# 1.26 determinants

$$\def\Real{\mathbb{R}}\def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}}\def\Field{\mathbb{F}}\def\Fun{\mathbf{Fun}}\def\e{\mathbf{e}} \def\f{\mathbf{f}}\def\bv{\mathbf{v}}\def\i{\mathbf{i}} \def\eye{\left(\begin{array}{cc}1&0\\0&1\end{array}\right)} \def\bra#1{\langle #1|}\def\ket#1{|#1\rangle}\def\j{\mathbf{j}}\def\dim{\mathrm{dim}} \def\ker{\mathbf{ker}}\def\im{\mathbf{im}}$$

1. Determinant is a function of square matrices (can be defined for any operator $$A:U\to U$$, but we’ll avoid this abstract detour). It can be defined as a function that has the following properties:

1. It is linear in columns (i.e. if in a matrix $$A$$ a column (say, $$k$$-th) is $$\lambda_1 e_1 +\lambda_2 e_2$$, then
$f(A)=\lambda_1f(A_1)+\lambda_2f(A_2),$
where $$A_i$$ obtained by replacing $$k$$-th column with $$e_i, i=1,2$$.
2. It is zero if two columns are the same, and
3. $$f(E)=1$$.

2. It is quite easy to see that these properties have the following important implication:
$\det(AB)=\det(A)\det(B).$

In particular, this means that if $$A$$ is invertible $$\Leftrightarrow$$ determinant is non-zero!

Conversely, if $$A$$ is not invertible, there exists a vector $$x\neq 0:Ax=0$$, and hence one of the columns can be represented as a linear combination of others. This, by the properties of the determinants, immediately implies that

$A \mathrm{\ is\ invertible\ }\Leftrightarrow \det A\neq 0. A \mathrm{\ is\ invertible\ }\Leftrightarrow \det A\neq 0.$

3. One can also derive the Kramer rule:
If the vector $$x$$ with coordinates $$x_1,\ldots, x_n$$ solves the system of linear equations
$$Ax=b$$ (with square matrix $$A$$), that is if

\begin{array}{cccc}
a_{1,1}x_1+&\ldots&+a_{1,n}x_n&=b_1\\
\vdots&&\vdots&\vdots\\
a_{n,1}x_1+&\ldots&+a_{n,n}x_n&=b_n\\
\end{array}

then for any $$k=1,\ldots,n$$,
$x_k\det(A)=\det(a_1,\ldots,a_{k-1},b,a_{k+1},\ldots,a_n),$
where $$(a_1,\ldots,a_{k-1},b,a_{k+1},\ldots,a_n)$$ is the matrix $$A$$ with $$k$$-th column $$a_k$$ replaced by the column-vector $$b$$.

4. This is an awfully useful result. In particular, it implies a formula for the inverse matrix: if one denote by $$M_{{k}{l}}$$ minor, the determinant of the matrix obtained from $$A$$ by deleting its $$k$$-th row and $$l$$-th column (called $$k,l$$-th minor), then the matrix with coefficients
$B_{kl}=(-1)^{k+l}M_{lk} \mathrm{(notice\ switched\ indices!)}$
satisfies
$AB=BA=\det(A)E.$

In other words, the inverse matrix to $$A$$ is a polynomial of its coefficients, divided by $$\det(A)$$…

5. One can also obtain
$\det(A)=\sum_\sigma (-1)^{s(\sigma)}a_{1\sigma_1}\cdot\ldots\cdot a_{n\sigma_n}.$

From here we can obtain that the determinant of block-triangular matrix is
$\det\left( \begin{array}{cc} A&B\\ 0&D\\ \end{array} \right)=\det(A)\det(D)$
for square-sized $$A,D$$.

6. There is an important way to reduce large determinants to smaller ones: it’s called Schur complement:

$\mathrm{If\ }\det(A)\neq 0, \det\left( \begin{array}{cc} A&B\\ C&D\\ \end{array} \right)=\det(A)\det(D-CA^{-1}B).$
(Notice that all the matrices make sense!)

7. One more general result (Binet-Cauchy formula) about determinants:

Consider two collections of functions, $$f_i,g_i, i=1,\ldots,n$$, and the matrix whose coefficients are integrals of products of these functions:
$A_{ij}=\int_a^b f_i(x)g_j(x)dx.$
Then
$\det(A)=\int\ldots\int_{a<x_1<x_2\ldots<x_n} \det(F(x_1,\ldots,x_n))\det(G(x_1,\ldots,x_n))dx_1\ldots dx_n.$
(Here \F((x_1,\ldots,x_n)\) is the matrix with the entries
$F_{ij}=f_i(x_j),$
and similarly for $$G$$.

8. There are way too many interesting and useful determinantal formulae… We’ll cover just a few.

A great compendium of results and methods can be found in this survey by Krattenthaler.

Vandermonde: well-known. It appears when one solves the Lagrange interpolation problem:
To find a polynomial $$p=a_0x^{n}+a_1x^{n-1}+\ldots+a_n$$ of degree $$n$$ which takes given values $$y_0, y_1,\ldots,y_n$$ at given points, $$x_0, x_1,\ldots,x_n$$.

The result is easy to obtain in different ways (it is
$\sum_k y_k\frac{(x-x_1)\ldots(x-x_{k-1})(x-x_{k+1}\ldots(x-x_n)}{(x_k-x_1)\ldots(x_k-x_{k-1})(x_k-x_{k+1}\ldots(x_k-x_n)},$
as one can verify easily), and this also gives many interesting identities…

Very useful is also the Cauchy determinant, for $$A_{kl}=\frac{1}{x_k+y_l}$$:\
$\det A=\frac{\prod_{k<l}(x_{k}-x_{l})(y_{k}-y_{l})}{\prod_{k,l}(x_k+y_l)}.$

Jacobi matrices appears in many problems:

$J_k=\left( \begin{array}{ccccc} a&b&0&\ldots&0\\ c&a&b&\ldots&0\\ 0&c&a&\ldots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\ldots&a\\ \end{array} \right)$
Their determinants satisfy the recursion
$j_{k+1}=aj_k-bc j_{k-1}.$

Exercises:

• Find the determinants of Jacobi matrices with $$a=1, bc=-1$$ and $$a=-1,bc=-1$$.
• Find
$\left| \begin{array}{ccccc} 1&1&1&\ldots&1\\ x_1&x_2&x_3&\ldots&x_n\\ x_1^2&x_2^2&x_3^2&\ldots&x_n^2\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ x_1^{n-2}&x_2^{n-2}&x_3^{n-2}&\ldots&x_n^{n-2}\\ x_1^n&x_2^n&x_3^n&\ldots&x_n^n\\ \end{array} \right|$