9 Determinants

Determinants are a fundamental concept in linear algebra, associated with square matrices. They have various applications, such as:

Determining the invertibility of matrices
Calculating the area and volume of geometric shapes

9.1 Axioms

Determinants are functions that assign a unique scalar value to each square matrix. We denote the determinant of a matrix $A$ as $\det(A)$ or $|A|$. The axioms that characterize determinants are:

Identity Axiom: The determinant of the identity matrix is 1.

\[|I_n| = 1,\quad\text{for every }n\in\mathbb{N},\]

where $I_n$ is the $n\times n$ identity matrix.

Linearity Axiom: The determinant is a linear function of each row separately.

(scalar multiplication) If a row is multiplied by a scalar, the determinant is multiplied by the same scalar.
(additivity) \[\det \begin{pmatrix} \leftarrow & \mathbf{r_1} & \rightarrow \\ & \vdots & \\ \leftarrow & \mathbf{x} + \mathbf{y} & \rightarrow \\ & \vdots & \\ \leftarrow & \mathbf{r_n} & \rightarrow \end{pmatrix} = \det \begin{pmatrix} \leftarrow & \mathbf{r_1} & \rightarrow \\ & \vdots & \\ \leftarrow & \mathbf{x} & \rightarrow \\ & \vdots & \\ \leftarrow & \mathbf{r_n} & \rightarrow \end{pmatrix} +\det \begin{pmatrix} \leftarrow & \mathbf{r_1} & \rightarrow \\ & \vdots & \\ \leftarrow & \mathbf{y} & \rightarrow \\ & \vdots & \\ \leftarrow & \mathbf{r_n} & \rightarrow \end{pmatrix}.\]

Alternating Property Axiom: If two rows are swapped, the determinant changes sign.

The three axioms uniquely define the determinant function, even though this may not be immediately apparent. It can be shown that these axioms lead to a formula for the determinant of an $n \times n$ matrix $A$: \[\det(A) = \sum_{\pi \in S_n} \text{sgn}(\pi) , a_{1\pi(1)} , a_{2\pi(2)} , \cdots , a_{n\pi(n)}\] where:

$S_n$ is the set of all permutations of the indices ${1, 2, \dots, n}$
$\text{sgn}(\pi)$ is the sign of the permutation $\pi$, which is +1 for even permutations and -1 for odd permutations
$a_{i\pi(i)}$ is the entry in the $i$-th row and $\pi(i)$-th column of $A$

However, this formula is not practical for calculating determinants, as there are $n!$ permutations in $S_n$, leading to a very large number of terms in the sum as $n$ increases.

Instead, we will see that the axioms are useful for keeping track of elementary row operations. This property allows us to compute determinants efficiently and prove various properties about them, without relying on the permutation formula directly.

We can also view the determinant on $n\times n$ matrices as a function that maps an ordered set of $n$ vectors in $\mathbb{R}^n$ to a real number: \[\det: \mathbb{R}^n \times \mathbb{R}^n \times \cdots \times \mathbb{R}^n \to \mathbb{R}\] Here, we interpret each element of $\mathbb{R}^n$ as a row of the matrix, meaning that the determinant takes $𝑛$ row vectors from $\mathbb{R}^n$ and returns a scalar. If an $n\times n$ matrix is represented in terms of its rows: \[A = \begin{bmatrix} \leftarrow &\mathbf{r}_1 &\rightarrow\\ \leftarrow &\mathbf{r}_2 & \rightarrow\\ &\vdots & \\ \leftarrow & \mathbf{r}_n & \rightarrow \end{bmatrix}\quad\text{then}\quad\det(A)=\det(\mathbf{r}_1,\mathbf{r}_2,\dots,\mathbf{r}_n).\]

Using this notation, the identity axiom states that: \[\det(\mathbf{e}_1,\mathbf{e}_2,\cdots,\mathbf{e}_n)=1\] where ${\mathbf{e}_1,\mathbf{e}_2,\cdots,\mathbf{e}_n}$ is the standard basis of $\mathbb{R}^n$, defined as: \[\mathbf{e}_1=\begin{bmatrix}1\\0\\ \vdots\\0\end{bmatrix},\quad \mathbf{e}_2=\begin{bmatrix}0\\1\\ \vdots\\0\end{bmatrix},\quad \cdots,\quad \mathbf{e}_n=\begin{bmatrix}0\\0\\ \vdots\\1\end{bmatrix}.\]

9.2 Properties of determinant

From the axioms we infer many properties of determinants:

If $A$ has a zero row, then $\det(A) = 0$

Let $A$ be a matrix with a zero row in position $i$. By the Linearity Axiom in row $i$, for any scalar $c$,

\[\begin{align} \det(A) &= \det(\mathbf{r}_1,\ldots,\mathbf{0},\ldots,\mathbf{r}_n)\\ & = \det(\mathbf{r}_1,\ldots,c\mathbf{0},\ldots,\mathbf{r}_n)\\ & = c\det(\mathbf{r}_1,\ldots,\mathbf{0},\ldots,\mathbf{r}_n) = c\det(A) \end{align}\] Since $c$ is arbitrary, $\det(A)=0$.

If $A$ has two equal rows, then $\det(A) = 0$

Let $A$ be a matrix where $\mathbf{r}_i = \mathbf{r}_j$ for some $i \neq j$. Then

\[\begin{align} \det(A) &= \det(\ldots,\mathbf{r}_i,\ldots,\mathbf{r}_j,\ldots)&&\text{definition}\\ &= -\det(\ldots,\mathbf{r}_j,\ldots,\mathbf{r}_i,\ldots)&&\text{Alternating Property Axiom}\\ &= -\det(\ldots,\mathbf{r}_i,\ldots,\mathbf{r}_j,\ldots)&&\text{because }\mathbf{r}_i=\mathbf{r}_j\\ &= -\det(A) \end{align}\] Therefore, $\det(A) = -\det(A)$, which implies $\det(A) = 0$.

For a scalar $c$, $\det(cA) = c^n\det(A)$

Note that $cA$ means multiplying each row by $c$. Using multilinearity $n$ times:

\[\begin{align} \det(cA) &= \det(c\mathbf{r}_1,c\mathbf{r}_2,\ldots,c\mathbf{r}_n)\\ &= c\det(\mathbf{r}_1,c\mathbf{r}_2,\ldots,c\mathbf{r}_n)\\ &= c^2\det(\mathbf{r}_1,\mathbf{r}_2,c\mathbf{r}_3\ldots,c\mathbf{r}_n)\\ &\;\;\vdots\\ &= c^n\det(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_n)\\ &= c^n\det(A) \end{align}\]

For diagonal matrices, $\det(A)$ is the product of the diagonal terms.

Let $A$ be a diagonal matrix and $\{\mathbf{e}_1,\dots,\mathbf{e}_n\}$ be the canonical basis of $\mathbb{R}^n$. Then using linearity $n$ times we get: \[\begin{align} \det(A)&=\det(a_{11}\mathbf{e}_1,a_{22}\mathbf{e}_2,\dots,a_{nn}\mathbf{e}_n)\\ &=a_{11}\det(\mathbf{e}_1,a_{22}\mathbf{e}_2,\dots,a_{nn}e_n)\\ &=a_{11}a_{22}\det(\mathbf{e}_1,\mathbf{e}_2,a_{33}\mathbf{e}_3\dots,a_{nn}\mathbf{e}_n)\\ &\;\;\vdots \\ &=a_{11}a_{22}\cdots a_{nn}\det(\mathbf{e}_1,\mathbf{e}_2,\dots,\mathbf{e}_n)\\ &=a_{11}a_{22}\cdots a_{nn}&& \text{By the identity axiom}\\ \end{align}\]

Suppose that the matrix $B$ is obtained from the matrix $A$ by applying the type III elementary row operation $c\mathbf{r}_j+\mathbf{r}_i\to\mathbf{r}_i$, where $i\neq j$ and $c$ is a scalar. This operation adds $c$ times the $j$-th row to the $i$-th row. Then, $\det(B)=\det(A)$.

\[\begin{align} \det(B) &= \det(\ldots,c\mathbf{r}_j+\mathbf{r}_i,\ldots,\mathbf{r}_j,\ldots)&&\text{definition of $B$}\\ &= c\det(\ldots,\mathbf{r}_j,\ldots,\mathbf{r}_j,\ldots) +\det(\ldots,\mathbf{r}_i,\ldots,\mathbf{r}_j,\ldots)&&\text{ by linearity}\\ &= 0+\det(\ldots,\mathbf{r}_i,\ldots,\mathbf{r}_j,\ldots)&&\text{two equal rows}\\ &= \det(A) \end{align}\]

These properties allow us to find the determinants of the elementary matrices. Suppose that $E$ is an elementary matrix. If $E$ multiplies row $i$ by a non-zero constant $c$, $E$ is diagonal with $c$ in the $i$-th term and 1’s elsewhere. Then $\det(E)=c$. If $E$ exchanges two rows, its determinant is -1. And if $E$ adds a multiple of a row to another row its determinant is 1.

We summarize these properties here:

Determinants of Elementary Matrices

Suppose that $E$ is an elementary matrix

If $E$ multiplies a row by a non-zero constant $c$, then $\det(E)=c$
If $E$ exchanges two rows, then $\det(E)=-1$
If $E$ adds a multiple of a row to another row, then $\det(E)=1$

Repeating the arguments on matrices and elementary row matrices, we obtain the following important result

Lemma 9.1 Suppose that $A$ is an $n\times n$ matrix and that $E$ is an $n\times n$ elementary matrix. Then \[\det(EA)=\det(E)\det(A)\]

Let’s illustrate this when $E$ multiplies the $i$-th row by a non-zero constant $c$. Then \[\begin{align} \det(EA) &= \det(\mathbf{r}_1,\cdots,\mathbf{r}_{i-1},c\mathbf{r}_i,\mathbf{r}_{i+1},\cdots,\mathbf{r}_n)\\ &= c\det(\mathbf{r}_1,\cdots,\mathbf{r}_{i-1},\mathbf{r}_i,\mathbf{r}_{i+1},\cdots,\mathbf{r}_n)\\ &=\det(E)\det(A) \end{align}.\] A similar argument is used for the other elementary matrices.

We use the Lemma to characterize invertibility of matrices in terms of determinants

Theorem 9.1 Let $A$ be an $n\times n$ matrix. Then $A$ is invertible if and only if $\det(A)\not=0$.

Proof. Let $A$ be an $n\times n$ matrix with row reduced echelon form $R$. Find elementary row operations $E_i$ such that \[E_k\cdots E_2E_1A=R.\] Applying Lemma 9.1 repeatedly we get \[\det(E_k)\cdots \det(E_2)\det(E_1)\det(A)=\det(R).\]

If $A$ is invertible, then $R=I_n$ and $\det(E_k)\cdots \det(E_2)\det(E_1)\det(A)=\det(I_n)=1,$ which implies that $\det(A)\not=0$.

On the other hand, if $A$ is not invertible, $R$ has a row of zeros and then $\det(R)=0$. Hence, $\det(E_k)\cdots \det(E_2)\det(E_1)\det(A)=\det(R)=0$. Since $\det(E_i)\not=0$ for $i\leq k$, we get that $\det(A)=0$. $\square$

Lemma 9.1 and Theorem 9.1 have several consequences:

Corollary 9.1 Let $A$ and $B$ be $n\times n$ matrices, then $\det(AB)=\det(A)\det(B)$

Indeed, if $A$ and $B$ are invertible, then each of them can be written as a product of elementary matrices, and the result follows from Lemma 9.1. On the other hand, if either $A$ or $B$ is not invertible, then $AB$ is not invertible either and $\det(AB)=0$. Since $\det(A)=0$ or $\det(B)=0$, we see that $\det(A)\det(B)=0$ as well.

If $A$ is invertible, $A^{-1}A=I_n$ and then $1=\det(I_n)=\det(A^{-1}A)=\det(A^{-1})\det(A)$ and we get:

Corollary 9.2 Suppose that $A$ is an invertible be $n\times n$ matrix. Then \[\det(A^{-1}) =\frac{1}{\det(A)}.\]

Lemma 9.1 and Theorem 9.1 also allow us to compute the determinant efficiently when the matrix is in echelon form, or if it is upper or lower triangular.

Corollary 9.3 Suppose that $A$ is an upper triangular (or lower triangular) matrix. Then \[\det(A)=a_{11}a_{22}\cdots a_{nn}\]

Proof. Let $A$ be an $n\times n$ upper triangular matrix. We consider two cases:

Case 1: If any of the diagonal entries $a_{ii}$ is zero, then $A$ has fewer than $n$ pivots and is not invertible. By Theorem 9.1, $\det(A)=0$. Since one of the diagonal entries is zero, the product $a_{11}a_{22}\cdots a_{nn}$ is also zero.

Case 2: If all of the diagonal entries $a_{ii}$ are non-zero, then $A$ is already in row echelon form. We can apply elementary row operations of type III (multiplying a multiple of a row to another row) to transform $A$ into a diagonal matrix $D$ with the same diagonal entries as $A$. By the properties of determinants, type III elementary row operations do not change the determinant. Therefore, $\det(A)=\det(D)$.

The determinant of a diagonal matrix $D$ is the product of its diagonal entries and since $A$ and $D$ have the same diagonals, \[\det(A)=\det(D)=a_{11}a_{22}\cdots a_{nn}.\quad \square\]