Let $\{A_i\}$ be a set of $d-$dimensional matrices. $A_1\otimes A_2$ refers to the tensor product or Kronecker product of $A_1$ and $A_2$. We now look at symmetric tensor products i.e. linear combinations of terms like $A_i\otimes A_i\otimes...$
I came across a result (just after (5) in this paper) that claims that the space of any linear combination of these i.e.
$$\left\{B: B=\sum_{i} p_{i} A_{i}^{\otimes n}, \quad p_{i} \in \mathbb{R}, A_{i} \in \mathbb{C}^{d \times d}\right\}$$
has a dimension bounded by $(n+1)^{d-1}$. Naively, I would have expected the dimension of $B$ to be $d^n$. How is this result obtained?
It is well known that for a $d^2$-dimensional space $V$, the dimension of $\operatorname{Sym}^n(V)$ is given by $$ \binom{n + d^2 - 1}{d^2-1} = \frac1{(d^2 - 1)!} \cdot (n + d^2 - 1)(n+d^2 -2) \cdots (n+1)\\ = \frac{n+d^2-1}{d^2 - 1} \cdot \frac{n+d^2 - 2}{d^2 - 2} \cdots \frac{n+1}{1} \\ \leq (n+1)\cdot (n+1) \cdots (n+1) = (n+1)^{d^2 - 1}. $$ This is the bound used in the paper.