Say,$~~f(x) = h(x)g(x)~~$ for all $x \in \mathbb{R}$, where $f(x), g(x), h(x)$ are polynomials with real coefficients.
Let $B$ be any non-negative square matrix. Can we always say $f(B) = h(B)g(B)?~$ Does it hold for any square matrix $B$? Can we formally prove this, if true?
Here $~f(B), g(B), h(B)$ are the matrix polynomials (https://en.wikipedia.org/wiki/Matrix_polynomial).
Any hints or guidance would be much appreciated. Thanks!
The problem is to see where the different formulas live. If you write $f(x)=h(x)g(x)$ the two members live in formal expressions but they are equal when they are evaluated as polynomials. Now, given a matrix $B$ there is a map (call it $\phi_B$) from the polynomials to the matrices such that $$ \phi_B(P+Q)=\phi_B(P)+\phi_B(Q)\ ;\ \phi_B(\alpha P)=\alpha\phi_B(P)\ ;\ \phi_B(PQ)=\phi_B(P)\phi_B(Q)\qquad (*) $$ and one sees that $\phi_B(P)=P(B)$. If fact, this map $\phi_B$ is the unique map $\phi : k[X]\to M(n,k)$ which satisfies $(*)$, $\phi(1)=I$ and $\phi(X)=B$. So, if two expressions (however different as expressions) evaluate as the same polynomial (as $X^2-1$ and $(X-1)(X+1)$) their value at $B$ are the same. Expressions are better understood when written with tree-like structures. For instance $X^2-1$ reads (sorry but I am a bit clumsy in Mathjax drawings) \begin{array}{cccccc} & & - & \\ &╱& &╲ \\ & * & & 1 & \\ ╱ & &╲ & & & \\ X & & X & & \end{array} whereas $(X-1)(X+1)$ reads \begin{array}{cccccc} & & * & \\ &╱& &╲ \\ & - & & & +\\ ╱ & &╲ & ╱ & &╲ \\ X & & 1 & X & & 1 \end{array} You can represent any polynomial expression like this. Identities are statements like $T_1=T_2$ where $T_i$ are formal expressions but are equal when evaluated within the realm of polynomials.
Hope it helps !