Determinants

In calculus, we can learn something about the behaviour of a function by computing the derivative. For example, if $f’$ is positive on an interval $[a, b]$ then $f$ is one-to-one on the interval. The determinant plays a similar role in linear algebra.

Definition. A determinant is a function $D: \mathbb{F}^n \times \cdots \times \mathbb{F}^n \to \mathbb{F}$ that assigns each $n$-tuple $\Set{a_1, …, a_n}$ of vectors in $\mathbb{F}^n$ to an element of $\mathbb{F}$, i.e. $D = D(a_1, …, a_n)$ such that

  • $D(…, a_i + a_j, …) = D(…, a_i, …)$ for $i \not= j$;

  • $D(…, \lambda a_i, …) = \lambda D(…, a_i, …)$;

  • $D(e_1, …, e_n) = 1$ for unit vectors $\Set{e_i}$.

Proposition. Let $D$ be a determinant function on $\mathbb{F}^n$. Then

  • $D(…, a_i, …, a_j, …) = - D(…, a_j, …, a_i, …)$ in which two of the vectors ($i \not= j$) are interchanged;

  • $D(…, a_i, …, a_i, …) = 0$ in which two of the vectors are equal;

  • $D(…, a_i + \sum_{i \not= j} \lambda_j a_j, …) = D(…, a_i, …)$;

  • $D = 0$ if $\Set{a_1, …, a_n}$ is linearly dependent;

  • $D(…, a_i + a_i’, …) = D(…, a_i, …) + D(…, a_i’, …)$.

Proof.

  • By the axioms,

    \[\begin{align*} D(..., a_i, ..., a_j, ...) &= -D(..., -a_i, ..., a_j, ...) \\ &= -D(..., -a_i, ..., -a_i + a_j, ...) \\ &= D(..., -a_i, ..., a_i - a_j, ...) \\ &= D(..., -a_j, ..., a_i - a_j, ...) \\ &= -D(..., a_j, ..., a_i - a_j, ...) \\ &= -D(..., a_j, ..., a_i, ...) \end{align*}\]

  • Interchanging the two rows we have $D(…, a_i, …, a_i, …) = - D(…, a_i, …, a_i, …)$ so $D = 0$.

  • For each $\lambda_j a_j$ with $j \not= i$ and $\lambda_j \not= 0$,

    \[\begin{align*} D(..., a_i, ..., a_j, ...) &= \lambda_j^{-1} D(..., a_i, ..., \lambda_j a_j, ...) \\ &= \lambda_j^{-1} D(..., a_i + \lambda_j a_j, ..., \lambda_j a_j, ...) \\ &= D(..., a_i + \lambda_j a_j, ..., a_j, ...) \end{align*}\]

  • Suppose that $a_i = \sum_{i \not= j} \lambda_j a_j$, then

    \[\begin{align*} D(..., a_i, ...) &= - D(..., -a_i, ...) \\ &= - D(..., -a_i + \sum_{i \not= j} \lambda_j a_j, ...) \\ &= - D(..., 0, ...) \\ &= 0 \end{align*}\]

  • Extend the vectors $\Set{a_1, … a_{i-1}, a_{i+1}, …, a_n}$ to a basis by adding $\hat{a_i}$ to them, then

    \[a_i + a_i' = (\lambda_i + \mu_i) \hat{a_i} + \sum_{i \not= j} (\lambda_j + \mu_j) a_j\]

    Therefore, we have

    \[\begin{align*} D(..., a_i + a_i', ...) &= (\lambda_i + \mu_i) D(..., \hat{a_i}, ...) \\ D(..., a_i, ...) &= \lambda_i D(..., \hat{a_i}, ...) \\ D(..., a_i', ...) &= \mu_i D(..., \hat{a_i}, ...) \end{align*}\]

    and $D(…, a_i + a_i’, …) = D(…, a_i, …) + D(…, a_i’, …)$ because $\mathbb{F}$ is distributive.

Volume Form

There is another set of axioms we can use to define determinant.

Definition. A volume form $D$ on $\mathbb{F}^n$ is a function $\mathbb{F}^n \times \cdots \times \mathbb{F}^n \to \mathbb{F}$, $(a_1, …, a_n) \mapsto D(a_1, …, a_n)$ such that

  • $D$ is multi-linear, i.e. for each $1 \le i \le n$,

    \[D(a_1, ..., a_{i-1}, -, a_{i+1}, ..., a_n) \in (\mathbb{F}^n)^\ast\]

  • $D$ is alternating, i.e. whenever $a_i = a_j$ for some $i \not= j$ then $D(…, a_i, …) = 0$.

We can see that the conditions correspond to the fifth and second properties proved above.

Proposition. Let $D$ be a volume form. Then

  • $D(…, a_i + a_j, …) = D(…, a_i, …)$ for $i \not= j$;

  • $D(…, \lambda a_i, …) = \lambda D(…, a_i, …)$.

Proof.

We have

\[\begin{align*} D(..., a_i + a_j, ..., a_j, ...) &= D(..., a_i, ..., a_j, ...) + D(..., a_j, ..., a_j, ...) \\ &= D(..., a_i, ..., a_j, ...) \end{align*}\]

and $D(…, \lambda a_i, …) = \lambda D(…, a_i, …)$ follows immediately from the multi-linearity.

Since the other properties proved above doesn’t rely on the third axiom so a volume form also satisfies them.

Existence and Uniqueness

Before defining the actual determinant function, we would like to show that if such a function exists, it must be unique.

Proposition. Let $D$ and $D’$ be two functions satisfying the axioms. Then for all $a_1, …, a_n$ in $\mathbb{F}^n$,

\[D(..., a_i, ...) = D'(..., a_i, ...)\]

Proof.

Consider the function

\[\Delta(..., a_i, ...) = D(..., a_i, ...) - D'(..., a_i, ...)\]

we have

  • $\Delta(e_1, …, e_n) = 0$;

  • $\Delta$ changes sign if two of the vectors are interchanged and $\Delta = 0$ if $a_i = a_j$ for $i \not= j$;

  • $\Delta(…, \lambda a_i, …) = \lambda \Delta(…, a_i, …)$;

  • $\Delta(…, a_i + a_i’, …) = \Delta(…, a_i, …) + \Delta(…, a_i’, …)$.

Suppose that $(e_1, …, e_n)$ is a basis of $\mathbb{F}^n$. Then

\[\begin{align*} \Delta(a_1, ..., a_n) &= \Delta \left( \sum_{j_1} \lambda_{1 j_1} e_{j_1}, a_2, ..., a_n \right) \\ &= \sum_{j_1} \lambda_{1 j_1} \Delta(e_{j_1}, a_2, ..., a_n) \\ &= \sum_{j_1} \sum_{j_2} \lambda_{1 j_1} \lambda_{2 j_2} \Delta(e_{j_1}, e_{j_2}, ..., a_n) \\ &= \cdots = \sum_{j_1, ..., j_n} \lambda_{1 j_1} \cdots \lambda_{2 j_n} \Delta(e_{j_1}, ..., e_{j_n}) \end{align*}\]

If any of $j_1, …, j_n$ are equal, then $\Delta(e_{j_1}, …, e_{j_n}) = 0$. If all $j_1, …, j_n$ are distinct, by interchanging rows repeatedly we have $\Delta(e_{j_1}, …, e_{j_n}) = \pm \Delta(e_1, …, e_n) = 0$. Hence, $\Delta(a_1, …, a_n) = 0$ as required.

A matrix $A \in \text{Mat}_n(\mathbb{F})$ can be viewed as having $n$-tuple of elements of $\mathbb{F}^n$ in its columns, i.e. $A = \begin{pmatrix} A^{(1)} & \cdots & A^{(n)} \end{pmatrix}$.

Definition. If $A \in \text{Mat}_n(\mathbb{F})$ then the determinant of $A$ is defined by

\[\det A = \sum_{\sigma \in S_n} \epsilon(\sigma) \left( \prod_{i=1}^n A_{i \sigma(i)} \right)\]

Proposition. $\det A = \det A^\intercal$.

Proof.

\[\begin{align*} \det A^\intercal &= \sum_{\sigma \in S_n} \epsilon(\sigma) \prod_{i=1}^n A_{\sigma(i) i} \\ &= \sum_{\sigma \in S_n} \epsilon(\sigma) \prod_{i=1}^n A_{\sigma(\sigma^{-1}(i)) \sigma^{-1}(i)} \\ &= \sum_{\sigma \in S_n} \epsilon(\sigma^{-1}) \prod_{i=1}^n A_{i \sigma^{-1}(i)} \\ &= \det A \end{align*}\]

Therefore, it doesn’t matter if we put the $n$-tuple of elements of $\mathbb{F}^n$ in columns or rows.

Proposition. $\det: \mathbb{F}^n \times \cdots \mathbb{F}^n \to \mathbb{F}$, $\begin{pmatrix} A^{(1)} & \cdots & A^{(n)} \end{pmatrix} \mapsto \det A$ is a volume form.

Proof.

For each product $\prod A_{i \sigma(i)}$, one term from each column appears in that so it is multi-linear. Since $\det$ is a sum of multi-linear functions, it is also multi-linear.

Suppose that $A^{(k)} = A^{(l)}$ for some $k \not= l$. Let $\tau$ be the transposition $(kl)$. Then $S_n$ is a disjoint union of the alternating group $A_n$ and $\tau A_n$. For each $\sigma’ = \tau \sigma \in \tau A_n$, since $A_{ik} = A_{il}$,

\[\prod A_{i \sigma'(i)} = \prod A_{i \sigma(i)}\]

but $\epsilon(\sigma’) = -\epsilon(\sigma)$ so the terms can be grouped in pair and each of them cancelled each other in the sum, i.e. $\det A = 0$.

Base on the proof above, together with the fact that $\det I_n = 1$, $\det$ is the unique determinant function as defined at the beginning.

References