Let the $m\times (n+1)$ rectangular Vandermonde matrix be $V$. More specifically, the matrix $V$ has the following form.

$V=\begin{pmatrix} 1 & a_1 & \cdots & a_1^{n} \\ 1 & a_2 & \cdots & a_2^{n}\\ \vdots& \vdots & \ddots &\vdots \\ 1 & a_m & \cdots & a_m^{n} \end{pmatrix}$ , where $m \geq n$.

I want a proof that $ \operatorname{rank}(V)=n$, if and only if the $a_i$ take exactly $n$ different values. Can you recommend me any paper or book that has a formal proof?

  • $\begingroup$ Thanks for the edit, but the matrix that I'm concerned about is $m\times (n+1)$, not $m\times n$. The last term is $a_i^n$, and I want a conclusion that $rank(V)=n$. $\endgroup$ – user568810 Jun 24 '18 at 8:11
  • $\begingroup$ I re-edited the question. Please let me know if my re-edition is wrong. $\endgroup$ – user568810 Jun 24 '18 at 8:13
  • $\begingroup$ I doubt that this is what you want. Can you clarify what "$n$ of $a_i$ distinct" means precisely for you? $\endgroup$ – Phira Jun 24 '18 at 8:16
  • $\begingroup$ Alternatively, you could write down the interpretation of your question fir $n=1$. $\endgroup$ – Phira Jun 24 '18 at 8:18
  • $\begingroup$ Sure. There are $n+1$ columns in $V$, including ones in the first column too. And there are $m$ elements of $a_i$, $a_1, a_2, \cdots a_m$. What I meant by "$n$ of $a_i$ distinct" is that the number of different element of $a_i$, $i=1,2,...,m$ are $n$. $\endgroup$ – user568810 Jun 24 '18 at 8:18

We will prove: If there are at least $k$ distinct values of the $a_i$ the rank is at least $k$.

Applying this for $k=n$ and $k=n+1$ gives the desired characterization of rank $n$.

Proof of the claim:

If there are less than $k$ distinct values there are less than $k$ distinct rows and the rank is clearly less than $k$.

If there are at least $k$ distinct values we can take the submatrix consisting of $k$ rows corresponding to distinct values and the first $k$ columns.

Its determinant is $$\prod_{1\le r < s \le k}(a_{i_r}-a_{i_s}) $$ which is nonzero so the rank is at least $k$.

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  • $\begingroup$ Thanks for the answer. I have a question: If $V$ is a rectangular matrix, then the determinant can be defined? $\endgroup$ – user568810 Jun 24 '18 at 8:16
  • $\begingroup$ No. It cannot be defined. $\endgroup$ – Phira Jun 24 '18 at 8:18
  • $\begingroup$ I have another question: The maximum rank that $V$ can take is $n+1$ if $(n+1)\leq m$. In this case, how can you still reach the conclusion that the rank is $n$? $\endgroup$ – user568810 Jun 24 '18 at 8:27
  • $\begingroup$ @user568810 In case you have not noticed: I edited my answer. $\endgroup$ – Phira Jun 24 '18 at 9:15
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    $\begingroup$ @user568810 Being identical is the strongest form of linear independence there is. You can eliminate all identical rows except one by forming the difference to an identical row, so the rank is clearly at most the number of distinct values of $a_i$. $\endgroup$ – Phira Jun 24 '18 at 16:59

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