Chapter 5 Determinants

In Linear Algebra I the determinant of a square matrix has been defined axiomatically (cf. Theorem 5.3 here). Here we begin with the following closed formula.

Definition 5.1 (Leibniz’ definition of determinant) Let \(F\) be a field. Let \(n \geq 1\) and \(A = (a_{ij})_{i,j=1, \dots,n} \in M_{n \times n}(F)\). Then \[ {\color{red}{\mathrm{det}(A)}} := \sum_{\sigma \in S_{n}} \mathrm{sgn}(\sigma) \prod_{i=1}^{n} a_{i, \sigma(i)} \in F \] is called the determinant of \(A\). We also write \(|A|\) for \(\mathrm{det}(A)\).

Example 5.2

Let \(n=1\) and \(A=(a_{11}) \in M_{1 \times 1}(F)\).
We have \(S_{1} = \{ \mathrm{id} \}\) and \[ \mathrm{det}(A) = \mathrm{sgn}(\mathrm{id}) a_{11} = a_{11}. \]
Let \(n=2\) and \(A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \in M_{2 \times 2}(F)\).
We have \(S_{2} = \{\mathrm{id}, \langle 1, 2 \rangle \}\) and \[ \mathrm{det}(A) = \mathrm{sgn}(\mathrm{id}) a_{11} a_{22} + \mathrm{sgn}(\langle 1,2 \rangle) a_{12} a_{21} = a_{11} a_{22} - a_{12} a_{21}. \] For example: if \(A = \begin{pmatrix} 1+2i & 3 +4i \\ 1-2i & 2-i \end{pmatrix} \in M_{2 \times 2}(\mathbb{C})\) then
\(\mathrm{det}(A) = (1 + 2i)(2 - i) - (3+4i)(1-2i)\) \(= (2 + 2 + 4i - i) - (3 + 8 + 4i - 6i)\) \(= -7 + 5i\).
Let \(n=3\) and \(A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \in M_{3 \times 3}(F)\).
We have \(S_{3} = \{\mathrm{id}, \langle 1,2,3 \rangle, \langle 1,3,2 \rangle, \langle 1,3 \rangle, \langle 2,3 \rangle, \langle 1,2 \rangle \}\) and \[\begin{align*} \mathrm{det}(A) &= \underbrace{\mathrm{sgn}(\mathrm{id})}_{= +1} a_{11}a_{22}a_{33} + \underbrace{\mathrm{sgn}( \langle 1,2,3 \rangle )}_{= +1} a_{12}a_{23}a_{31} + \underbrace{\mathrm{sgn}( \langle 1,3,2 \rangle )}_{= +1} a_{13}a_{21}a_{32} \\ & + \underbrace{\mathrm{sgn}( \langle 1,3 \rangle )}_{= -1} a_{13}a_{22}a_{31} + \underbrace{\mathrm{sgn}( \langle 2,3 \rangle )}_{= -1} a_{11}a_{23}a_{32} + \underbrace{\mathrm{sgn}(\langle 1,2 \rangle )}_{= -1} a_{12}a_{21}a_{33}. \end{align*}\] Trick to memorise: \(\begin{array}{ c c c c c c c c c c} a_{11} & & a_{12} & & a_{13} & & a_{11} & & a_{12} \\ & \diagdown & & \times & & \times & & \diagup & \\ a_{21} & & a_{22} & & a_{23} & & a_{21} & & a_{22} \\ & \diagup & & \times & & \times & & \diagdown & \\ a_{31} & & a_{32} & & a_{33} & & a_{31} & & a_{32} \end{array}\)
(Rule of Sarrus)
For example, let \(A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{pmatrix} \in M_{3 \times 3}(\mathbb{F}_{2})\).
\(\implies\) \(\mathrm{det}(A) = (0+0+0) - (1+0+0) = 1\) (because \(-1=1\) in \(\mathbb{F}_{2}\)).
Let \(A = (a_{ij})\) be an upper (or lower) triangular matrix. So \(A\) is of the form \(\begin{pmatrix} a_{11} & \ldots & \ldots & a_{1n} \\ 0 & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \vdots \\ 0 & \ldots & 0 & a_{nn} \end{pmatrix}\). In other words, we have \(a_{ij} = 0\) if \(i > j\).
Then \(\mathrm{det}(A) = a_{11} a_{22} \cdots a_{nn}\), i.e. \(\mathrm{det}(A)\) is the product of the entries on the main diagonal. For example, \(\mathrm{det}(I_{n})= 1\).

Proof:
Let \(\sigma \in S_{n}, \sigma \neq \mathrm{id} \implies \exists i_{0} \in \{ 1, \ldots , n \}\) such that \(i_{0} > \sigma(i_{0})\);
\(\implies\) \(a_{i_{0}, \sigma(i_{0})} = 0 \implies \prod_{i=1}^{n} a_{i, \sigma(i)} = 0\).
\(\implies \mathrm{det}(A) = \mathrm{sgn}(\mathrm{id}) \prod_{i=1}^{n} a_{i, \mathrm{id}(i)} = a_{11} a_{22} \cdots a_{nn}\). \(\square\)

Theorem 5.3 (Weierstrass’ axiomatic description of the determinant map) See Defn 4.1 of L.A.I. Let \(F\) be a field. Let \(n \geq 1\). The map \[\mathrm{det}: \ M_{n \times n}(F) \rightarrow F, A \mapsto \mathrm{det}(A),\] has the following properties and is uniquely determined by these properties:

\(\mathrm{det}\) is linear in each column:
\(\mathrm{det} \begin{pmatrix} & a_{1s} + b_{1s} & \\ C & \vdots & D \\ & a_{ns} + b_{ns} & \end{pmatrix} = \mathrm{ det} \begin{pmatrix} & a_{1s} & \\ C & \vdots & D \\ & a_{ns} & \end{pmatrix} + \mathrm{det} \begin{pmatrix} & b_{1s} & \\ C & \vdots & D \\ & b_{ns} & \end{pmatrix}\).
Multiplying any column of a matrix \(A \in M_{n \times n}(F)\) with a scalar \(\lambda \in F\) changes \(\mathrm{det}(A)\) by the factor \(\lambda\):
\(\mathrm{det} \begin{pmatrix} & \lambda a_{1s} & \\ C & \vdots & D \\ & \lambda a_{ns} & \end{pmatrix} = \lambda \cdot \mathrm{det} \begin{pmatrix} & a_{1s} & \\ C & \vdots & D \\ & a_{ns} & \end{pmatrix}\).
If two columns of \(A \in M_{n \times n}(F)\) are equal, then \(\mathrm{det}(A) = 0\).
\(\mathrm{det}(I_{n}) = 1\).

Proof: Omitted here (please see the extended notes).

Remark 5.4 Theorem 5.3 and the following Corollary 5.5 also hold when “columns” are replaced with “rows” (similar proofs).

Corollary 5.5 Let \(F\) be a field. Let \(A \in M_{n \times n}(F)\). Then:

For all \(\lambda \in F\) we have \(\mathrm{det}(\lambda A) = \lambda^{n}\mathrm{det}(A)\).
If a column of \(A\) is the zero column then \(\mathrm{det}(A) = 0\).
Let \(B\) be obtained from \(A\) by swapping two columns of \(A\). Then \(\mathrm{det}(B) = -\mathrm{det}(A)\).
Let \(\lambda \in F\) and let \(B\) be obtained from \(A\) by adding the \(\lambda\)-multiple of the \(j^{th}\) column of \(A\) to the \(i^{th}\) column of \(A\) \((i \neq j)\). Then \(\mathrm{det}(B) = \mathrm{det}(A)\).

Proof:

Apply Theorem 5.3(b) \(n\) times.
Apply Theorem 5.3(b) with \(\lambda = 0\).
Let \(\underline{a}\), \(\underline{b}\) denote the two columns of \(A\) to be swapped.
\(\implies\) \(\mathrm{det}(A) + \mathrm{det}(B) = \mathrm{det}(\ldots \underline{a} \ldots \underline{b} \ldots) + \mathrm{det}(\ldots \underline{b} \ldots \underline{a} \ldots)\)
            \(= \mathrm{det}(\ldots \underline{a} \ldots \underline{b} \ldots) + \mathrm{det}(\dots \underline{b} \ldots \underline{a} \ldots)\)
                 \(+ \underbrace{\mathrm{det}(\ldots \underline{a} \ldots \underline{a} \ldots)}_{=0} + \underbrace{\mathrm{det}(\ldots \underline{b} \ldots \underline{b} \ldots)}_{=0}\) (using 5.3(c))
            \(= \mathrm{det}(\ldots \underline{a} \ldots \underline{a}+\underline{b} \ldots) + \mathrm{det}(\ldots \underline{b} \ldots \underline{a} + \underline{b} \ldots)\) (by 5.3(a))
            \(= \mathrm{det}(\ldots \underline{a} + \underline{b} \ldots \underline{a}+\underline{b} \ldots)\) (by 5.3(a))
            \(= 0\) (by 5.3(c))
\(\mathrm{det}(B) = \mathrm{det}(\ldots \underline{a} + \lambda \underline{b} \ldots \underline{b} \ldots)\)
\(= \mathrm{det}(\ldots \underline{a} \ldots \underline{b} \ldots) + \lambda\, \underbrace{\mathrm{det}(\ldots \underline{b} \ldots \underline{b} \ldots)}_{=0}\) (by 5.3(a),(b))
\(= \mathrm{det}(A)\). (by 5.3(c))\(\square\)

Theorem 5.6 Let \(F\) be a field. Let \(A \in M_{n \times n}(F)\). Then the following are equivalent:

\(A\) is invertible.
The columns of \(A\) span \(F^{n}\).
\(N(A) = \{ \underline{0} \}\).
\(\mathrm{det}(A) \neq 0\).

For (a) \(\iff\) (d), compare Thm 4.14 of L.A.I.

Before the proof, recall that we denote the standard basis vectors of \(F^n\) as \(\underline{e}_{1} = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \dots , \underline{e}_n = \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix}\). So, in particular, \(I_n=(\underline{e}_{1}, \dots , \underline{e}_{n})\).

Proof:

“(a) \(\implies\) (b)”: We’ll prove the following more precise statement:
\((\star)\) \(\exists B \in M_{n \times n}(F)\) such that \(AB = I_{n}\) \(\iff\) The columns of A span \(F^n\).
Proof of (\(\star\)): Let \(\underline{a}_{1}\ \ldots , \underline{a}_{n}\) denote the columns of \(A\). Then:
LHS \(\iff \exists \underline{b}_{1}, \ldots, \underline{b}_{n} \in F^{n}\) such that \(A \underline{b}_{i} = \underline{e}_{i}\) for all \(i=1, \dots, n\)
(we can use the columns of \(B=(\underline{b}_{1}, \dots, \underline{b}_{n})\))
\(\iff\) \(\exists \underline{b}_{1}, \ldots , \underline{b}_{n} \in F^{n}\) such that \(\underline{a}_{1}b_{i1} + \dots + \underline{a}_{n}b_{in} = \underline{e}_{i}\) for all \(i=1, \dots, n\)
\(\iff\) \(\underline{e}_{1}, \dots, \underline{e}_{n} \in \mathrm{Span}(\underline{a}_{1}, \dots, \underline{a}_{n})\)
\(\iff\) RHS.

“(b) \(\iff\) (c)”: The columns of \(A\) span \(F^{n}\)
\(\iff\) The columns of \(A\) form a basis of \(F^{n}\) (by 3.15)
\(\iff\) The columns of \(A\) are linearly independent (by 3.15)
\(\iff\) \(N(A)= \{ \underline{0} \}\) (by definition of \(N(A)\))

“(b) \(\implies\) (a)”: \(\exists B \in M_{n \times n}(F)\) such that \(AB = I_{n}\) (by \((\star)\))
\(\implies\) \(A(BA) = (AB)A = I_{n}A = A = A I_{n}\)
\(\implies\) \(A(BA - I_{n}) = \underline{0}\)
\(\implies\) Every column of \(BA - I_{n}\) belongs to \(N(A)\)
\(\implies\) \(BA - I_{n} = \underline{0}\) (because \(N(A) = \{ \underline{0} \}\) by (b) \(\iff\) (c))
\(\implies\) \(BA = I_{n}\)
\(\implies\) \(A\) is invertible (as both \(AB = I_{n}\) and \(BA = I_{n}\))

“(b) \(\iff\) (d)”: We apply column operations to the matrix \(A\) until we arrive at a lower triangular matrix \(C\). Then:
The columns of \(A\) span \(F^{n}\)
\(\iff\) the columns of \(C\) span \(F^{n}\) (by 3.3(b))
\(\iff\) all the diagonal elements of \(C\) are non-zero (because \(C\) is triangular)
\(\iff\) \(\mathrm{det}(C) \neq 0\) (by 5.2(d))
\(\iff\) \(\mathrm{det}(A) \neq 0\) (because \(\mathrm{det}(C) = \lambda \mathrm{det}(A)\) for some non-zero \(\lambda \in F\) by 5.5)\(\square\)

Theorem 5.7 (See also Thm 4.21 of L.A.I.) Let \(F\) be a field. Let \(A, B \in M_{n \times n}(F)\). Then: \[ \mathrm{det}(AB) = \mathrm{det}(A) \cdot \mathrm{det}(B) \]

Proof: Omitted. (See the proof of Thm 4.21 in L.A.I., it works for any field \(F\).) \(\square\)

Example 5.8 For each \(m \in \mathbb{N}\) compute \(\mathrm{det}(A^{m}) \in \mathbb{C}\), here \(A = \begin{pmatrix} 1 + 4i & 1 \\ 5 + i & 1 - i \end{pmatrix} \in M_{2 \times 2}(\mathbb{C})\).

Solution: \(\mathrm{det}(A) = (1+4i)(1-i) - (5+i) = (1+4+4i-i) - (5+i) = 2i\).
\(\implies \mathrm{det}(A^{m}) = \mathrm{det}(A)^{m}\) \(= (2i)^{m} = \begin{cases} 2^{m} & \text{if } m \text{ is of the form } 4k; \\ 2^{m}i & \text{if } m \text{ is of the form } 4k+1; \\ -(2^{m}) & \text{if } m \text{ is of the form } 4k+2; \\ -(2^{m})i & \text{if } m \text{ is of the form } 4k+3. \end{cases}\) (by 5.7)