---
title: Determinant 2
tags: Linear algebra
GA: G-77TT93X4N1
---

# Determinant 2

> Proof of properties.

---

## Notation

$$
det(A) = |A| = D({\bf v}_1, \cdots, {\bf v}_n)
$$

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## Definition

* Determinant of identity matrix is one
$$\tag{1}
det(I_n) = 1
$$
* Antisymmetry
$$\tag{2}
D(\cdots, {\bf v}_j, \cdots, {\bf v}_k,\cdots) = -D( \cdots, {\bf v}_k, \cdots, {\bf v}_j,\cdots)
$$
* Linearity in each argument:
$$\tag{3}
D(\cdots, \alpha{\bf v}_k, \cdots) = \alpha D( \cdots, {\bf v}_k, \cdots)
$$
and
$$\tag{4}
D(\cdots, {\bf u}_k+{\bf v}_k, \cdots) = D( \cdots, {\bf u}_k, \cdots) + D( \cdots, {\bf v}_k, \cdots)
$$

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## Properties

1. Two equal columns, then determinant equals to zero.
* pf: By antisymmetry, Eq.~(2),
$$
D( \cdots, {\bf v}, \cdots, {\bf v},\cdots) = -D( \cdots, {\bf v}, \cdots, {\bf v},\cdots).
$$
Therefore
$$\tag{5}
D( \cdots, {\bf v}, \cdots, {\bf v},\cdots) = 0.
$$
2. A column of zero, then determinant equals to zero.
* pf: By linearity, Eq.~(3),
$$
\alpha D( \cdots, {\bf 0}, \cdots) = D( \cdots, \alpha{\bf 0}, \cdots) = D( \cdots, {\bf 0}, \cdots).
$$

Therefore (choose $\alpha=2$),

$$\tag{6}
D( \cdots, {\bf 0}, \cdots) = 0.
$$
3. Preservation under “column replacement”
$$\tag{7}
D(\cdots, {\bf v}_j+\alpha{\bf v}_k, \cdots, {\bf v}_k,\cdots) = D( \cdots, {\bf v}_j, \cdots, {\bf v}_k,\cdots).
$$
* pf: By linearity, Eq.~(4),
$$
D(\cdots, {\bf v}_j+\alpha{\bf v}_k, \cdots, {\bf v}_k,\cdots) = D( \cdots, {\bf v}_j, \cdots, {\bf v}_k,\cdots) + D(\cdots, \alpha{\bf v}_k, \cdots, {\bf v}_k,\cdots),
$$
where (use (3) and (5))
$$
D(\cdots, \alpha{\bf v}_k, \cdots, {\bf v}_k,\cdots) = \alpha D(\cdots, {\bf v}_k, \cdots, {\bf v}_k,\cdots) =0.
$$
4. If $A$ is singular, then $\text{det}(A) =0$.
* pf: If $A$ is singular, one of ${\bf v}_j$ is linear combination of other columns, then we can use "column replacement" to zero-out this column, i.e.,
$$
D(\cdots, {\bf v}_j, \cdots) = D(\cdots, {\bf 0}, \cdots) =0.
$$
5. Determinant of upper triangular matrix is the product of its diagonal elements, i.e.,
$$\tag{8}
\left|\begin{matrix}
d_1 & * & \cdots & *\\
0 & d_2 & \cdots & *\\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & d_n \\
\end{matrix}\right| = d_1d_2\cdots d_n.
$$
* pf:
* If $d_1=0$, we have a column of zero and $|A|=0$ by (6). So LHS equals to RHS, true.
* If $d_1\ne 0$, we can apply "column replacement" to have
$$
\left|\begin{matrix}
d_1 & * & \cdots & *\\
0 & d_2 & \cdots & *\\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & d_n \\
\end{matrix}\right| =
\left|\begin{matrix}
d_1 & 0 & \cdots & 0\\
0 & d_2 & \cdots & *\\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & d_n \\
\end{matrix}\right|
$$
* Use induction, either $d_k=0$ for some $k$ and $|A|=0$, or we have
$$
\left|\begin{matrix}
d_1 & * & \cdots & *\\
0 & d_2 & \cdots & *\\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & d_n \\
\end{matrix}\right| =
\left|\begin{matrix}
d_1 & 0 & \cdots & 0\\
0 & d_2 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & d_n \\
\end{matrix}\right|=d_1d_2\cdots d_n \left|\begin{matrix}
1 & 0 & \cdots & 0\\
0 & 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 \\
\end{matrix}\right|=d_1d_2\cdots d_n.
$$
In either case, we have LHS equals to RHS. So the statement is true.
* Remark: Same conclusion applied to lower triangular matrices.
6. $\text{det}(AB) = \text{det}(A)\text{det}(B)$.
* pf:
* If $A$ is singular, $AB$ is singular, and $\text{det}(AB)=\text{det}(A)=0$. So the equality is true.
* If $A$ is non-singular, $|A|\ne 0$, we define a function $\tilde{D}:M_{n\times n}\to \mathbb{R}$ as
$$\tag{9}
\tilde{D}(B) = \frac{|AB|}{|A|}.
$$
* Properties of $\tilde{D}$:
a.
$$
\tilde{D}(I_n) = \frac{|AI_n|}{|A|}= 1
$$
b. Let $B = [\cdots, {\bf v}_j, \cdots, {\bf v}_k,\cdots]$, then $[\cdots, {\bf v}_k, \cdots, {\bf v}_j,\cdots] = BP$, where $P$ is the permutation matrix that exchanges columns $j$ and $k$. Also we have
$$
|A(BP)| = |(AB)P| = -|AB|.
$$
So
$$
\tilde{D}(\cdots, {\bf v}_k, \cdots, {\bf v}_j,\cdots) = \tilde{D}(BP) =-\tilde{D}(B)= -\tilde{D}(\cdots, {\bf v}_j, \cdots, {\bf v}_k,\cdots).
$$
c. Let $B = [\cdots, {\bf v}_k, \cdots]$, then $[\cdots, \alpha{\bf v}_k, \cdots ] = BJ$, where $J$ is the identity matrix except $J_{k,k} = \alpha$. Then
$$
\tilde{D}(\cdots, \alpha{\bf v}_k, \cdots) = \tilde{D}(BJ) = \frac{|ABJ|}{|A|}= \frac{\alpha|AB|}{|A|} =\alpha\tilde{D}(B)= \alpha D( \cdots, {\bf v}_k, \cdots).
$$
Finally, recall that we have $A[\cdots, {\bf u}_k+{\bf v}_k, \cdots] =[\cdots, A{\bf u}_k+A{\bf v}_k, \cdots]$, so
$$
\text{det}(A[\cdots, {\bf u}_k+{\bf v}_k, \cdots]) = \text{det}([\cdots, A{\bf u}_k+A{\bf v}_k, \cdots]) \\= \text{det}([\cdots, A{\bf u}_k, \cdots]) + \text{det}([\cdots, A{\bf v}_k, \cdots]).
$$
It is then easy to show that
$$
\tilde{D}(\cdots, {\bf u}_k+{\bf v}_k, \cdots) = \tilde{D}(\cdots, {\bf u}_k, \cdots) + \tilde{D}(\cdots, {\bf v}_k, \cdots).
$$
* Summary: It turns out that $\tilde{D}$ satisfies exactly the same $3$ definitions as $D$, so $\tilde{D}=D$ and $\tilde{D}(B) = |B|$. Therefore, $|AB|=|A||B|$.
7. $\text{det}(A^T) = \text{det}(A)$.
* pf:
* Observation: If $P$ is a permutation matrix, then $|P| = |P^T|$ and $|P|=\pm 1$, i.e., either both are $1$ or both are $-1$.
* Let $PA = LU$, where $P$ is the permutation matrix, $L$ is lower triangular that has all the diagonals being $1$ and $U$ is upper triangular. Then $|L|=|L^T|=1$, $|U|=|U^T|$, and
$$
|A| = |P^TLU| = |P^T||L||U| = |U^T||L^T||P| = |U^TL^TP| = |A^T|.
$$

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## References
* [Linear Algebra Done Wrong: Ch2](https://www.math.brown.edu/streil/papers/LADW/LADW.html)