathbf{a}_j\) respectively. The \(-\) sign can be taken outside determinant by linearity.

4. Determinant vanishes if any two columns are equal.

   • Make the columns adjacent via repeated exchange of adjacent matrices, in which only the sign of determinant changes.

   • Subtract one column from the other. By axiom 2., the determinant vanishes.

5. Determinant is invariant when a scalar multiple of a column is added to any other column.

   • Make the columns adjacent via repeated exchange of adjacent matrices, in which only the sign of determinant changes.

   • By property 2, adding does not change determinant.

   • Now perform in reverse order all the previous exchanges of adjacent columns. Any sign change is cancelled out.

6. Determinant changes sign if any two columns are exhanged.

   • Similar to the above argument, use property 3.

7. Let \(\sigma\) be a permutation of \(\{ 1, 2, \ldots , n \}\). Let \(\mathrm{sgn}(\sigma)\) denote the sign of the permutation \(\sigma\). Then,

   \begin{equation*} \det(\mathbf{a}_{\sigma(1)}, \mathbf{a}_{\sigma(2)}, \ldots , \mathbf{a}_{\sigma(n)} = \mathrm{sgn}(\sigma) \det(\mathbf{a}_1, \mathbf{a}_2, \ldots , \mathbf{a}_n). \end{equation*}

   • This is clear from property 3 along with a \(2\)-cycle decomposition of the permutation \(\sigma\).

Leibniz Expansion and Multiplication Theorem

Replace \(\mathbf{a}_j\) with the column vector

where \(b_{ij}\) are scalars.

Note, here the \(j\) in \(\mathbf{a}_j\) denotes the column of \(\mathbf{A}\) and its \(i^\mathrm{th}\) row is \(a_{ij}\). Consider the \(i^\mathrm{th}\) of the above expression for \(\mathbf{a}^\prime_j\). We get,

Therefore the matrix \(\mathbf{A}^\prime\) is nothing other than the matrix product \(\mathbf{A}\mathbf{B}\).

Now we compute the determinant \(\det(\mathbf{a}^\prime_1, \mathbf{a}^\prime_2 , \ldots , \mathbf{a}^\prime_n)\). Using axiom 1.,

Here \(S_n\) is the group of all permutations on \(n\) symbols. Commutativity of the ring \(R\) was used in moving from equation ref{eq:detexp1} to ref{eq:detexp2}.

!!! note "Verify that"

For the statements proved so far, only the axioms (1) of linearity and (2) of vanishing determinant of matrices with equal adjacent rows, were used.

Now, let \(\mathbf{A}\) be given by the Kronecker delta and \(\mathbf{a}_j\) be the corresponding unit column vectors. Therefore \(\mathbf{a}^\prime_j = \mathbf{b}_j\), the column vector of the matrix \(\mathbf{B} = (b_{ij})\). Now by equation eqref{eq:colexp} and by axiom (3) for determinants,

This is the Leibnitz expansion of the determinant of the matrix \(\mathbf{B} = (b_{ij})\).

Plugging this back into eqref{eq:colexp}, we get,

This is called the multiplication theorem for the determinant.

Uniqueness of Determinant

The Leibniz expansion shows that determinants are unique if they exist, since they must equal the right side of equation \(\eqref{eq:leibexp}\).

Moreover, let \(F\) be any function of the determinant satisfying the axioms (1) and (2), but not necessarily (3). Then, using equation eqref{eq:colexp} which is still valid for \(F\), and substituing the Leibniz expansion for \(D\), we get:

Letting \(\mathbf{a}_j\) be equal to \(\mathbf{u}_j\), the unit vectors corresponding to the Kronecker delta \(\mathbf{U}\), this becomes:

where \(f = F(\mathbf{u}_1, \mathbf{u}_2, \ldots , \mathbf{u}_n )\) is the determinant of the Kronecker delta.

Thus, when axiom (3) is not satisfied, there will only be a possible change of a scalar factor. Therefore axiom (3) normalizes the determinant function.

Determinant of Transpose

Fix two adjacent indices \(i\) and \(i+1\), with \(0 \leq i \leq n-1\). Now let \(\mathbf{A}\) be the matrix of a row transformation defined in the following way. Let \(\mathbf{u}_j\) denote the \(j^\mathrm{th}\) row of the matrix given by the Kronecker delta.

Then, let \(\mathbf{a}_j = \mathbf{u}_j\) when \(j \neq i, i+1\), let :math:` mathbf{a}_i = mathbf{u}_i + mathbf{u}_{i+1}` and let \(\mathbf{a}_{i+1} = 0\).

Left multiplication by \(A\) replaces row \(i+1\) with row \(i\), not changing the other rows, and \(\det(\mathbf{A}) = 0\).

Using this in the multiplication theorem eqref{eq:detproduct}, we see that:

  8. Determinant vanishes if two adjacent rows are identical.

Each term on the right side of the Leibniz expansion eqref{eq:leibexp} is a product with exactly one term from each row. Thus, as a function of each row, the determinant is linear and homogenous. Moreover, the determinant of the matrix where the rows are given by the unit row vectors specified by the Kronecker delta, is \(1\), as the Kronecker delta is symmetric.

Therefore, the determinant considered as a function of the row vectors satisfies all three axioms. This is equivalent to taking the transpose, and, therefore, by uniqueness,

  9. Determinant is unchanged under transpose operation.

It follows that:

  10. Determinant is zero if any two rows are identical. Determinant changes sign if any two rows are switched. Determinant is invariant when a scalar multiple of a row is added to another.

Similarly, property 7 can be restated in terms of row permutations, and Leibniz expansion can be rewritten where in the products the second index is being permuted instead of the first.

Cofactor expansion and Existence of Determinant

We now prove the existence of determinants by induction, and using cofactor expansion.

For \(n=1\), the matrix \(a_{11}\) has determinant \(a_{11}\).

Suppose determinants exist for \((n-1 )\times (n-1)\) matrices over \(R\). Let \(\mathbf{A}=(a_{ij})\) be an \(n \times n\) matrix. To each element \(a_{ij}\), we associate an \((n-1 )\times (n-1)\) matrix by cancelling out the \(i^\mathrm{th}\) row and \(j^\mathrm{th}\) column of \(\mathbf{A}\), and its determinant is called the minor \(M_{ij}\).

From the minor, the cofactor \(A_{ij}\) is defined as \((-1)^{i+j}M_{ij}\).

Now fix a row \(i\) and define a function \(D\) as the sum of the products of row and cofactors:

Now fix a column \(k\) and consider \(D\) as a function of the \(k^\mathrm{th}\) column \(\mathbf{a}_k\). In the sum in equation eqref{eq:cofexp}, for \(j \neq k\), we have the product :math:` a_{ij}A{ij}` with \(a_{ij}\) not depending at all on \(\mathbf{a}_k\) and the cofactor \(A_{ij}\) linearly depending on each of the columns of the \((n-1) \times (n-1)\)-matrix obtained by removing the \(i^\mathrm{th}\) row and \(j^\mathrm{th}\) column from \(\mathbf{A}\); i. e., it depends linearly on \(\mathbf{a}_k\). Similarly, when \(k = j\), in the product :math:` a_{ij}A{ij}`, the first term \(a_{ij}\) depends linearly on \(\mathbf{a}_k\) and the second term \(A{ij}\) does not depend at all on \(\mathbf{a}_k\), the \(k^\mathrm{th}\) row having been removed. Therefore \(D\) satisfies axiom 1 for determinants.

Now suppose two adjacent columns \(k\) and \(k+1\) are equal for a matrix \(\mathbf{A}\). Then, \(a_{i\ k} = a_{i\ k+1}\) and the minors \(M_{i\ k} = M_{i\ k+1}\), but the cofactor \(A_{i\ k} = - A_{i\ k+1}\), so we obtain \(D = 0\) in equation eqref{eq:cofexp}. Therefore \(D\) satisfies axiom 2 for determinants.

Setting \(\mathbf{A}\) equal to a unit matrix and explicitly computing, we see that \(\det (\mathbf{A}) =1\), so \(D\) satisfies axiom 3 also, and thus we have shown the existence of the \(n \times n\) determinant and of formula eqref{eq:cofexp}, which is also called the row expansion, or Laplace expansion of the determinant

Linear equations and Cramer's Rule

Now instead of the right side of equation eqref{eq:cofexp}, consider the sum:

where \(j \neq i\). As before, this sum satisfies axioms \(1\) and \(2\). Letting \(\mathbf{A} =\mathbf{1}\), the unit matrix, the only nonzero term in this sum is \(a_{jj} A{ij}\). With \(i \neq j\), when removing the \(i^\mathrm{th}\) row and \(j^\mathrm{th}\) column from an \(n \times n\) unit matrix, we remove both \(a_{ii}=a_{jj}=1\) and have to find the determinant of an \((n-1) \times (n-1)\) matrix with only \(n-2\) nonzero terms. But each product in the Leibniz expansion for an \(n\times n\) matrix has \(n\) terms, and so this determinant is \(0\). Therefore,

So far in equations eqref{eq:cofexp} and eqref{eq:cofinj}, we might as well have expanded by the columns instead of by the rows, by the symmetry of determinant under transpose. Doing so, we get:

Define the adjugate matrix \(\mathrm{adj}(\mathbf{A})\) of a matrix \(\mathbf{A}\) as follows:

That is, write the cofactor matrix and take the transpose.

Then the equations eqref{eq:cofexp}, eqref{eq:cofinj} and eqref{eq:cofcol} mean:

where \(\mathbf{1}\) is the unit matrix.

This leads to the following lemmas.

Lemma A matrix \(\mathbf{A}\) over a ring \(R\) is invertible if and only if \(\det(\mathbf{A})\) is.

Proof Suppose \(\mathbf{A}\) is invertible with inverse \(\mathbf{A}^{-1}\). Then by the multiplication theorem, \(\det(\mathbf{A}).\det(\mathbf{A}^{-1}) = 1\). Now suppose \(\det(\mathbf{A})\) is invertible. Then, by equation eqref{eq:adjdet},

Q. E. D.

Lemma If the \(n\) column vectors \(\mathbf{a}_1, \mathbf{a}_2, \ldots , \mathbf{a}_n\) can generate any column vector \(\mathbf{b} \in R^n\) via linear combinations, then \(\det(\mathbf{A})\) is invertible.

Proof The assumption of the lemma means that in equation eqref{eq:colsubn}, we can choose \(\mathbf{a}^\prime_j = \mathbf{u_j}\), the unit vector. So, we have \(\mathbf{1} = \mathbf{A} \mathbf{B}\) for some \(\mathbf{B}\), and by the multiplication formula, \(\det(\mathbf{A})\) is invertible.

Q. E. D.

Theorem: Cramer's rule

Suppose \(\det(\mathbf{A})\) is invertible. Then the set of linear equations

where \(b_i\) are arbitary scalars, has a unique solution, which can be computed via determinants involving \(a_{ij}\) and \(y_i\).

Proof For a given \(j\), multiply the equation for \(y_i\) with \(A_{ij}\) and add, and apply equation eqref{eq:cofcol}, and examine the coeffients of \(x_j\):

The right side is also the determinant of the matrix where the \(j^\mathrm{th}\) column of \(\mathbf{A}\) is replaced by the column vector \(\mathbf{y} = (y_1, \ldots , y_n)\).

Q. E. D.

Bibliography

!!! note "Amazon links"

Symbols of the form {% amazon 1294823477 %} are links to Amazon, when relevant.