determinant

Notation: $det(A)=|A|$
$\hat{i},\hat{j}$ denote the unit vectors of the $x,y$ plane: ${\left[\begin{array}{cc}1& 0\end{array}\right]}^T,{\left[\begin{array}{cc}0& 1\end{array}\right]}^T$, they form the unit square with area $\hat{i}\hat{j}=1\cdot 1=1$.

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Properties of the determinant.

Properties 4-10 are derived from the first three.
1)$\det (I)=1$
2) Exchanging two rows or columns changes the sign of the determinant
3a) $\left|\begin{array}{cc}ta& tb\\ c& d\end{array}\right|=t\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$ … multiplying a row by a scalar is the same as multiplying the entire determinant by that scalar.
3b) $\left|\begin{array}{cc}a+a^{\prime }& b+b^{\prime }\\ c& d\end{array}\right|=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|+\left|\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c& d\end{array}\right|$ … determinants are linear: adding elements in a row is the same as adding the determinants. Note: $\det (A+B)\ne \det (A)+\det (B)$)
4) Two equal rows → $\det =0$
5) Subtracting a multiple of one row from another does not change the determinant.
6) Row of zeros → $\det (A)=0$
7) $\text{Let }U=\left|\begin{array}{cccc}{d}_1& \ast & \ast & \ast \\ 0& d_2& \ast & \ast \\ 0& 0& \ddots & \ast \\ 0& 0& 0& d_n\end{array}\right|=(d_1,d_2,\dots ,d_n)$ … the determinant of a triangular matrix is the product of its pivots.
8) $\det (A)=0$ when $A$ is singular (not invertible).
9) $\det (AB)=\det (A)\det (B)$
10) $\det (A^T)=\det (A)$

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From 1) and 2): The determinant of every permutation matrix (permutation of the identity matrix) is either $1$ or $-1$, depending on whether the number of permutations is even or odd.
From rule 3a) we can see how the determinant behaves like volume: $\det (2A)=2^n\det (A)$.
We can easily prove 4): Exchange two equal rows → sign changes, but the matrix is still the same, so the determinant must be zero (and confirmed by the fact that the rank of the matrix is less than $n$).
Proof for 5): $\left|\begin{array}{cc}a& b\\ c-\lambda a& d-\lambda b\end{array}\right|\stackrel{(3)}{=}\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|+\left|\begin{array}{cc}a& b\\ -\lambda a& -\lambda b\end{array}\right|=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|-\lambda \left|\begin{array}{cc}a& b\\ a& b\end{array}\right|=0$
Proof for 6): $\left|\begin{array}{cc}0& 0\\ c& d\end{array}\right|=0\cdot \left|\begin{array}{cc}0& 0\\ c& d\end{array}\right|=0$
Proof for 7): We can simply eliminate the stars by multiplying the rows with the corresponding $d_i$. This leavues us with a diagonal matrix. By rule 3a) we can pull out the $d_i$ from the determinant, leaving us with the product of the pivots times the $\det (I)$, which is $1$.
Proof for 8): The determinant is zero if one of the $d_i$ is zero, which means we can construct a row of zeros → not linearly independent.
→ So the algorithm for finding the determinant, when $A$ is invertible is $A\to U\to D\to d_1{d}_2\dots d_n$.
For a $2\times 2$ matrix: $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|\to \left|\begin{array}{cc}a& b\\ 0& d-\frac{c}{a}b\end{array}\right|=ad-bc$
How can we calculate $\det (A^{-1})$? With the help of 9), easy to see when thinking about diagonal matrices: $\det (A^{-1}A)=\det (I)⟹\det (A^{-1})=\frac{1}{\det (A)}$
$⟹\det (A^n)=\det (A{)}^n$
$⟹\det (2A)=2^n\det (A),A\in {\mathbb{R}}^{n\times n}$

There is nothing special about “row1”, because we can exchange rows (rule 2) and there is nothing special about rows in general, because we can exchange rows for columns (rule 10).
The proof for 10) is also about bringing the matrices into triangular forms.

Intuition

The absolute value of the determinant is the Area/Volume/… in ${\mathbb{R}}^2/{\mathbb{R}}^3$/… etc.

It is easiest to grasp the changes of a matrix multiplied with another matrix / set of vectors does, by visualizing how it transforms the unit vectors of of the vectorspace.

The matrix $A=\left[\begin{array}{ccc}3& & 0\\ 0& & 2\end{array}\right]$ multiplies $\hat{i}$ by 2 and $\hat{j}$ by 3, scaling the area of the unit square by $\det (A)=6$:

A shear matrix like $A=\left[\begin{array}{ccc}1& & 1\\ 0& & 1\end{array}\right]$ tilts the coordinate system and the unit square into a paralellogram.
For this particular $A$, the Area of the resulting parallelogram will stay the same
($P_x\cdot P_h=1\cdot 1=1$):

The determinant tells us how every area on the entire space changes in size.
The elements of the transformation matrix tell us about the shape. As we are dealing with linear transformations, the grid lines always remain parallel and evenly spaced.

If $\det (A\vec{v})=0$, then the transformation changes the space in a way that it collapses to a lower dimension (for the case of ${\mathbb{R}}^3$ it may be a plane, line or point: anything with 0 volume).


If the determinant is zero, the matrix is not invertible.

Negative elements mean that the positions of $\hat{i}$ and $\hat{j}$ have been switched.
$A=\left[\begin{array}{ccc}-2& & 0\\ 0& & 1\end{array}\right]$ for example puts $\hat{i}$ to the left and multiplies it by two. The resulting determinant/area will be $-2$:

Computation

For $2\times 2$ matrices:

$$det(\left[\begin{array}{ccc}a& & b\\ c& & d\end{array}\right])=ad-bc$$

Intuition:
$a$ scales $\hat{i}$, $d$ scales $\hat{j}$. Multiplying them gives the area of the bigger / smaller rectangle.
If $b$ or $c$ is $0$, we have a parallelogram, still with area $ad$. (If only $b$ or $c$ changes in general, the paralellogram gets more / less stretched out, but area stays $1$.
See: Parallelogram example

Non square matrices don’t have a determinant

Rule of Sarrus

May be used for matrices up to $3\times 3$:

$$A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$$ $$\det (A)=aei+bfg+cdh-ceg-bdi-afh$$

Bigger than 3x3 matrices

Honestly why are you not using computers?

Laplace / cofactor expansion

This is a recursive method, simple to do by hand for small matrices but inefficient for larger matrices:

$$\det (A)=\sum _{j=1}^n(-1{)}^{i+j}{a}_{i,j}{m}_{i,j}$$

Transclude of Minor-(of-a-matrix)#^definition

For example:

$$A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 5& 0\\ 7& 8& 9\end{array}\right]$$

First choose a row ($i$), preferrably one with many $a_{i,j}$ being zero. $i=2$ in this case.
Create $n$ minor matrices by crossing out row $i$ and columns $1-n$:

Transclude of Minor-(of-a-matrix)#^example

Multiply the determinant of each minor by the corresponding element of the original matrix $a_{i,j}$:

$$0\cdot \left|\begin{array}{cc}2& 3\\ 8& 9\end{array}\right|+5\cdot \left|\begin{array}{cc}1& 3\\ 7& 9\end{array}\right|+0\cdot \left|\begin{array}{cc}1& 2\\ 7& 8\end{array}\right|=-60=\det (A)$$

Now we see the benefit of choosing the row with the most zeros.

Should there be more zeros in a col rather than a row, you can simply flip $i$ for $j$ (make the chosen col static and loop over the rows).

If you have bigger than $3\times 3$ matrices, apply the process recursively, i.e. split the matrices until you get to $2\times 2$ matrices.

$(-1{)}^{i+j}\cdot m_{i,j}$ is called the cofactor.

Elimination

We can use Gaussian Elimination to obtain an upper triangular matrix. The product of its pivots (elements on the main diagonal) is the determinant.

If you interchange (swap positions) of two rows, you need to multiply the resulting determinant by $-1$.
So for an even number of swaps, the sign stays the same.

Example computations

squished rectangle / parallelogram:

$$\begin{gathered} \text{displaylines}A=\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}\right] \\ \vec{v}=\left[\begin{array}{c}2\\ 3\end{array}\right] \\ {\vec{v}}_{new}=A\vec{v}=\left[\begin{array}{c}-4\\ 3\end{array}\right]=(A\left[\begin{array}{cc}\hat{i}& \hat{j}\end{array}\right])\vec{v} \end{gathered}$$

$det(A)=1\cdot 1-(-2)\cdot 0=1$
→ Area stays the same.
→ The unit vectors of the original vector space form the identity matrix.
→ The new unit vectors equal the columns of the transformation matrix:

References

3b1b: The determinant | Chapter 6, Essence of linear algebra
18. Properties of Determinant (Gilbert)