Matrix

Determinant

The determinant is a function mapping matrices to real scalars. It characterizes some properties of the matrix and the linear map represented by the matrix.

What actually is the determinant?

I’ve learned it so many times, but I’ve only really learned to apply formulas. Never got a fundamental understanding of how it works.

2025-01-22: hopefully getting the intuition by reading the deep learning textbook.

Thinking about determinants

The absolute value of the determinant can be thought of as a measure of how much a multiplication by the matrix expands or contracts space.

determinant = 0 <→ singular matrix <→ linearly dependent columns or rows

Matrix Inverse

The inverse is of $A$ is equal to $A^{-1}=\frac{1}{\det A}adjA$ Thus, $A$ is invertible if and only if $\det A\ne 0$

For $2\times 2$ Matrix

We can just apply a simple formula for determinants and adjugates. Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\in M_{2\times 2}(\mathbb{R})$ The determinant of $A$ is $\det A=ad-bc$ and the adjugate of $A$ is $adjA=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

See below for the more generalized definitions.

For $n\times n$ matrix

Cofactor

We need to use the Cofactor expansion to calculate the determinant. Let’s first define cofactor.

Let $A\in M_{n\times n}(\mathbb{R})$ and let $A(i,j)$ be the $(n-1)\times (n-1)$ matrix obtained from $A$ by deleting the ith row and jth column of $A$. The (i, j)-cofactor of $A$, denoted by $Cij$ , is $C_{ij}=(-1{)}^{i+j}\det A(i,j)$ so the cofactor is just a number.

Determinant

Let $A\in M_{n\times n}(\mathbb{R})$. For any $i=1,\dots ,n$, we define the determinant of $A$ as the cofactor expansion of $A$ along ANY row or column $i$ of $A$. $\det A=a_{i1}{C}_{i1}+a_{i2}{C}_{i2}+\cdots +a_{in}{C}_{in}$

Cofactor Matrix and Adjugate

The cofactor matrix of $A$ is $cofA=[C_{ij}]\in M_{n\times n}(\mathbb{R})$

The adjugate of $A$ is $adjA=[C_{ij}{]}^T\in M_{n\times n}(\mathbb{R})$

Example

1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}$$ We compute the determinant: $$\det(A) = 1(1 \cdot 0 - 4 \cdot 6) - 2(0 \cdot 0 - 4 \cdot 5) + 3(0 \cdot 6 - 1 \cdot 5) = 1$$ Compute the cofactor matrix of $A$: $$\text{Cof}(A) = \begin{bmatrix} ei - fh & -(di - fg) & dh - eg \\ -(bi - ch) & ai - cg & -(ah - bg) \\ bf - ce & -(af - cd) & ae - bd \end{bmatrix}$$ So we have that $$\text{Cof}(A) = \begin{bmatrix} (1 \cdot 0 - 4 \cdot 6) & -(0 \cdot 0 - 4 \cdot 5) & (0 \cdot 6 - 1 \cdot 5) \\ -(2 \cdot 0 - 3 \cdot 6) & (1 \cdot 0 - 3 \cdot 5) & -(1 \cdot 6 - 2 \cdot 5) \\ (2 \cdot 4 - 3 \cdot 1) & -(1 \cdot 4 - 3 \cdot 0) & (1 \cdot 1 - 2 \cdot 0) \end{bmatrix} = \begin{bmatrix} -24 & 20 & -5 \\ 18 & -15 & 2 \\ -4 & 5 & -1 \end{bmatrix}$$ Step 3: Transpose the cofactor matrix to get $\text{adj}(A)$

\text{adj}(A) = \text{cof}(A)^T = \begin{bmatrix} -24 & 18 & -4 \ 20 & -15 & 5 \ -5 & 2 & -1 \end{bmatrix}$$ Therefore,

-24 & 18 & -4 \\ 20 & -15 & 5 \\ -5 & 2 & -1 \end{bmatrix}$$ ### Properties of determinants Let $A,B \in M_{n\times n}(\mathbb{R})$ and $k \in \mathbb{R}$. $$\det(kA) = k^n \det A$$ $$\det(A^k) = (\det A)^k$$ $$\det(AB) = (\det A)(\det B) = (\det B)(\det A) = \det(BA)$$ $$\det(A^T) = \det(A)$$ > [!danger] Danger > > > $$\det(A+B) \neq \det(A) + \det(B)$$ ### Determinants and Area Somehow the determinant is related to the area? Also volume?? ![[attachments/Pasted image 20211124004027.png]] ### Determinants and Volume Very similar derivation as area. ![[attachments/Pasted image 20211124004000.png]]