In The Symmetric Group, we are presented with the following theorem
Let $\chi$ and $\psi$ be irreducible characters of a group $G$. Then $$\langle\chi,\psi\rangle=\delta_{\chi,\psi}\tag{1} $$
In the proof he writes the following: Suppose $\chi,\psi$ are the characters of matrix representations $A,B$ of degrees $d,f$, respectively and let $X=(x_{i,j})$ be a $d\times f$ matrix of indeterminates and consider the matrix $$\tag{2} Y=\frac{1}{|G|}\sum_{g\in G}A(g)XB(g^{-1}) $$ The author then shows that $$\tag{3} Y=\begin{cases} 0, & \text{if } A \not\cong B, \\ c I_d, & \text{if } A \cong B. \end{cases} $$ When we consider $\chi\neq \psi$, then eq. (3) implies that $$\tag{4} \frac{1}{|G|} \sum_{k, l} \sum_{g \in G} a_{i, k}(g) x_{k, l} b_{l, j}\left(g^{-1}\right)=0 $$ for all $i,j$. The author writes that "If this polynomial is to be zero, the coefficient of each $x_{k,l}$ must also be zero, so $$\tag{5} \frac{1}{|G|} \sum_{g \in G} a_{i, k}(g) b_{l, j}\left(g^{-1}\right)=0 $$ for all $i,j,k,l$". What I do not understand about this is that it assumes that $x_{k,l}$ are nonzero (because if they are zero, then eq. (4) says nothing about $a_{i,k}$ and $b_{l,j}$). So the question is
why does he assume that $x_{k,l}\ne 0?$
From my understanding, it comes from the fact that the matrix $X=(x_{i,j})$ is a matrix of indeterminants, so these $x_{i,j}$ are just variables, and the quantity $Y$ will be a polynomial in the variables $x_{i,j}$, so what is being said is if $\chi\neq \psi$, then we have that the polynomial $Y=0$, and the only way for a polynomial to be $0$ is for each of the coefficients to be $0$ which is how equation $5$ is derived. It is like saying let us assume we have a quadratic polynomial $ax^2+bx+c$, but if we assume that this polynomial is identically $0$ for any value of $x$, then we must conclude that $a=b=c=0$.