The Determinant

Transformations tend to stretch or shrink (squish) space. So how do we measure by how much it shrinks? We can measure the factor by which a given area increases or decreases.

Notation
Let $\mu_{ij}$ be the $(n-1)\times(n-1)$ submatrix of $A$ obtained by removing the $i$th row and $j$th column of $A$.

Determinant
The determinant is the factor by which a given transformation changes any area. The determinant is $0$ if the transform transforms space into a lower dimension (for example, $\mathbb{R}^3$ is transformed into a plane, or even a line). Well... ish. Sorry. The determinant can be negative! This is when space is "inverted".

Ex. $$\det\left({\begin{bmatrix}3 & 0\\0 & 2\end{bmatrix}}\right)=6.$$ Do you see why? We stretch the $y$-axis by $3$, and the $x$-axis by $2$, so the area of any shape is stretched by $6$.

The determinant ($\det(A)=|A|$) is more formally defined recursively:

1. If $A$ is the $1\times 1$ matrix $A=[a_{11}]$, then $\det(A)=a_{11}$.
2. If $A$ is an $n\times n$ matrix (where $n > 1$), then $$\det(A)=\sum_{j=1}^n (-1)^{j + 1}a_{1j} \det(\mu_{1j}(A)).$$

For a matrix $A=[a_{ij}]$, the scalar quantity $\det(\mu_{ij}(A))$ is called a minor of $A$, and $(-1)^{j+i}\det(\mu_{ij}(A))$ is called a cofactor. To compute the determinant of $A$, we can

1. Compute the cofactor of each element in the first row.
2. Multiply each element in the first row by its cofactor and sum the results.

We can do this for any row or column, not just the first one. This obviously means that if a matrix has a zero row, the determinant is zero. Similarly, for an upper triangular matrix, the determinant is the product of the diagonal elements. The same is true for diagonal and lower triangular matrices.

Properties

• $(u_{ij}(A))^T=u_{ji}(A^T)$. Therefore, for any square matrix $A$, $\det(A)=\det(A^T)$.
• If $A$ has two identical rows or columns, $\det(A)=0$. That is, if $r(A) < n$, $\det(A)=0$.
• If $E$ is the elementary matrix corresponding to interchanging two rows of $A$, then $\det(E)=-1$ and $\det(EA)=-\det(A)=\det(E)\det(A)$.
• If $E$ is the elementary row matrix that corresponds to multiplying a row of $A$ by a scalar $\lambda$, then $\det(E)=\lambda$ and $\det(EA)=\lambda\det(A)$.
• If $E$ is the elementary row matrix corresponding to adding $\lambda$ times row $i$ of $A$ to row $j$, then $\det(E)=1$ and $\det(EA)=\det(A)$.
• For any two $n\times n$ matrices $A$ and $B$, $\det(AB)=\det(A)\det(B)$.
  • From this theorem, it follows that if $A$ is an invertible matrix, $\det(A^{-1})=(\det(A))^{-1}$.
  • Also, similar matrices have the same determinants (proven by taking determinants of $A=P^{-1}BP$).
    • Note that this means if $\mathbb{V}$ is a finite-dimensional vector space and $T:\mathbb{V}\to\mathbb{V}$ is linear, then $\det(T)$, defined as the determinant of any matrix representation of $T$, is a well-defined scalar, independent of the choice of basis.

A square matrix $A$ is nonsingular if and only if its determinant is nonzero. There's a proof, but think of it this way: a zero determinant decreases the dimension of space (and therefore the kernel is nontrivial). That means that the transformation isn't invertible. The proof (vaguely) relies on the facts that no elementary row operation has a zero determinant, and therefore if $A$ is singular, it can be transformed to an upper triangular matrix $B$ with at least one zero on its diagonal (meaning it has a zero determinant), so the determinant of $A$ is $0$.