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· 12 min read
· by Xianbin
What is Determinant?
In Wiki, it says that
In mathematics the determinant is a scalar-valued function of the entries of a square matrix.
In Geometry, Determinant is a Volume
Since it is a volume, the output must be a value!
Let us put it more simple. Determinant is function \(f\): \(\mathbb{R}^n\times \cdots \times \mathbb{R}^n \to \mathbb{R}\).
Let \(\textbf{a}_1,\textbf{a}_2\) be vectors of size two.
\[f(\textbf{a}_1,\textbf{a}_2) = \textup{Det}(\textbf{a}_1,\textbf{a}_2)\]
If we increase \(\textbf{a}_1\) to be \(c\textbf{a}_1\) where \(c > 1\), then the volume increases \(c\) times.
\[1): f(c\textbf{a}_1,\textbf{a}_2) = cf(\textbf{a}_1,\textbf{a}_2)\]
That is simple, right?
By similar tricks, we have that
\[2): f(\textbf{a}_1+\textbf{v},\textbf{a}_2) = f(\textbf{a}_1,\textbf{a}_2) +f(\textbf{v},\textbf{a}_2)\]
Now, let us consider the exchange of two vectors in \(f\).
\[f(\textbf{a}_1,\ldots, \textbf{a}_k, \ldots,\textbf{a}_i,\ldots,\textbf{a}_n)\]
We know that the absolute value cannot change. Notice that due to the definition of determinant, the notion sign will change.
Then,
\[3): f(\textbf{a}_1,\ldots, \textbf{a}_k, \ldots,\textbf{a}_i,\ldots,\textbf{a}_n) = -
f(\textbf{a}_1,\ldots, \textbf{a}_i, \ldots,\textbf{a}_k,\ldots,\textbf{a}_n)\]
Finally, we know that if the vectors all unit vectors, we have
\((4) f(\textbf{e}_1,\ldots, \textbf{e}_i, \ldots,\textbf{e}_k,\ldots,\textbf{e}_n)=1\)
where \(\textbf{e}_i \perp \textbf{e}_j\) for any \(i\not = j\).
That is because, the volume can only be one.
Now, let us see \(\textup{Det}(\textbf{a}_1,\textbf{a}_2)\). By property (1),(2),(3),(4).
\[\textup{Det}(\textbf{a}_1,\textbf{a}_2) = \textup{Det}\left( {v_1 \choose v_2}, {v_3 \choose v_4} \right)\\=
\textup{Det}\left( {v_1 \choose 0} + {0 \choose v_2}, {v_3 \choose v_4} \right)\\=
\textup{Det}\left( {v_1 \choose 0} , {v_3 \choose v_4} \right) + \textup{Det}\left( {0 \choose v_2}, {v_3 \choose v_4} \right)\\=
v_1\textup{Det}\left( {1 \choose 0} , {v_3 \choose v_4} \right) + v_2\textup{Det}\left( {0 \choose 1}, {v_3 \choose v_4} \right)\\ =
v_1v_3 \textup{Det}\left( {1 \choose 0}, {1 \choose 0}\right)
+v_1v_4\textup{Det}\left( {1 \choose 0}, {0 \choose 1}\right) \\+v_2v_3\textup{Det}\left( {0 \choose 1}, {1 \choose 0}\right) + v_2v_4\textup{Det}\left( {0 \choose 1}, {0 \choose 1}\right)
\\=
v_1 v_4 - v_2v_3\]
Let us calculate \(\textbf{Det}(\textbf{a}_1,\ldots,\textbf{a}_n)\).
\[\textup{Det}\left(\left(
\begin{matrix}
v_{11}\\
\vdots\\
v_{n1}
\end{matrix}\right)
\left(
\begin{matrix}
v_{12}\\
\vdots\\
v_{n2}
\end{matrix}\right),
\cdots,
\left(
\begin{matrix}
v_{n1}\\
\vdots\\
v_{n1}
\end{matrix}\right)
\right) =
\\
\textup{Det}\left(
v_{11}e_1+\ldots,v_{n1}e_n, \Delta_1
\right)=\\
v_{11}\textup{Det}(e_1,\Delta_1)+\ldots+v_{n1}\textup{Det}(e_n,\Delta_1)\\=
\sum_{j_1=1}^n v_{j_1,1}\textup{Det}(e_{j_1},\Delta_1)\]
where
\(\Delta_1 = \left(\left(
\begin{matrix}
v_{n2}\\
\vdots\\
v_{n2}
\end{matrix}\right),
\cdots,
\left(
\begin{matrix}
v_{n1}\\
\vdots\\
v_{n1}
\end{matrix}\right)\right)\)
Now, we obtain a recursion equality.
\[\textup{Det}(\textbf{a}_1,\Delta_1) = \sum_{j_1=1}^n v_{j_1,1}\textup{Det}(e_{j_1},\Delta_1)\]
We can see that the relationship is this:
First, we swap \(\textbf{a}_2\) to \(e_{j1}\), then, we have
\[\textup{Det}(\textbf{a}_1,\textbf{a}_2,\Delta_2) = \\
\sum_{j_1=1}^n v_{j_1,1} \textup{Det}(\textbf{a}_2,e_{j_1},\Delta_2) \cdot (-1) \\
=\sum_{j_1=1}^n \sum_{j_2=1}^n v_{j_1,1} v_{j_2,2} \textup{Det}(e_{j2},e_{j_1},\Delta_2) \cdot (-1)\]
Then, we swap it back.
\[\textup{Det}(\textbf{a}_1,\Delta_1) =\\ \textup{Det}(\textbf{a}_1,\textbf{a}_2,\Delta_2) = \sum_{j_1=1}^n \sum_{j_2=1}^n v_{j_1,1} v_{j_2,2} \textup{Det}(e_{j_1}, e_{j_2},\Delta_2) \cdot (-1) \cdot(-1) \\=
\sum_{j_1=1}^n \sum_{j_2=1}^n v_{j_1,1} v_{j_2,2} \textup{Det}(e_{j_1}, e_{j_2},\Delta_2)\]
We construct \(\Delta_{n-1}\subset\cdots\subset\Delta_{1}\) such that \(\Delta_{i+1} \cup \textbf{a}_{i+1} = \Delta_{i}\), where \(\Delta_{n-1} = e_{jn}\) and \(i\in [n-2]\).
Then, we have
\[\textup{Det}(\textbf{a}_1, \textbf{a}_2,\ldots, \textbf{a}_n) = \sum_{j_1=1}^n \sum_{j_2=1}^n \ldots \sum_{j_n=1}^nv_{j_1,1}\ldots v_{j_n,n}\textup{Det}(e_{j_1},\ldots,\Delta_{n-1})\\=
\sum_{\delta = (j_1,\ldots,j_n)\in \pi_n}\text{sgn}(j_1,\ldots,j_n) \prod_{i=1}^n v_{j_{\delta(i)}, i}\]
where \(\pi_n\) is a permutation of \(\{1,\ldots,n\}\).
Reference
[1] The Bright Side of Mathematics. Linear Algebra 45,46.