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$$(c_i-c_j)^k = \sum_{h=0}^k \binom{k}{h} (-1)^{k-h}c_i^h c_j^{k-h}$$

and each summand is a rank-1 matrix (since it's a function of $i$ times a function of $j$). To prove that the rank is not lower than that, consider that the vectors $\mathbf{v}_h = (c_i^h)_{i=1}^n$ are independent because they form a Vandermonde matrix, and so are the $\mathbf{w}_h = (c_j^h)_{j=1}^n$.

$$(c_i-c_j)^k = \sum_{h=0}^k \binom{k}{h} (-1)^{k-h}c_i^h c_j^{k-h}$$

and each summand is a rank-1 matrix (since it's a function of $i$ times a function of $j$). To prove that the rank is not lower than that, consider that the vectors $\mathbf{v}_h = (c_i^h)_{i=1}^n$ are independent because they form a Vandermonde matrix, and so are the $\mathbf{w}_h = (c_j^h)_{j=1}^n$.

Sometimes these concepts appears in tensor / matrix approximation problems as "separability" of a two-variable function; in your case, we just showed that the polynomial $(x-y)^k$ has separation rank $k+1$, i.e., it can be written as the sum of $k+1$ terms of the form $f(x)g(y)$.

$$(c_i-c_j)^k = \sum_{h=0}^k \binom{k}{h} (-1)^{k-h}c_i^h c_j^{k-h}$$

and each summand is a rank-1 matrix (since it's a function of $i$ times a function of $j$). To prove that the rank is not lower than that, consider that the vectors $\mathbf{v}_h = (c_i^h)_{i=1}^n$ are independent because they form a Vandermonde matrix, and so are the $\mathbf{w}_h = (c_j^h)_{j=1}^n$.

$$(c_i-c_j)^k = \sum_{h=0}^k \binom{k}{h} (-1)^{k-h}c_i^h c_j^{k-h}$$

and each summand is a rank-1 matrix (since it's a function of $i$ times a function of $j$). To prove that the rank is not lower than that, consider that the vectors $\mathbf{v}_h = (c_i^h)_{i=1}^n$ are independent because they form a Vandermonde matrix, and so are the $\mathbf{w}_h = (c_j^h)_{j=1}^n$.

$$(c_i-c_j)^k = \sum_{h=0}^k \binom{k}{h} (-1)^{k-h}c_i^h c_j^{k-h}$$

and each summand is a rank-1 matrix (since it's a function of $i$ times a function of $j$). To prove that the rank is not lower than that, consider that the vectors $c_i^h$ are independent because they form a Vandermonde matrix.

$$(c_i-c_j)^k = \sum_{h=0}^k \binom{k}{h} (-1)^{k-h}c_i^h c_j^{k-h}$$

and each summand is a rank-1 matrix (since it's a function of $i$ times a function of $j$). To prove that the rank is not lower than that, consider that the vectors $\mathbf{v}_h = (c_i^h)_{i=1}^n$ are independent because they form a Vandermonde matrix, and so are the $\mathbf{w}_h = (c_j^h)_{j=1}^n$.