# What is the principle to calculate the determinant of Vandermonde matrix with $f_j(t)$?

Let $$V = \left( \begin{array}{cccc}{1} & {x_{1}} & {\cdots} & {x_{1}^{n-1}} \\ {1} & {x_{2}} & {\cdots} & {x_{2}^{n-1}} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {1} & {x_{n}} & {\cdots} & {x_{n}^{n-1}}\end{array}\right)$$ Compute $$\det(V)$$.

If we assume:

$$f_{1}(t)=1, f_{j}(t)=\prod_{i=1}^{j-1}\left(t-x_{i}\right)=\sum_{i=1}^{j} a_{i j} t^{i-1}, \quad 2 \leqslant j \leqslant n$$

Vandermonde Matrix can write into:

$$\left( \begin{array}{cccc}{1} & {x_{1}} & {\cdots} & {x_{1}^{n-1}} \\ {1} & {x_{2}} & {\cdots} & {x_{2}^{n-1}} \\ {\vdots} & {\vdots} & {\cdots} & {\vdots} \\ {1} & {x_{n}} & {\cdots} & {x_{n}^{n-1}}\end{array}\right) \left( \begin{array}{cccc}{1} & {a_{12}} & {\cdots} & {a_{1 n}} \\ {0} & {1} & {\cdots} & {a_{2 n}} \\ {\vdots} & {\ddots} & {\ddots} & {\vdots} \\ {0} & {\cdots} & {0} & {1}\end{array}\right)=\left( \begin{array}{cccc}{f_{1}\left(x_{1}\right)} & {0} & {\cdots} & {0} \\ {f_{1}\left(x_{2}\right)} & {f_{2}\left(x_{2}\right)} & {\ddots} & {\vdots} \\ {\vdots} & {\vdots} & {\ddots} & {0} \\ {f_{1}\left(x_{n}\right)} & {f_{2}\left(x_{n}\right)} & {\cdots} & {f_{n}\left(x_{n}\right)}\end{array}\right)$$

then $$det(V)det(I)=det(*)$$ because that they all $$\in\mathbb{F}^{n\times n}$$.

but I cannot understand how to manufacture the element of *, on the other word, $$f_j(t)$$.

Why can we set up a function $$f_j(t)$$? How does this function be assumed?

What is the principle behind this $$f_j(t)$$?

further more： $$C=\left( \begin{array}{cccc}{\frac{1}{s_{1}-t_{1}}} & {\frac{1}{s_{1}-t_{2}}} & {\cdots} & {\frac{1}{s_{1}-t_{n}}} \\ {\frac{1}{s_{2}-t_{1}}} & {\frac{1}{s_{2}-t_{2}}} & {\dots} & {\frac{1}{s_{2}-t_{n}}} \\ {\vdots} & {\vdots} & {\cdots} & {\vdots} \\ {\frac{1}{s_{n}-t_{1}}} & {\frac{1}{s_{n}-t_{2}}} & {\cdots} & {\frac{1}{s_{n}-t_{n}}}\end{array}\right)=\left(\frac{q_{j}\left(s_{i}\right)}{p\left(s_{i}\right)}\right)=\\\left( \begin{array}{cccc}{\frac{1}{p(s_{1})}} & { } & { } & { } \\ { } & {\frac{1}{p\left(s_{2}\right)}} \\ { } & { } & { \cdots } \\{ }& { }& { }&{\frac{1}{p\left(s_{n}\right)}}\end{array}\right)\left( \begin{array}{cccc}{1} & {s_{1}} & {\cdots} & {s_{1}^{n-1}} \\ {1} & {s_{2}} & {\cdots} & {s_{2}^{n-1}} \\ {\vdots} & {\vdots} & {\cdots} & {\vdots} \\ {1} & {s_{n}} & {\cdots} & {s_{n}^{n-1}}\end{array}\right)\left( \begin{array}{cccc}{1} & {t_{1}} & {\cdots} & {t_{1}^{n-1}} \\ {1} & {t_{2}} & {\cdots} & {t_{2}^{n-1}} \\ {\vdots} & {\vdots} & {\cdots} & {\vdots} \\ {1} & {t_{n}} & {\cdots} & {t_{n}^{n-1}}\end{array}\right)^{-1}\left( \begin{array}{cccc}{q_1(t_1)} & { } & { } & { } \\ { } & {q_2(t_2)} \\ { } & { } & { \cdots } \\{ }& { }& { }&{q_n(t_n)}\end{array}\right)$$

Assumed $$p(x)=\prod_{i=1}^{n}\left(x-t_{i}\right), q_{j}(x)=\frac{p(x)}{x-t_{j}}=\sum_{i=1}^{n} a_{i j} x^{i-1}​$$

like $$f_j(t)$$ in Vandermonde Matrix

Perhaps the following line of thinking is clearer.

Suppose you pick some numbers $$a_{ij}$$ for $$i \le j$$ and form the matrix $$U := \begin{bmatrix}a_{11} & {a_{12}} & {\cdots} & {a_{1 n}} \\ {0} & {a_{22}} & {\cdots} & {a_{2 n}} \\ {\vdots} & {\ddots} & {\ddots} & {\vdots} \\ {0} & {\cdots} & {0} & {a_{nn}}\end{bmatrix}.$$ Then you can show (by directly doing the matrix multiplication) that $$VU = \begin{bmatrix}{f_{1}\left(x_{1}\right)} & {f_2(x_1)} & {\cdots} & {f_n(x_1)} \\ {f_{1}\left(x_{2}\right)} & {f_{2}\left(x_{2}\right)} & {\cdots} & {f_n(x_2)} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {f_{1}\left(x_{n}\right)} & {f_{2}\left(x_{n}\right)} & {\cdots} & {f_{n}\left(x_{n}\right)}\end{bmatrix}$$ where $$f_j(t) := \sum_{i=1}^j a_{ij} t^{i-1}$$.

Now, what if we choose the $$a_{ij}$$ in a particular way? Specifically, choose them to satisfy $$\sum_{i=1}^j a_{ij} t^{i-1} = \prod_{i=1}^{j-1} (t-x_i)$$ for $$j \ge 2$$ and $$a_{11} = 1$$.

Is this even possible? Well, $$\prod_{i=1}^{j-1} (t-x_i)$$ is a polynomial of degree $$j-1$$, so we can choose $$a_{1j}, a_{2j}, \ldots, a_{jj}$$ to be the coefficients. No problem.

Further note that the leading coefficient (coefficient of $$t^{j-1}$$) is $$a_{jj}=1$$. So, in this case the above $$U$$ matrix has $$1$$s on the diagonal, like in your post.

Finally, note that $$f_j(x_i) = 0$$ for $$i < j$$ by our particular choice of $$a_{ij}$$, so the product matrix is lower triangular, as it appears in your post.

• Thank you, I found $f_j(t)$! but I don't understand if a square matrix can't be LU decomposed how should I know which matrix I can time on the right of the square matrix. For example, Cauchy Matrix is decomposed into the product of Vandemonde, the progress also assume some functions like $f_j(t)$. – Mcnutt Ryan Librix May 28 at 5:52