# Lecture 006

$3 \times 3$-Determinant: $\det \begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i\\ \end{vmatrix} = a \det \begin{vmatrix} e & f\\ h & i\\ \end{vmatrix} - b \det \begin{vmatrix} d & f\\ g & i\\ \end{vmatrix} + c \det \begin{vmatrix} d & e\\ g & h\\ \end{vmatrix}$

Submatrix notation: $A_{ij}$ is the submatrix after remove $i$-th row and $j$-th column. $(i, j)$-minor is $\det A_{ij}$.

$n \times n$-Determinant:

• first row: $\det A = \sum_{j = 1}^n (-1)^{j + 1} a_{ij} \det A_{ij}$

• along $i$-th row: $\det A = \sum_{j = 1}^n (-1)^{i+j} a_{ij} \det A_ij$

• along $j$-th column: $\det A = \sum_{i = 1}^n (-1)^{j + j} a_{ij} \det A_{ij}$

• $\det I_n = 1$

• matrix of all $1$: $\begin{cases} 0 & \text{if } n \geq 2\\ 1 & \text{if } n = 1 \text{ (single element)}\\ \end{cases}$

• permutation matrix: $\pm 1$

Properties of Determinant:

• if $A$ has a row/column of all $0$, then $\det A = 0$.

• by swapping two rows / columns, the determinant gets multiplied by $-1$.

• if $A$ has two identical row/column, then $\det A = 0$.

• multiply one row/column by $k$, then $\det A \rightarrow k\det A$.

• if for some rows $A_i, B_i, C_i$ in matrix $A, B, C$ such that $A_i + B_i = C_i$ (and everything else are the same), then $\det A + \det B = \det C$. (same is true for columns)

• Multiplying a row by $k$ results determinant multiply by $k$

• Adding a multiple of a row/column to another row/column doesn't change the determinant.

• gaussian process does not change determinant. (except swaping two rows changes the sign)

• $\det A \neq 0 \iff A \text{ is invertible}$

• $\det A = \det A^T$

• $\det (AB) = \det A \cdot \det B$ // TODO: proof

• If $A$ invertible, then $\det A^{-1} = \frac{1}{\det A}$

Orthogonal Matrix: a square matrix such that columns (and rows) are orthogonal and unit length.

• Example: identity, rotation, reflection, permutation matrix.

For $Q \in \mathbb{R}^{n \times n}$, the followings are equivalent

• $A \text{ is orthogonal}$

• Transpose Identity: $Q Q^T = I$

• Transformation preserve length of vector: $(\forall x \in \mathbb{R}^n)(\|Qx\| = \|x\|)$

• Transformation preserve relative angle of vector: $(\forall x, y \in \mathbb{R}^n)(Qx \cdot Qy = x \cdot y)$

Determinant rank: $\text{det-rank}(A) = \max(\{k | B^{k \times k} = \text{submatrix}(A) \land \det B \neq 0\})$

• Theorem: for any $A \in \mathbb{R}^{n \times n}$, $\text{row-rank}(A) = \text{column-rank}(A) = \text{determinant-rank}(A)$

• // TODO: proof

Cramer's rule: $\{\frac{\det A_i(b)}{\det A} | 1 \leq i \leq n\}$ is the unique solution to any system $Ax = b$ if $A$ is invertible // TODO: proof

• Notation: $A_i(b)$ means matrix obtained by replacing $i$-th column of matrix $A$ with column vector $b$

• $A = (a_{ij})_{1 \leq i, j \leq n}$

• $(i, j)$-cofactor of $A$ is $C_{ij} = (-1)^{i+j} \det A_{ij}$

• Adjoint Matrix: \begin{align*} \text{adj} A = &(C_{ij})_{1 \leq i, j \leq n}^T\\ = &(C_{ji})_{1 \leq i, j \leq n}\\ = &\begin{pmatrix} C_{11} & C_{21} & ... & C_{n1}\\ C_{12} & C_{22} & ... & C_{n2}\\ ... & ... & ... & ...\\ C_{1n} & C_{2n} & ... & C_{nn}\\ \end{pmatrix} \end{align*}

• Theroem: If $A \in \mathbb{n \times n}$ is invertible, then $A^{-1} = \frac{1}{\det A} \text{adj} A$ // TODO: proof

Geometric Volume:

• In 2D: area of parallelogram spanned by 2 vectors

• In 3D: volume of parallelotope spanned by 3 vectors

• In nD: hyper-volume spanned by $v_1, ..., v_n$ is $\left| \det \begin{pmatrix} v_1\\ ...\\ v_n\\ \end{pmatrix}\right|$

Geometric Colinear

• 2D colinear test: $(a_1, a_2), (b_1, b_2), (c_1, c_2)$ are collinear iff $\begin{vmatrix} a_1 & a_2 & 1\\ b_1 & b_2 & 1\\ c_1 & c_2 & 1\\ \end{vmatrix} = 0$

• 3D colinear test: $(a_1, a_2, a_3), (b_1, b_2, b_3), (c_1, c_2, c_3), (d_1, d_2, d_3)$ are in the same plane (coplanar) iff $\begin{vmatrix} a_1 & a_2 & a_3 & 1\\ b_1 & b_2 & b_3 & 1\\ c_1 & c_2 & c_3 & 1\\ d_1 & d_2 & d_3 & 1\\ \end{vmatrix} = 0$

• 2D line: line from $(b_1, b_2), (c_1, c_2)$ can be expressed as $\begin{vmatrix} x & y & 1\\ b_1 & b_2 & 1\\ c_1 & c_2 & 1\\ \end{vmatrix} = 0$

• 3D line: line from $(b_1, b_2, b_3), (c_1, c_2, c_3), (c_1, c_2, d_3)$ can be expressed as $\begin{vmatrix} x & y & z & 1\\ b_1 & b_2 & b_3 & 1\\ c_1 & c_2 & c_3 & 1\\ d_1 & d_2 & d_3 & 1\\ \end{vmatrix} = 0$

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