determinant.md

tags
math

Determinant

The determinant of a matrix $M \in \mathbb{R}^{n \times n}$ is defined as:

$$ \DeclareMathOperator{\sgn}{sgn}

\det M = |M| = \sum_{\sigma \in S_n} \sgn(\sigma) \prod_{i = 1}^n a_{i\sigma(i)} $$

where:

• $\sigma$ is a permutation of $n$ elements
• $S_n$ is a permutation group -- set of all $\sigma$
• $\sgn$ is a sign of a permutation

Geometric meaning

If we view the matrix $M$ as a linear transformation, then its determinant would give us the amount of scaling the transformation does. To elaborate, imagine a hypercube in the domain (e.g. 1x1 square for 2D), then the determinant of transformation $M$ would tell us how much $M$ increases the volume of the hypercube. Since any shape can be approximated with several of these hypercubes, we can safely assume that any volume would get scaled $|M|$ times.

There is a little lie in the paragraph above. When the transformation flips axes, the determinant may become negative. So scaling is given by the absolute value of the determinant $|\det(M)|$.

The proof is not something that is intuitive, but we can build a little bit of intuition by viewing the simple cases in small dimension. Let's imagine following 2d transformations, for which the determinant is defined as:

$$ |M| = \begin{vmatrix} a & c \\ b & d \\ \end{vmatrix} = ad - bc $$

$$ M^1 = \begin{bmatrix} a & 0 \\ 0 & d \\ \end{bmatrix} $$

$M^1$ scales axis $x$ by $a$ and axis $y$ by $d$. Its determinant is $|M^1| = ad$. If we imagine 1x1 square, it would get projected to a $a\times d$ square, ergo its volume would grow $ad$ times.

We can skew the square's image by skewing either the projection of $x$ axis:

$$ M^2 = \begin{bmatrix} a & 0 \\ b & d \\ \end{bmatrix} $$

or the $y$ axis:

$$ M^3 = \begin{bmatrix} a & c \\ 0 & d \\ \end{bmatrix} $$

Either way, the determinant is still the same. This is accordance with the fact that skewing of a square doesn't change its volume.

When we skew both axes, we get a different volume. We can rewrite determinant as:

$$ \begin{aligned} \det(M) &= (a + c)(b + d) - ab - dc - 2bc\\ &= ab + ad + bc + cd - ab - dc - 2bc\\ &= ad - bc\\ \end{aligned} $$

Which corresponds to the following image:

So in essence, determinant computes the encompassing volume of the hypercube's image, minus those parts, which are really not taken up by the hypercube's image due to the projection's skewness.