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# 29 Determinants: Part 1.
For any fixed positive integer $n$, the determinant $\det:M_{n\times n}(F)\to F$ is the unique scalar function on all $n\times n$ matrices over the field[^field] $F$ that is (1) multilinear in the rows (and columns), (2) antisymmetric in the rows (and columns) (3) and normalizes at the identity matrix, namely $\det(I_{n})=1$. For us, the field $F$ is either $\mathbb{R}$ or $\mathbb{C}$.
If we have an $n\times n$ matrix $A$ whose $(i,j)$-th entry is $a_{ij}$, then we have an entry-wise formula for determinant $$
\det(A)=\sum_{\sigma \in S_{n}}\text{sign}(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}\cdots a_{n\sigma(n)}
$$where the sum ranges over all permutations $\sigma$ on $n$ numbers, $\{1,2,\ldots,n\}$. However, we will compute determinants in a combination of three ways (1) by patterns and properties of determinants (2) recursively ala Laplace row (column) expansion, and (3) elementary row operations.
The original development of the determinant came largely from solving linear systems, via Cramer's rule. And since then determinant has led to many applications in geometry, algebra, analysis (calculus), and combinatorics, and we will see some of it. We will, however, primarily use determinant as an additional tool to determine[^objection] invertibility of a square matrix. Namely, a square matrix is invertible if and only if its determinant is not zero.
Before we get too far, let us write down the formulas for determinants of small square matrices.
## Determinants of orders 1, 2, and 3.
Let us record some results first. You may have been familiar with these already.
If we have a $1\times 1$ matrix $A=[\ a\ ]$, then
> $$
\det [\ a \ ] = a .
$$
If we have $2\times 2$ matrix $A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$, then we have
> $$
\det \begin{pmatrix} a & b \\ c & d\end{pmatrix} = ad - bc.
$$
If we have $3\times 3$ matrix $A = \begin{pmatrix}a & b & c\\d & e &f \\ g & h & i\end{pmatrix}$, then we have
> $$
\det \begin{pmatrix}a & b & c\\d & e &f \\ g & h & i\end{pmatrix} = aei+bfg+cdh-ceg-bdi-afh.
$$
There is a mnemonic called **Rule of Sarrus** (or I call it "basket weaving") for $3\times 3$ determinants, **and only $3\times 3$**. It goes as follows, first duplicate the first two columns and place it on the right, then draw three sets of diagonals going from top left to bottom right, these terms are the terms with the positive sign. Then draw three sets of diagonals going from top right to bottom left, these are the terms that receive an extra minus sign:
![[smc-spring-2024-math-13/linear-algebra-notes/---files/29-determinants 2024-04-01 18.52.36.excalidraw.svg]]
%%[[smc-spring-2024-math-13/linear-algebra-notes/---files/29-determinants 2024-04-01 18.52.36.excalidraw.md|🖋 Edit in Excalidraw]]%%
Or, an alternative (this is how I do it):
![[smc-spring-2024-math-13/linear-algebra-notes/---files/29-determinants 2024-04-01 18.59.15.excalidraw.svg]]
%%[[smc-spring-2024-math-13/linear-algebra-notes/---files/29-determinants 2024-04-01 18.59.15.excalidraw.md|🖋 Edit in Excalidraw]]%%
Nevertheless, heed this warning: **This does not apply to $4\times 4$ or any higher!** We will develop methods to deal with that. Ok. We now dive into determinants.
## Characteristic definition of the determinant.
There are many ways to "define" the determinant. We will take a more axiomatic approach, asserting certain function exists and is unique, which is the determinant.
> **Definition / Proposition.**
> Fix any positive integer $n$, the determinant on $n\times n$ matrices over field $F$ is the unique scalar function $\det : M_{n\times n}(F)\to F$ such that it is
> (1) **multilinear** in each row (column, resp.), with other rows (columns, resp.) fixed,
> (2) **antisymmetric** in the rows (columns, resp.), and
> (3) **normalizing at the identity**, where $\det (I_{n})=1$.
Let us believe that there is such a function first, and that it is unique. So it makes sense to speak of **the** determinant. But first, what do the words multilinear and antisymmetric means?
### Multilinearity.
In general, a function $f=f(x_{1},x_{2},\ldots,x_{n})$ is **multilinear** if it is linear in each entry with other entries fixed. That is, if we pick the $i$-th entry, say, then $f$ preserves sum and scaling at the $i$-th entry, with other entries unchanged: $$
\begin{align*}
f(x_{1},\ldots,cx_{i},\ldots,x_{n}) & = cf(x_{1},\ldots,x_{i},\ldots,x_{n}) \\
f(x_{1},\ldots,x_{i}+y,\ldots,x_{n}) & = f(x_{1},\ldots,x_{i},\ldots,x_{n}) + f(x_{1},\ldots,y,\ldots,x_{n})
\end{align*}
$$
By the way, to check preserving sum and scaling we can check both at once, by showing $$
f(x_{1},\ldots,cx_{i}+ y,\ldots,x_{n}) = cf(x_{1},\ldots,x_{i},\ldots,x_{n}) + f(x_{1},\ldots,y,\ldots,x_{n})
$$
**Example.**
The function $f:\mathbb{R}^{3}\to\mathbb{R}$ with $f(x,y,z)=xyz$ is multilinear. $\blacktriangleright$ Indeed, say we check the first component, then $$
f(cx,y,z)=(cx)yz=c(xyz)=cf(x,y,z),
$$and $$
f(x,y+y',z)=x(y+y')z=xyz+xy'z=f(x,y,z)+f(x,y',z).
$$So we see that $f$ preserves sum and scaling at the first component. Check that $f$ also preserves sum and scaling in the second and third component. Hence $f$ is multilinear in each of the components. $\blacklozenge$
**Example.**
The function $f:\mathbb{R}^{3}\to \mathbb{R}$ with $f(x,y,z)=xy(z+2)$ is not multilinear -- it is only linear in the first two components, but not the third. $\blacktriangleright$ Let us check the first to components: $$
f(cx+x',y,z)=(cx+x')y(z+2)=cxy(z+2)+x'y(z+2)=cf(x,y,z)+f(x',y,z)
$$Hence $f$ is linear in the first component. Similarly for the second component: $$
f(x,cy+y',z)=x(cy+y')(z+2)=x(cy)(z+2)+xy'(z+2)=cf(x,y,z)+f(x,y',z)
$$
But it is **not** linear in the third component. Just like a linear function, if it is linear in the third component, then $f(x,y,0)=0$ no matter what $x,y$ is. But $$
f(1,1,0)=2\neq0
$$Hence $f$ not linear in the third component. Note you can also show $f$ does not preserve sum or scaling in the third component as well. $\blacklozenge$
So when we say determinant is multilinear in each row, this means if we pick a special row, the determinant is a linear function in that row, with all other rows fixed and not changed. Namely we have $$
\det \begin{pmatrix} \text{row 1} \\
\text{row 2} \\
\vdots \\
c (\text{row \(i\)})+ \text{another row} \\
\vdots \\
\text{row \(n\)}\end{pmatrix}
=
c\cdot \det \begin{pmatrix} \text{row 1} \\
\text{row 2} \\
\vdots \\
\text{row \(i\)} \\
\vdots \\
\text{row \(n\)}\end{pmatrix} +
\det \begin{pmatrix} \text{row 1} \\
\text{row 2} \\
\vdots \\
\text{another row} \\
\vdots \\
\text{row \(n\)}\end{pmatrix}
$$Note in above, the rows other than $i$ are not changed. Also, determinant is multilinear in the columns as well, namely $$
\begin{align*}
& \det \begin{pmatrix}\vec v_{1} & \vec v_{2} & \cdots & c\vec v_{i} + \vec w & \cdots & \vec v_{n} \end{pmatrix} \\
= & c\det \begin{pmatrix}\vec v_{1} & \vec v_{2} & \cdots & \vec v_{i}& \cdots & \vec v_{n} \end{pmatrix}+\det \begin{pmatrix}\vec v_{1} & \vec v_{2} & \cdots & \vec w & \cdots & \vec v_{n} \end{pmatrix}
\end{align*}
$$where $\vec v_{k}$ and $\vec w$ are column vectors.
**Example.** Observe the following equality illustrating $2\times 2$ determinant preserve sum and scaling in the second row $$
\underbrace{\det \begin{pmatrix}4 & 3 \\
\color{blue}7 & \color{blue} 14\end{pmatrix}}_{35} =7\cdot \underbrace{\det \begin{pmatrix}4 & 3 \\
\color{blue}1 & \color{blue}2\end{pmatrix}}_{5}
$$and $$
\underbrace{\det \begin{pmatrix}4 & 3 \\
\color{blue}7 & \color{blue} 14\end{pmatrix}}_{35} = \det \begin{pmatrix}4 & 3 \\
\color{blue}10-3 & \color{blue} 10+4\end{pmatrix}= \underbrace{\det \begin{pmatrix}4 & 3 \\
\color{blue}10 & \color{blue}10\end{pmatrix}}_{10} + \underbrace{\det \begin{pmatrix}4 & 3 \\
\color{blue}-3 & \color{blue}4\end{pmatrix}}_{25}. \quad\blacklozenge
$$
**Example.** Suppose we are dealing with $2\times 2$ determinants, and $\vec a,\vec b,\vec c,\vec d$ denote $2\times 1$ column vectors. Then $$
\begin{align*}
\det [\ \vec a +\vec b\ | \ \vec c + \vec d\ ] & = \det [\ \vec a \ | \ \vec c + \vec d\ ] + \det [\ \vec b\ | \ \vec c + \vec d\ ] \\
& =\det [\ \vec a \ | \ \vec c \ ] +\det [\ \vec a \ | \ \vec d\ ]+ \det [\ \vec b\ | \ \vec c \ ] +\det [\ \vec b\ | \ \vec d\ ] ,
\end{align*}
$$by applying multilinearity in one column at a time. $\blacklozenge$
### Antisymmetry.
In general, a function $f=f(x_{1},\ldots,x_{n})$ is **antisymmetric** if switching one pair of inputs negates the value, that is, for two different indices $i\neq j$, $$
f(x_{1},\ldots,x_{i},\ldots,x_{j},\ldots,x_{n})= - f(x_{1},\ldots,x_{j},\ldots,x_{i},\ldots,x_{n})
$$
**Example.**
The function $f:\mathbb{R}^{2}\to \mathbb{R}$ where $f(x,y)=x-y$ is antisymmetric. $\blacktriangleright$ Indeed, observe that $$
f(y,x)=y-x=-(x-y)=-f(x,y). \quad\blacklozenge
$$
**Example.**
In vector algebra, cross product is antisymmetric in the following sense: Let $f:\mathbb{R}^{3}\times \mathbb{R^{3}}\to \mathbb{R}^{3}$ to be the function $f(\vec x,\vec y)=\vec x\times \vec y$, where $\times$ is cross product. Then from vector algebra we know that $$
f(\vec y,\vec x)=\vec y\times \vec x=-\vec x \times \vec y= - f(\vec x,\vec y).\quad\blacklozenge
$$
So when we say determinant is antisymmetric in the rows, we mean that swapping a pair of rows negates the value of the determinant. And determinant is also antisymmetric in the columns as well.
**Example.** Observe the following equalities: $$
\underbrace{\det \begin{pmatrix}\color{blue}4 & \color{blue}2 \\ \color{red}8 &\color{red} 3\end{pmatrix}}_{-4} = -\ \underbrace{\det \begin{pmatrix} \color{red}8 &\color{red} 3 \\\color{blue}4 & \color{blue}2\end{pmatrix}}_{4}
$$and $$
\underbrace{\det \begin{pmatrix}\color{blue}3 & 4 & \color{red}1 \\ \color{blue}1 & 0 & \color{red}3\\\color{blue}2 & 2 & \color{red}1\end{pmatrix}}_{4} = -\ \underbrace{\det \begin{pmatrix} \color{red}1& 4 & \color{blue}3 \\ \color{red}3 & 0 & \color{blue}1\\ \color{red}1& 2 & \color{blue}2\end{pmatrix}}_{-4}. \quad\blacklozenge
$$
### Alternating property and linear dependent columns.
Using these defining properties of the determinants, we can actually deduce several properties of determinants, before even knowing how to compute them in general.
First, determinants are **alternating in the rows (columns, resp.)**, meaning that if two rows (columns, resp.) are the same, then the determinant is zero.
> **Proposition.** If $A$ is a square matrix with two identical rows (columns, resp.), then $\det(A)=0$. This property is called **alternating**.[^alt]
$\blacktriangleright$ Proof. Let us think of determinant as a function of $n$ rows (or $n$ columns, resp.), where $\det (A)= \det(r_{1},r_{2},\ldots,r_{n})$, where $r_{i}$ are the rows (columns, resp.) of $A$. Without loss, let us assume that $r_{1}=r_{2}=r$, and $\det(A)=d$. Now by antisymmetry, $$
\det(r_{1},r_{2},\ldots ,r_{n}) = -\det(r_{2},r_{1},\ldots ,r_{n})
$$so we get $$
\det(r,r,\ldots ,r_{n}) = -\det(r,r,\ldots ,r_{n})
$$But this says $d=-d$, which is true when $d=0$ (provided that $2\neq 0$). One can generalize to other pairs of indices. $\blacksquare$
**Example.**
Note $\det \begin{pmatrix}\color{blue}3 & \color{blue}2 & \color{blue}1 \\ 1&7&4 \\ \color{blue}3 & \color{blue}2 & \color{blue}1\end{pmatrix} =0$, without doing much work. $\blacklozenge$
If we combine multilinearity and antisymmetry together, we can say something even more:
> **Proposition.** If $A$ is a square matrix with linearly dependent rows (or columns, resp.), then $\det(A) = 0$.
In particular,
> **Corollary.** If $A$ is a square matrix with a row of zeros (or a column of zeros), then $\det(A) = 0$.
$\blacktriangleright$ Proof. Let $A$ be a matrix with rows (or columns, resp.) $r_{1},r_{2},\ldots,r_{n}$, and interpret $\det(A)$ as a function of the rows (columns, resp.), namely $\det(A)=\det(r_{1},r_{2},\ldots,r_{n})$. Suppose that $r_{1},\ldots,r_{n}$ are linearly dependent. Then by a result before we know that one of the rows (columns, resp.) is a linear combination of the other rows (columns, resp.). Say without loss, that $r_{1}$ is a linear combination of $r_{2},\ldots,r_{n}$. That is, $$
r_{1}=c_{2}r_{2}+\cdots +c_{n}r_{n}=\sum_{k=2}^{n}c_{k}r_{k}
$$for some scalars $c_{k}$. Then by multilinearity, we have $$
\begin{align*}
\det (A) & = \det (r_{1},r_{2}, \ldots , r_{n}) \\
& = \det ( \sum_{k=2}^{n}c_{k}r_{k} , r_{2} , \cdots , r_{n}) \\
& = \sum_{k=2}^{n} c_{k} \det(r_{k},r_{2},\ldots,r_{n}) = 0.
\end{align*}
$$Note above is a sum of determinants where the first row (column, resp.) is a repeat of one of the following rows (columns, resp.), as $k$ ranges from $2$ to $n$. So by the alternating property, repeating rows (columns, resp.) makes the determinant zero. Hence the sum is zero. $\blacksquare$
**Example.**
By careful observation, note that $$
\det \begin{pmatrix}3 & 1 & 4 \\
-2 & 1 & -1 \\
5 & 1 & 6\end{pmatrix} = 0
$$because the third column is a linear combination of the first two columns.
By the way, recall we have a computational way of deciding linear dependence and independence of a set of column vectors from [[smc-spring-2024-math-13/linear-algebra-notes/19-linear-independence-a-first-look|19-linear-independence-a-first-look]]. Computationally, a set of column vectors is linearly independent if and only if when we put them together as a matrix, it is row equivalent to an echelon from with pivot every columns. In other words we have
> **Proposition.** Let $A$ be a square matrix. If $A$ is row equivalent to an echelon from where there is a free column, then the columns are linearly dependent, and hence $\det(A)=0$.
And we can re-phrase it in terms of the rank of the matrix:
> **Proposition.** If $A$ is a square matrix of size $n\times n$, and that $\text{rank}(A) < n$, then $\det(A)=0$.
And of course its **contrapositive statement:**
> **Proposition.** If $A$ is an $n\times n$ matrix with $\det(A)\neq0$, then $\text{rank}(A)=n$. Namely, $A$ is a matrix that is row equivalent to an echelon form with $n$ pivots.
**Example.**
Consider the matrix $A= \begin{pmatrix}1&0&5 \\ 1 & 1 &1\\ 2 & 3 &-2\end{pmatrix}$. This matrix $A$ has $\det(A)=0$ because when we row-reduce $A$, we see that $$
A= \begin{pmatrix}1&0&5 \\ 1 & 1 &1\\ 2 & 3 &-2\end{pmatrix} \stackrel{\text{row}}\sim \begin{pmatrix}1&0&5 \\ 0 & 1 &-4\\ 0 & 0 &0\end{pmatrix}
$$where $\text{rank}(A) = 2 < 3$, hence by proposition $\det(A)=0$.
**Remark.** Now, it is not yet clear why the converse is also true: If $A$ is an $n\times n$ matrix where $\text{rank}(A)=n$, then $\det(A)\neq0$. But it is also true in fact, and we will show that later.
**Example.** With these properties we can compute determinants that are simple looking enough (and in fact works in general.) Observe the following steps to compute this determinant:$$
\begin{align*}
\det \begin{pmatrix}0 & 5 & 3 & 0 \\
0 & 0 & 0 & 8 \\
0 & 0 & 7 & 0 \\
9 & 0 & 0 & 0\end{pmatrix} & = \det \begin{pmatrix}0 & 5 & 3 & 0 \\
0 & 0 & 0 & 8 \\
0 & 0 & 0 & 0 \\
9 & 0 & 0 & 0\end{pmatrix}+\det \begin{pmatrix}0 & 5 & 0 & 0 \\
0 & 0 & 0 & 8 \\
0 & 0 & 7 & 0 \\
9 & 0 & 0 & 0\end{pmatrix} \\
& = 5\cdot8\cdot7\cdot9\cdot\det \begin{pmatrix}0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0\end{pmatrix} \\
& = 5\cdot8\cdot7\cdot9\cdot(-1)\cdot(-1)\cdot\det \begin{pmatrix}1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1\end{pmatrix} \\
= 2520. \quad\blacklozenge
\end{align*}
$$
## Leibniz formula: An entrywise formula of determinant.
So far, we defined determinant as a scalar function on matrices that is multilinear, antisymmetric, and $\det(I_{n})=1$. But does such a determinant function actually exist? Of course earlier we just "used" determinants of $2\times 2$ or $3\times 3$ formulas, **as if** they are actually determinants, but we did not justify they are indeed determinant functions. So here we show that yes, determinant function exist, and we can write down an **entrywise formula** (sometimes called **Leibniz formula**) for the determinant. And the $2\times 2$ and $3\times 3$ determinant formula we gave above can be obtained from Leibniz formula.
> **Leibniz entrywise formula for determinant.**
> Let $A$ be an $n\times n$ matrix with the $(i,j)$-th entry $[A]_{ij}=a_{ij}$. Then $$
\det(A)=\sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma) a_{1 \sigma(1)}a_{2 \sigma(2)}\cdots a_{n \sigma(n)} = \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma) \prod_{k=1}^{n}a_{k\sigma(k)}
$$That is, we claim this is a function that satisfies what defines a determinant (multilinear, antisymmetric, normalizing at the identity matrix).
Note, here $a_{ij}$ denotes the $i$-th row $j$-th column entry, and so $a_{k\sigma(k)}$ is the $k$-th row $\sigma(k)$-th column entry.
We will leave the verification of this entrywise formula indeed satisfies the properties of a determinant later / as an exercise. But first, let us demonstrate that the formulas we gave for the $2\times 2$ and $3\times 3$ determinants indeed can be obtained by this entrywise formula.
**Example. $2\times 2$ determinant formula from Leibniz entrywise formula.**
Let $A = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}$ be a $2 \times 2$ with $(i,j)$-th entry $a_{ij}$ as usual. Then by Leibniz entrywise formula $$
\det (A) = \sum_{\sigma \in \mathfrak S_{2}} \text{sign}(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}.
$$Note $\mathfrak S_{2} = \{12,21\}$, so there are two terms in that sum, when $\sigma = 12$ and when $\sigma = 21$. So we have $$
\det(A) = \underbrace{\text{sign}({\color{blue}12})a_{1{\color{blue}1}} a_{2{\color{blue}2}}}_{\sigma={\color{blue}12}, \sigma(1)={\color{blue}1},\sigma(2)={\color{blue}2}} + \underbrace{\text{sign}({\color{blue}21})a_{1{\color{blue}2}}a_{2{\color{blue}1}}}_{\sigma={\color{blue}21},\sigma(1)={\color{blue}2},\sigma(2)={\color{blue}1}} = +a_{11}a_{22}-a_{12}a_{21}.
$$This is the familiar $2\times 2$ determinant formula!
**Example. $3\times 3$ determinant formula from Leibniz entrywise formula.**
Let $A = \begin{pmatrix}a_{11} & a_{12}& a_{13} \\ a_{21} & a_{22}& a_{23}\\ a_{31} & a_{32}& a_{33}\end{pmatrix}$ be a $3 \times 3$ with $(i,j)$-th entry $a_{ij}$ as usual. Then by Leibniz entrywise formula we have $$
\det (A) = \sum_{\sigma \in \mathfrak S_{3}} \text{sign}(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}a_{3\sigma(3)}.
$$ Note $\mathfrak S_{3} = \{123, 132, 213, 231, 312, 321\}$, so there are six terms in that sum. Writing them out we get $$
\begin{align*}
\det(A) =& \text{sign}({\color{blue}123})a_{1{\color{blue}1}}a_{2{\color{blue}2}}a_{3\color{blue}{3}} \\
& +\text{sign}({\color{blue}132})a_{1{\color{blue}1}}a_{2{\color{blue}3}}a_{3\color{blue}{2}} \\
& +\text{sign}({\color{blue}213})a_{1{\color{blue}2}}a_{2{\color{blue}1}}a_{3\color{blue}{3}} \\ & +\text{sign}({\color{blue}231})a_{1{\color{blue}2}}a_{2{\color{blue}3}}a_{3\color{blue}{1}} \\ & +\text{sign}({\color{blue}312})a_{1{\color{blue}3}}a_{2{\color{blue}1}}a_{3\color{blue}{2}} \\ & +\text{sign}({\color{blue}321})a_{1{\color{blue}3}}a_{2{\color{blue}2}}a_{3\color{blue}{1}} \\ \\
\implies \det(A)=& + a_{1{\color{blue}1}}a_{2{\color{blue}2}}a_{3\color{blue}{3}} \\
& - a_{1{\color{blue}1}}a_{2{\color{blue}3}}a_{3\color{blue}{2}} \\
& - a_{1{\color{blue}2}}a_{2{\color{blue}1}}a_{3\color{blue}{3}} \\
& + a_{1{\color{blue}2}}a_{2{\color{blue}3}}a_{3\color{blue}{1}} \\
& + a_{1{\color{blue}3}}a_{2{\color{blue}1}}a_{3\color{blue}{2}} \\
& - a_{1{\color{blue}3}}a_{2{\color{blue}2}}a_{3\color{blue}{1}},
\end{align*}
$$which is the basket-weaving formula that we have for $3\times 3$ determinants! $\blacklozenge$
As an exercise, as excruciating as it may be, one should try to write out the $4\times 4$ determinant formula using Leibniz entrywise formula, involving 24 terms.
**Remark.** The Leibniz entrywise formula may not seem very "efficient", but it is useful in giving insights of certain properties of determinants. Also technically we should also verify that this formula does indeed satisfy all the properties of what defines the determinant.
Let us illustrate how we can use this formula to prove determinant is invariant under transposes.
> **Proposition.**
> If $A$ is a square matrix, then $\det(A)=\det(A^{T})$.
$\blacktriangleright$ Proof.
Let $A$ to be an $n\times n$ matrix with $(i,j)$-th entry $a_{i,j}$. Then $A^{T}$ has $(i,j)$-th entry $[A^{T}]_{i,j}=a_{j,i}$. Let us apply Leibniz entrywise formula on $A^{T}$, we get $$
\begin{align*}
\det (A^{T}) = & \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma)[A^{T}]_{1,\sigma(1)} [A^{T}]_{2,\sigma(2)}\cdots [A^{T}]_{n,\sigma(n)} \\
= & \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma)a_{ \sigma(1),1} a_{ \sigma(2), 2}\cdots a_{ \sigma(n), n}
\end{align*}
$$Note when the pair $(\sigma(k),k)=(\ell,k)$, we have $k=\sigma^{-1}(\ell)$, so the pair $(\sigma(k),k)=(\ell,\sigma^{-1}(\ell))$. And since the numbers $\sigma(1),\sigma(2),\ldots,\sigma(n)$ are just an arrangement of the numbers $1,2,\ldots,n$, one of the $\sigma(k)$ equals $1$, one of them equals 2, etc. Hence we can re-arrange the product in each term as follows: $$
\begin{align*}
\det (A^{T}) = & \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma)a_{ \sigma(1),1} a_{ \sigma(2), 2}\cdots a_{ \sigma(n), n} \\
= & \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma)a_{1, \sigma^{-1}(1)} a_{ 2,\sigma^{-1}(2)}\cdots a_{n, \sigma^{-1}(n)}
\end{align*}
$$
Next, summing over the permutations $\mathfrak S_{n} = \{\sigma:\sigma\in \mathfrak S_{n}\}$ is the same as summing over their inverses, the set $\{\sigma^{-1}: \sigma\in \mathfrak S_{n}\}$ (see [[1 teaching/smc-spring-2024-math-13/linear-algebra-notes/28-permutations|28-permutations]], the last section on involution), so we have re-write the sum as $$
\begin{align*}
\det (A^{T})
= & \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma)a_{1, \sigma^{-1}(1)} a_{ 2,\sigma^{-1}(2)}\cdots a_{n, \sigma^{-1}(n)} \\
= & \sum_{\sigma^{-1} \in \mathfrak S_{n}} \text{sign}(\sigma^{-1})a_{1, \sigma(1)} a_{ 2,\sigma (2)}\cdots a_{n, \sigma (n)} \\
= & \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma^{-1})a_{1, \sigma(1)} a_{ 2,\sigma (2)}\cdots a_{n, \sigma (n)}
\end{align*}
$$But as $\text{sign}(\sigma)=\text{sign}(\sigma^{-1})$ (again, see [[1 teaching/smc-spring-2024-math-13/linear-algebra-notes/28-permutations|28-permutations]]), we get $$
\begin{align*}
\det (A^{T})
= & \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma^{-1})a_{1, \sigma(1)} a_{ 2,\sigma (2)}\cdots a_{n, \sigma (n)} \\
= & \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma)a_{1, \sigma(1)} a_{ 2,\sigma (2)}\cdots a_{n, \sigma (n)} \\
= & \det(A)
\end{align*}
$$as claimed! $\blacksquare$
So we see that this formula is useful in proving properties of determinants. However, there are more efficient ways of computing determinants. If we loosely use the word "flop" to denote an addition or multiplication operation is carried, then using Leibniz entrywise formula to compute an $n\times n$ determinant uses $\approx n$ multiplications per summand, but with $\approx n!$ summands, which comes out to $\approx n\cdot n !$ flops. This gets huge as $n$ increases! Later as it turns out, determinants can be computed by performing row-reduction, and it would take about $\approx n^{3}$ flops. And in the next part we will show other methods of computing determinants.
Nevertheless, Leibniz entrywise formula offers many insights, for example:
> **Observation.** The computation of the determinant of a square matrix $A$ can be done using only addition, subtraction, and multiplication, and **never** division.
We end this section on verifying Leibniz entrywise formula is indeed a determinant.
### Leibniz formula is determinant.
We show for a square matrix $A$ with $(i,j)$-th entry $a_{ij}$, the Leibniz formula $$
\det(A) = \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}\cdots a_{n\sigma(n)}
$$ satisfies the properties defining a determinant on $A$, namely it is multilinear, antisymmetric and normalizing at the identity matrix. We will, without ambiguity, write $\det$ for Leibniz formula in this part. This proof is one of those activities where you do it once in your life, and then never have to again (well, unless forced to).
$\blacktriangleright$ Proof.
**It is multilinear.**
Recall multilinear means linear in each row (column, resp.) with other rows (columns, resp.) fixed. We will show that it is linear in the first row, and let you generalize to other rows. Suppose the first row of $A$ is some linear combination of some two rows, where $$
a_{1k}=x_{1k}+cy_{1k}
$$for each $k=1,\ldots,n$. Then $$
\begin{align*}
\det(A) & = \sum_{\sigma \in \mathfrak S_{n}}\text{sign}(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}\cdots a_{n\sigma(n)} \\
& = \sum_{\sigma \in \mathfrak S_{n}}\text{sign}(\sigma)(x_{1\sigma(1)}+cy_{1\sigma(1)})a_{2\sigma(2)}\cdots a_{n\sigma(n)} \\
& =\sum_{\sigma \in \mathfrak S_{n}}\text{sign}(\sigma)x_{1\sigma(1)}a_{2\sigma(2)}\cdots a_{n\sigma(n)} + c\sum_{\sigma \in \mathfrak S_{n}}\text{sign}(\sigma)y_{1\sigma(1)}a_{2\sigma(2)}\cdots a_{n\sigma(n)} \\
& = \det A' + c\det A''
\end{align*}
$$where $A'$ is the matrix $A$ whose first row is replaced by $x_{1k}$, while $A''$ is the matrix $A$ with the first row replaced by $y_{1k}$. This shows $\det$ is linear in the first row. One can check with same calculation to show $\det$ is linear in any of its $k$-th row. Hence Leibniz formula is multilinear.
**It is antisymmetric.**
Recall antisymmetric means switching a pair of rows (columns, resp.) will negate the value. Let us demonstrate this on just the first and second row, and one can generalize to an arbitrary pair of rows.
Denoting $A'$ to be the matrix obtained by swapping row 1 and row 2 of the matrix $A$, then $[A']_{1k}=a_{2k}$ and $[A']_{2k}=a_{1k}$, where $a_{ij}$ is the $(i,j)$-th entry of $A$, and that $[A']_{ik}=a_{ik}$ for all $i\neq1,2$. Then we have $$
\begin{align*}
\det (A') & = \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma) [A']_{1\sigma(1)}[A']_{2\sigma(2)}\cdots [A']_{n\sigma(n)} \\
& = \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma) a_{2\sigma(1)}a_{1\sigma(2)}\cdots a_{n\sigma(n)} \\
& = \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma) a_{1\sigma(2)}a_{2\sigma(1)}\cdots a_{n\sigma(n)}
\end{align*}
$$Notice that this is almost the determinant of $A$, except we are looking at the indices $(1,\sigma(2))$ and $(2,\sigma(1))$ instead of $(1,\sigma(1))$ and $(2,\sigma(2))$. That is, it is "off by a swap". Intuitively, this is where we are going to gain an extra negative sign.
Notice, for any permutation $\sigma$, there exists a permutation $\tau$ such that $\tau(1)=\sigma(2)$ and $\tau(2)=\sigma(1)$, and $\tau(k)=\sigma(k)$ for all $k\neq1,2$. This $\tau$ is almost $\sigma$ but with an extra swap, so $\text{sign}(\tau) = -\text{sign}(\sigma)$. One can check that for each $\sigma$ this $\tau$ can be uniquely obtained, and that summing over $\sigma$ amounts to summing over $\tau$. Hence we have $$
\begin{align*}
\det (A') & = \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma) a_{1\sigma(2)}a_{2\sigma(1)}\cdots a_{n\sigma(n)} \\
& = \sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\sigma) a_{1\tau(1)}a_{2\tau(2)}\cdots a_{n\tau(n)} \\
& = \sum_{\sigma \in \mathfrak S_{n}} -\text{sign}(\tau) a_{1\tau(1)}a_{2\tau(2)}\cdots a_{n\tau(n)} \\
& = -\sum_{\sigma \in \mathfrak S_{n}} \text{sign}(\tau) a_{1\tau(1)}a_{2\tau(2)}\cdots a_{n\tau(n)} \\
& = -\sum_{\tau \in \mathfrak S_{n}} \text{sign}(\tau) a_{1\tau(1)}a_{2\tau(2)}\cdots a_{n\tau(n)} \\
& =-\det(A).
\end{align*}
$$As this can be generalized to any pair of rows, Leibniz formula is indeed antisymmetric.
**It normalizes at the identity matrix.**
Note the identity matrix $I_{n}$ has entries $a_{ii}=1$ and $a_{ij}=0$ whenever $i\neq j$. So applying Leibniz formula gives $$
\det(I_{n}) = \sum_{\sigma\in \mathfrak S_{n}} \text{sign}(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}\cdots a_{n\sigma(n)}.
$$But using our third-eye we see that if a permutation $\sigma$ is **not** the identity permutation, then there will be some index $k$ such that $\sigma(k)\neq k$, which makes product involving this $\sigma$ zero. Hence there is only one term left in the sum, the term with the identity permutation $\text{id}$, so $$
\begin{align*}
\det I_{n} & = \text{sign}(\text{id})a_{1,\text{id}(1)}a_{2,\text{id}(2)}\cdots a_{n,\text{id}(n)} \\
& = +a_{1,1}a_{2,2}\cdots a_{n,n} \\
& = + (1)(1)\cdots (1) \\
\implies \det(I_{n}) & =1,
\end{align*}
$$as claimed.
This proves that indeed, Leibniz entrywise formula satisfies all properties of determinant, and hence the formula is worthy of being labeled with the name $\det$, huzzah! $\blacksquare$
**Remark.** It remains to show, that determinant is uniquely defined. We will establish this later.
---
[^field]: Just think a commutative and associative number system with addition, subtraction, multiplication, and division by nonzero, like the reals $\mathbb{R}$ and the complex numbers $\mathbb{C}$.
[^objection]: There are, however, people who think this approach is "bad", that there is a "linear algebra done right" approach, where we develop linear algebra largely independent of using the determinant as a gadget. I cannot say I subscribe to this philosophy. Nevertheless, there is mathematical value in trying to develop X without using Y as a good mathematical inquiry.
[^alt]: Okay, provided that the scalar field we are working with is not of characteristic 2, meaning $2$ is not zero.