1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $f_j(t)$. $\endgroup$ – Mcnutt Ryan Librix May 28 at 5:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.