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For a $n \times n$ Vandermonde matrix $$V:=\begin{bmatrix}1 & c_1 & c_1^2 & \cdots & c_1^{n-1} \\ 1 & c_2 & c_2^2 & \cdots & c_2^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & c_n & c_n^2 & \cdots & c_n^{n-1}\end{bmatrix}$$ we know that it is nonsingular if and only if $c_i \ne c_j$ for $i\ne j$.

I am curious if this property can be generalized for non-integer degrees. Suppose I am given a $n \times n$ matrix $$W:=\begin{bmatrix}c_1^{d_1} & c_1^{d_2} & c_1^{d_3} & \cdots & c_1^{d_n} \\ c_2^{d_1} & c_2^{d_2} & c_2^{d_3} & \cdots & c_2^{d_n} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ c_n^{d_1} & c_n^{d_2} & c_n^{d_3} & \cdots & c_n^{d_n}\end{bmatrix}.$$ Is it true that $W$ is nonsingular if and only if $c_i \ne c_j$ for $i\ne j$? If it matters, then I consider complex values for $c_i$ and real values for $d_i$.

If (as I guess) it is something well-known, could you, please, give me a reference?

Update: Obviously, I assume that $c_i \ne 0$ for all $i$. Let us also assume that $d_i \ne d_j$ for $i \ne j$.

Update 2: I have tried the following condition: for all $d_k\ne 0$ we have $c_i^{d_k} \ne c_j^{d_k}$ for $i \ne j$. It does not work. Actually, for $n=2$ the condition is $c_1^{d_2-d_1} \ne c_2^{d_2-d_1}$. This is satisfied, particularly, for $|c_1| \ne |c_2|$.

Update 3: I have the following intuition. Let $\Delta_{ij}:=d_i-d_j$. The hypothesis: if for all $i\ne j$ we have $c_k^{\Delta_{ij}} \ne c_l^{\Delta_{ij}}$ for $k\ne l$, then $\det{W} \ne 0$. For integer $d$ we have exactly the Vandermond condition.

Arastas
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2 Answers2

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I think I have a partial solution: suppose that all the $c_k$ are distincts and $>0$, and that the $d_k$ are distinct and real. Then the generalized determinant is not $0$. I leave to you the case $n=2$, then we proceed by induction.

The key point is the following: if we have an expression of the form

$$f(x)=\sum_{k=1}^n u_k\exp(v_kx)$$ with all the $v_k$ distincts, and if $f$ has $n$ distinct zeros $w_1<..<w_n$, then $f=0$ (and all the coefficients $u_k$ are zero). To see why, the case $n=1$ is trivial, and proceed by induction, multiplying $f$ by $\exp(-v_1x)$, and using that the derivative of $f$ has a zero in $]w_k, w_{k+1}[$ for $1\leq k\leq n-1$.

Now we return to the determinant; We write the terms on the last line in the form $\exp(d_k\log c_n)$, we replace $\log c_n$ by $x$, and developping with respect to the last line, we get an expression of the form $$W(x)=\sum_{k=1}^n b_k\exp(d_kx)$$

Note that $b_n$ is the determinant constructed with the $c_k$ and the $d_k$ with $k\leq n-1$. By the induction hypothesis, we have $b_n\not =0$.

Now obviously, $W(\log c_k)=0$ for $k=1,\cdots,n-1$. Suppose in addition that $W(\log c_n)=0$. This a contradiction with the property above, and we are done.

Kelenner
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  • Thanks. But it seems that some extra conditions are missing. Note that Stephen has presented a negative example for $n=2​$. Indeed, if I chose $u_1=1$, $v_1=z$, $u_2=-a^z$ and $v_2=0$, then I have $f(x)=\exp(z,x)-a^z$. The first root is $x=\log(a)​$, but for $x=\log(b)$ we have $b^z-a^z$ that has many solutions for complex $a$ and $b$. E.g. choose $a=\exp(\frac{\mathrm{i}\pi}{3})​$, $b=\exp(\frac{\mathrm{i}\pi}{6})​$ and $z=12​$. – Arastas Oct 11 '17 at 22:27
  • It seems that the proof of the property of $f(x)$ works only for real-valued functions of reals? – Arastas Oct 11 '17 at 22:46
  • @Arasras Yes, i think that my hypothesis that the $c_k$ are real positive and the $d_k$ real is necessary for the proof to work. Simply think of $f(x)=\exp(ix)+\exp(-ix)-2$, $f(x)=0$ has an infinity of distincts solutions. – Kelenner Oct 12 '17 at 15:59
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Already in case $n=2$ the answer is no: for $l>1$

$$\mathrm{det}\left(\begin{matrix} 1 & a^l \\ 1 & b^l \end{matrix} \right)=b^l-a^{l}$$ is zero whenever $b/a$ is an $l$th root of $1$.

There is no reason to expect the answer to be yes for any class of examples very much larger than the Vandermonde determinants. For instance, already for examples coming from monomial reflection groups (the Vandermonde corresponding to the symmetric group) the answer is no.

Stephen
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  • I guess that in my question we have $k=m$ and $l=n$, thus $p=q$ and $\det$ is nonzero for $b \ne a$? – Arastas Oct 11 '17 at 16:05
  • Ah, ok I misunderstood. I edited the post. – Stephen Oct 11 '17 at 16:06
  • @Arastas ...without assuming $a$ and $b$ are real and $l$ is odd I don't think the result as you claimed is true even for $n=2$. – Stephen Oct 11 '17 at 16:13
  • Thanks, it is very useful. Well, I have to work with complex-valued elements $c_i$. Is it true that the regular Vandermond matrix property holds for complex elements? I.e. for $d_i=i-1$ $\det V \ne 0$ for $c_i \ne c_j$ for $i\ne j$? – Arastas Oct 11 '17 at 16:19
  • @Arastas The determinant of the usual Vandermonde matrix is the product of all the differences $c_i-c_j$ for pairs $i$ strictly less than $j$, so yes. – Stephen Oct 12 '17 at 13:18