T1: What is Determinant

July 12, 2024

• What is Determinant?
• Leibniz Formula for Determinants
• Reference

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).

$$\textup{Det}(\textbf{a}_1,\textbf{a}_2) = \textup{Det}\left( {v_1 \choose v_2}, {v_3 \choose v_4} \right)\\= \textup{Det}\left( {v_1 \choose 0} + {0 \choose v_2}, {v_3 \choose v_4} \right)\\= \textup{Det}\left( {v_1 \choose 0} , {v_3 \choose v_4} \right) + \textup{Det}\left( {0 \choose v_2}, {v_3 \choose v_4} \right)\\= v_1\textup{Det}\left( {1 \choose 0} , {v_3 \choose v_4} \right) + v_2\textup{Det}\left( {0 \choose 1}, {v_3 \choose v_4} \right)\\ = v_1v_3 \textup{Det}\left( {1 \choose 0}, {1 \choose 0}\right) +v_1v_4\textup{Det}\left( {1 \choose 0}, {0 \choose 1}\right) \\+v_2v_3\textup{Det}\left( {0 \choose 1}, {1 \choose 0}\right) + v_2v_4\textup{Det}\left( {0 \choose 1}, {0 \choose 1}\right) \\= v_1 v_4 - v_2v_3$$

Leibniz Formula for Determinants

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.