---- Let [[field]] be a [[field|field]]. > [!definition] Definition. ([[determinant of a matrix]]) > For a square matrix $A \in \ff^{n \times n}$, the [[determinant]] of $A$ can be expressed in terms of [[permutation]]s of its elements $\{ a_{ij} \}_{i,j=1}^{n}$: $\det A = \sum_{\tau \in S_{n}} \text{sign}(\tau) \prod_{i=1}^{n} a_{i, \tau(i)}= \sum_{\tau \in S_{n}} \text{sign}(\tau) a_{1\tau(1)}a_{2\tau(2)}\cdots a_{n \tau(n)}. $ > > **Phrased in words:** > - An $n$-by-$n$ [[matrix]]'s [[determinant]] is given by a sum of $n!$ terms: > - one for each [[permutation]] $\tau$ of $S_{n}:=\{1,2,\dots,n\}$. > - Each term is a product of $n$ entries: > - each contributed by a distinct row and distinct column (e.g., a term can't include $2$ entries from column $3$). > - The *rows* 'come in order' in each term — $1,2,\dots,n$. The columns correspond to *[[permutation]]s* — $\tau(1),\tau(2),\dots,\tau(n)$. > - The sign of each term corresponds to the sign of the [[permutation]] $\tau$: > - negative if $\tau$ is odd, positive if $\tau$ is even. > [!intuition] > This formula is overwhelming at first; let's think about a $3 \times 3$ case. Consider $A=\begin{pmatrix}a_{11} & a_{12} & a_{13} \\ > a_{21} & a_{2 2} & a_{23} \\ > a_{31} & a_{32} & a_{33}\end{pmatrix}.$ > Using [[Laplace expansion]] we easily compute $\det A$ to be $ \begin{align} > \det A= & a_{11}a_{2 2}a_{33} - a_{11}a_{23}a_{32} \\ > - & a_{12}a_{21}a_{33} +a_{12}a_{23}a_{31} \\ > + & a_{13}a_{21}a_{32} - a_{13}a_{2 2}a_{31}. \end{align} $ > > If we name the first subscript along the entries of this sum, it would sound like a waltz with one measure per term: $'1 2 3 ~ 1 2 3 ~ 1 2 3 ~ 1 2 3 ~ 1 2 3 ~ 1 2 3'.$ > The interesting part happens with the second subscript (denoting column) in each entry— these are scrambled. In fact... they're *as scrambled as possible!* Every [[permutation]] of $1,2,3$ is represented (hence why there are $6 = 3!$ terms). Noting that the negative terms correspond to *odd* [[permutation]]s of $1,2,3$ — for example, $3,2,1$ is obtained from $1,2,3$ by making $3$ swaps — we see that the Leibniz formula is satisfied. > [!basicproperties] > $\det$ is continuous: Let $A \in \mathbb{R}^{m \times m}$. Consider the map $A \to \det A$ from $\mathbb{R}^{m \times m}$. Since the [[determinant#Leibniz Formula]] consists of [[polynomial]] entries, this map is continuous as a algebraic composition of [[continuous]] functions. In fact it is [[continuously differentiable]] ($C^\infty$ to be precise) since all polynomials are [[smooth]]. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```