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Martin Sleziak

14:16

Q: Why are Vandermonde matrices invertible?

A Vandermonde-matrix is a matrix of this form: $$\begin{pmatrix} x_0^0 & \cdots & x_0^n \\ \vdots & \ddots & \vdots \\ x_n^0 & \cdots & x_n^n \end{pmatrix} \in \mathbb{R}^{(n+1) \times (n+1)}$$. condition ☀ : $\forall i, j\in \{0, \dots, n\}: i\neq j \Rightarrow x_i \neq x_j$ Why are Vandermond...

A: Why are Vandermonde matrices invertible?

This is not entirely dissimilar to the answer already posted by Chris Godsil, but I'll post this anyway, maybe it can provide slightly different angle for someone trying to understand this. We want to show that the matrix $$\begin{pmatrix} x_0^0 & \cdots & x_0^n \\ \vdots & \ddots & \vdots \\ x...

Sorry but I cannot understand where $k$ arrived. Up until $c_i \cdot v_i=$ its ok. But — Eric_ 5 mins ago

@Eric_ I am not exactly sure which place you mean. Are you asking about the part after $c_0v_0+c_1v_1+\dots+c_nv_n=\vec 0=(0,0,\dots,0)$? Feel free to ping me in this chatroom and we can discuss this there - to avoid having a long comment thread here. — Martin Sleziak 22 secs ago

> So let us assume that $c_0v_0+c_1v_1+\dots+c_nv_n=\vec 0=(0,0,\dots,0)$, where $v_j=(x_0^j,x_1^j,\dots,x_n^j)$ is the $j$-the column written as a vector and $c_0,\dots,c_n\in\mathbb R$.

So we have $c_0(x_0^0,x_1^0,\dots,x_n^0)+c_1(x_0^1,x_1^1,\dots,x_n^1)+\dots+c_n(x_0^n,x_1^n,\dots,x_n^n)=(0,0,\dots,0)$.

This gives us $n+1$ equations - one for each coordinate.

\begin{align*}
c_0x_0^0+c_1x_0^1+c_2x_0^2+\dots+c_nx_0^n&=0\\
c_0x_1^0+c_1x_1^1+c_2x_1^2+\dots+c_nx_1^n&=0\\
&\vdots\\
c_0x_n^0+c_1x_n^1+c_2x_n^2+\dots+c_nx_n^n&=0
\end{align*}

So we can write this in short as $$c_0x_k^0+c_1x_k^1+c_2x_k^2+\dots+c_nx_k^n=0$$ for $k=0,1,\dots,n$.

Naturally, $x_k^0=1$ and $x_k^1=x_k$, so this is the same as $$c_0+c_1x_k+c_2x_k^2+\dots+c_nx_k^n=0$$ for $k=0,1,\dots,n$.

@Eric_ In case this is what you're after - we have $n+1$ equations - one for each coordinate: \begin{align*} c_0x_0^0+c_1x_0^1+c_2x_0^2+\dots+c_nx_0^n&=0\\ c_0x_1^0+c_1x_1^1+c_2x_1^2+\dots+c_nx_1^n&=0\\ &\vdots\\ c_0x_n^0+c_1x_n^1+c_2x_n^2+\dots+c_nx_n^n&=0 \end{align*} So we can write this in short as $$c_0x_k^0+c_1x_k^1+c_2x_k^2+\dots+c_nx_k^n=0$$ for $k=0,1,\dots,n$. (So $k$ is just a notation used for an integer variable that runs through $0,1,\dots,n$.) — Martin Sleziak 17 secs ago

Eric_

Hello. I will be reading your reply.

Martin Sleziak

14:33

> Then we get on the $k$-th coordinate
$$c_0+c_1x_k+c_2x_k^2+\dots+c_nx_k^n=0,$$
which means that $x_k$ is a root of the polynomial $p(x)=c_0+c_1x+c_2x^2+\dots+c_nx^n$

^ Is this the part you were asking about?

Eric_

Yes. I just want to read exactly after to make sure I dont miss anything else

Martin Sleziak

Well, I think my comment explains this part.

If have edited the post to add "for $k=0,1,\dots,n$".

Eric_

Yes, thank you! It is all clear now :-)

Martin Sleziak

I am glad to hear that. Have a nice day!

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