Agreement of two polynomials How can I prove that if two polynomials (of matrix coefficients) agree in a neighborhood of $0$, then they are identical?
Thank you!
 A: Note that this reduces to checking that any polynomial which is $0$ in a neighborhood of $0$ is the zero polynomial. Suppose $A_nX^n+\cdots+A_1X+A_0=0$ for all $X\in U$ where $U\subset M_n(\mathbb C)$ is an open set containing $0$. Then we have some open ball $B_r(0)$ on which $A_nX^n+\cdots+A_0=0$, and in fact since polynomials are continuous $A_nX^n+\cdots+A_0=0$ on $\overline{B_r(0)}=\{X: \|X\|\leq r\}$. Consider the matrices $X_{kj}$ which have only one nonzero entry, $\frac{jr}{n}$, which is the $k^{th}$ entry along the diagonal. Observe that for $1\leq k,j\leq n$ each such matrix is in $\overline{B_r(0)}$, so $A_nX_{kj}^n+\cdots+A_0=0$. Furthermore, $$\{v_j\}_{j=1}^n=\left\{\left(\left(\frac{jr}{n}\right)^n,\left(\frac{jr}{n}\right)^{n-1},\ldots,1\right)\right\}_{j=1}^n$$
is a basis for $\mathbb C^n$ over $\mathbb C$, and for all $j$
$$0=A_nX_{kj}^n+\cdots+A_0=a_{nk}\left(\frac{jr}{n}\right)^n+\cdots+a_{0k}=(a_{nk}|\cdots|a_{0k})v_{j}$$
where $a_{xy}$ denotes the $y^{th}$ column vector of $A_x$. Thus for all $k$, $(a_{nk}|\cdots|a_{0k})=0$ so each column of each matrix $A_x$ must be $0$. Thus $A_n=\cdots=A_0=0$.
Remark: The same proof works in $\mathrm{GL}_n(\mathbb C)$ if we replace the diagonal entries of $X_{kj}$ which are $0$ with $1$.
