kevinlui's site

5.1/5.2 - Determinants

Let $T:\mathbb{R}^n\to\mathbb{R}^n$ be a linear transform and $A$ be the square matrix defined so that $T(x)=Ax$. Then the absolute value of the determinant of $A$, denoted det($A$), measures the change in volume under $T$. This means that if $S$ is a shape of volume V then $T(S)=\{T(s):s\in \mathbb{R}^n\}$ has volume $|\det(A)|$V.

Definition: For any $n\in \mathbb{N}$, The determinant is the unique function from the set of square matrices of size $n$ so the real numbers with the following 3 properties:

• $\det(I_n)=1$,

• When viewing a square matrices as a list of $n$ column vectors, the determinant is $n$-linear. This means $\det([a_1,\ldots,ca_i+db_i,\ldots,a_n])=c\det([a_1,\ldots,a_i,\ldots,a_n])+d\det([a_1,\ldots,b_i,\ldots,a_n])$.

• If this there is a column of zero, then the determinant is zero.

The determinant of a linear transform $T:\mathbb{R}^n\to\mathbb{R}^n$ is the determinant of its associated matrix.

Properties:

• $\det(AB)=\det(A)\det(B)$,

• $\det(A^t)=\det(A)$,

• If $A$ is upper (or lower) triangular, then $\det(A)$ is the product of the diagonal,

• $\det(cA)=c^n\det(A)$,

• If $A$ has positive nullity, then $\det(A)=0$.

Computation: There is a thing called cofactor expansion. It is terrible but we will learn it. In practice, another method is used like $LU$-decomposition. There $LU$ decompositon of a matrix $A$ is $A=LU$ where $L$ is lower trianguluar and $U$ is upper triangular. Then $\det(A)=\det(L)\det(U)$, where $\det(L)$ and $\det(U)$ is the product of the diagonal. Give $n=2$ and $n=3$ shortcuts and the general cofactor formula. Note that you can expand along any row or column.

Notation: Often you denote the determinant of a matrix by replacing the square brackets by straight lines.

Examples: Compute the determinant in the following cases:

• $T(x,y)=(x+y,y)$

• $T(x,y)=(3x+y,-y)$. Also what is the determinant of $T^{-1}$ here? (demonstrate that $\det(A^{-1})=\det(A)^{-1}$ here)

• Here is a $3\times 3$ example

• Here is a $4\times 4$ example that I won't actually work out fully

Theorem: Let $S=\{a_1,\ldots,a_n\}$ be a set of vectors in $\mathbb{R}^n$, let $A=[a_1\; \ldots \; a_n]$ and $T:\mathbb{R}^n\to\mathbb{R}^n$ be given by $T(x)=Ax$. Then the following are equivalent:

• $S$ spans $\mathbb{R}^n$,

• $S$ is linearly independent,

• $S$ is a basis for $\mathbb{R}^n$,

• $Ax=b$ has a unique solution for every $b\in \mathbb{R}^n$,

• $T$ is onto,

• $T$ is one-to-one,

• $\ker T=\{0\}$,

• $range(T)=\mathbb{R}^n$,

• $col(A)=\mathbb{R}^n$,

• $row(A)=\mathbb{R}^n$,

• $rank(A)=n$,

• $nullity(A)=0$,

• Any echelon form of $A$ has no zero entries on the diagonal,

• The reduced echelon form of $A$ is the identity matrix,

• $\det(A)\neq 0$,

tldr: Determinant tells you about change of volume. It is a test of invertibility.