Let $(\omega_1, \eta_1） \dots (\omega_n, \eta_n)$ be $n$ pairs of complex numbers where $\omega_i \ne \omega_j$ for all $1 \leq i \ne j \leq n$. We define the following polynomial space $$ Z_n^d(\eta, \omega) := \{ p(x) \in \mathbb C[x] : \deg( p ) \leq d, p'( \omega_i ) = \eta_i p( \omega_i ) \; \forall \; 1 \leq i \leq n \} $$ I am interested in computing the dimension of $\mathbb C$-vector space $Z_n^d( \eta, \omega)$. Note if $d \geq 2n - 1$, $$\dim_{\mathbb C} [Z_n^d( \eta, \omega )] = d +1 - n$$ Suppose $d \geq 2n - 1$, we can use Hermite interpolation to claim the evaluation homomorphism $\text{ev}_n: Z_n^d( \eta, \omega) \rightarrow \mathbb C^n$ defined via $$ \text{ev}_n( p) = (p(\omega_1), \dots, p( \omega_n) ) $$ is surjective. In addition, it's easy to obtain $$ \ker ( \text{ev}_n ) = \{ g(x) \Omega^2(x) \mid \deg( g ) \leq d - 2n \} $$ where $\Omega( x ) = \prod_{i = 1}^n ( x - \omega_i)$. Therefore from the isomorphism theorem $Z_n^d( \eta, \omega) \cong \ker( \text{ev}_n ) \oplus \mathbb C^n$ which implies $$ \dim_{\mathbb C} [ Z_n^d( \eta, \omega) ] = \dim_{\mathbb C} [ \ker( \text{ev}_n ) ] + \dim_{\mathbb C} \mathbb C^n = (d - 2n + 1) + n = d + 1 - n $$ I am interested in what restrictions we have to put on $\eta_i$s in terms of $\omega_i$s so that a similar dimension formula would hold when $n \leq d < 2n - 1$. For each $p(x) \in Z_n^d( \eta, \omega)$, if we consider $p(x) = \sum_{i = 0}^d a_i x^i$ in the standard monomial expansion, the prescribed condition $p'( \omega_i ) = \eta_i p( \omega_i ) \; \forall \; 1 \leq i\leq n$ in the definition of $Z_n^d( \eta, \omega)$ says $(a_0, a_1, \dots, a_d) \in \mathbb C^{d + 1}$ solves the matrix equation $Ax = 0$ where $$ A = \begin{pmatrix} \eta_1 & \omega_1 \eta_1 - 1 & \dots & \omega_1^d \eta_1 - d \omega_1^{d - 1} \\ \eta_2 & \omega_2 \eta_2 - 1 & \dots & \omega_2^d \eta_2 - d \omega_1^{d - 1} \\ \vdots & \vdots & \ddots &\vdots\\ \eta_{n} & \omega_{n} \eta_{n} - 1 & \dots & \omega_{n}^d \eta_{n} - k \omega_{n}^{d - 1} \\ \end{pmatrix} $$ Since $d \geq n$, $A$ has more columns then rows. Using basic linear algebra we get $$ \dim_{\mathbb C} [ Z_n^d( \eta, \omega) ] = d + 1 - \text{rank } A $$ This gives an equivalent way to restate our problem: For what $\eta_i$s as functions of distinct complex numbers $\omega_1, \dots, \omega_n$ does $\text{rank } A = n$? I found a counterexample when $d = n = 2$. Put $\eta_1 = \omega_1 = 1, \eta_2 = \omega_2 = -1$, then $$ A = \begin{pmatrix} 1 & 0 & -1 \\ -1 & 0 & 1 \\ \end{pmatrix} \Longrightarrow \text{rank } A = 1 $$ In general, we can prove by induction for all $n \geq 2$ with $d = 2n - 2$, $$ \text{rank } A < n \Longrightarrow \eta_i = \sum_{j \ne i}^n \frac{ 2 }{ \omega_i - \omega_j } = \frac{ \Omega''( \omega_i ) }{ \Omega'( \omega_i ) } \; \forall \; 1 \leq i \leq n $$ However I don't know how to proceed when $n \leq d \leq 2n - 3$, the matrix algebra used in case $d = 2n-2$ is much more complicated here.
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