In linear algebra, an alternant matrix is a matrix formed by applying a finite list of functions pointwise to a fixed column of inputs. An alternant determinant is the determinant of a square alternant matrix.

Generally, if ${\displaystyle f_{1},f_{2},\dots ,f_{n))$ are functions from a set ${\displaystyle X}$ to a field ${\displaystyle F}$, and ${\displaystyle {\alpha _{1},\alpha _{2},\ldots ,\alpha _{m))\in X}$, then the alternant matrix has size ${\displaystyle m\times n}$ and is defined by

${\displaystyle M={\begin{bmatrix}f_{1}(\alpha _{1})&f_{2}(\alpha _{1})&\cdots &f_{n}(\alpha _{1})\\f_{1}(\alpha _{2})&f_{2}(\alpha _{2})&\cdots &f_{n}(\alpha _{2})\\f_{1}(\alpha _{3})&f_{2}(\alpha _{3})&\cdots &f_{n}(\alpha _{3})\\\vdots &\vdots &\ddots &\vdots \\f_{1}(\alpha _{m})&f_{2}(\alpha _{m})&\cdots &f_{n}(\alpha _{m})\\\end{bmatrix))}$

or, more compactly, ${\displaystyle M_{ij}=f_{j}(\alpha _{i})}$. (Some authors use the transpose of the above matrix.) Examples of alternant matrices include Vandermonde matrices, for which ${\displaystyle f_{j}(\alpha )=\alpha ^{j-1))$, and Moore matrices, for which ${\displaystyle f_{j}(\alpha )=\alpha ^{q^{j-1))}$.

## Properties

• The alternant can be used to check the linear independence of the functions ${\displaystyle f_{1},f_{2},\dots ,f_{n))$ in function space. For example, let ${\displaystyle f_{1}(x)=\sin(x)}$, ${\displaystyle f_{2}(x)=\cos(x)}$ and choose ${\displaystyle \alpha _{1}=0,\alpha _{2}=\pi /2}$. Then the alternant is the matrix ${\displaystyle \left[{\begin{smallmatrix}0&1\\1&0\end{smallmatrix))\right]}$ and the alternant determinant is ${\displaystyle -1\neq 0}$. Therefore M is invertible and the vectors ${\displaystyle \{\sin(x),\cos(x)\))$ form a basis for their spanning set: in particular, ${\displaystyle \sin(x)}$ and ${\displaystyle \cos(x)}$ are linearly independent.
• Linear dependence of the columns of an alternant does not imply that the functions are linearly dependent in function space. For example, let ${\displaystyle f_{1}(x)=\sin(x)}$, ${\displaystyle f_{2}=\cos(x)}$ and choose ${\displaystyle \alpha _{1}=0,\alpha _{2}=\pi }$. Then the alternant is ${\displaystyle \left[{\begin{smallmatrix}0&1\\0&-1\end{smallmatrix))\right]}$ and the alternant determinant is 0, but we have already seen that ${\displaystyle \sin(x)}$ and ${\displaystyle \cos(x)}$ are linearly independent.
• Despite this, the alternant can be used to find a linear dependence if it is already known that one exists. For example, we know from the theory of partial fractions that there are real numbers A and B for which ${\displaystyle {\frac {A}{x+1))+{\frac {B}{x+2))={\frac {1}{(x+1)(x+2)))}$. Choosing ${\displaystyle f_{1}(x)={\frac {1}{x+1))}$, ${\displaystyle f_{2}(x)={\frac {1}{x+2))}$, ${\displaystyle f_{3}(x)={\frac {1}{(x+1)(x+2)))}$ and ${\displaystyle (\alpha _{1},\alpha _{2},\alpha _{3})=(1,2,3)}$, we obtain the alternant ${\displaystyle {\begin{bmatrix}1/2&1/3&1/6\\1/3&1/4&1/12\\1/4&1/5&1/20\end{bmatrix))\sim {\begin{bmatrix}1&0&1\\0&1&-1\\0&0&0\end{bmatrix))}$. Therefore, ${\displaystyle (1,-1,-1)}$ is in the nullspace of the matrix: that is, ${\displaystyle f_{1}-f_{2}-f_{3}=0}$. Moving ${\displaystyle f_{3))$ to the other side of the equation gives the partial fraction decomposition ${\displaystyle A=1,B=-1}$.
• If ${\displaystyle n=m}$ and ${\displaystyle \alpha _{i}=\alpha _{j))$ for any ${\displaystyle i\neq j}$, then the alternant determinant is zero (as a row is repeated).
• If ${\displaystyle n=m}$ and the functions ${\displaystyle f_{j}(x)}$ are all polynomials, then ${\displaystyle (\alpha _{j}-\alpha _{i})}$ divides the alternant determinant for all ${\displaystyle 1\leq i. In particular, if V is a Vandermonde matrix, then ${\textstyle \prod _{i divides such polynomial alternant determinants. The ratio ${\textstyle {\frac {\det M}{\det V))}$ is therefore a polynomial in ${\displaystyle \alpha _{1},\ldots ,\alpha _{m))$ called the bialternant. The Schur polynomial ${\displaystyle s_{(\lambda _{1},\dots ,\lambda _{n})))$ is classically defined as the bialternant of the polynomials ${\displaystyle f_{j}(x)=x^{\lambda _{j))}$.