Can composition between linear operator and polynomial be possible? Can composition between linear operator and polynomial be possible?
I'm studying linear algebra and I found weird notation
Let T be a linear operator and let g(t) be a polynomial.
and what g(T)(x) means?
If i let g(t) be like t^n+t^(n-1)+t^(n-2) .........+t+1
then g(T)(x) must be (T^n+..............T+1)(x).
but the last part of g(T)(x),  1(x) is not a function.
what is this ??
 A: It's a standard abuse of notation in abstract algebra to write "$1$" to mean the multiplicative unit in any ring, which here is the ring of linear operators. As egreg says in the comments, this means "$1$" here is standing for the identity operator $I$.
This is good notation for several reasons. For example, if $g(t) = \sum g_n t^n$ is a polynomial and we define
$$g(T) = \sum g_n T^n$$
where $T : V \to V$ is a linear operator, then the map $g \mapsto g(T)$ turns out to be a ring homomorphism, which is useful for lots of reasons, e.g. it gets you all the way to a proof of the Jordan normal form theorem.
A: A polynomial is always of the form
$$
g(X) = a_n X^n + a_{n-1} X^{n-1} +\cdots + a_1 X^1 + a_0 X^0.
$$
We often ommit the $X^0$ and just write $a_0$ for the last term instead, since $X^0$ is the multiplicative unit in the polynomial ring, also written as $1$. However, when evaluating $g$ for an operator $T$, it pays off to leave it there: then you get
$$
g(T) = a_n T^n + a_{n-1} T^{n-1} +\cdots + a_1 T^1 + a_0 T^0,
$$
where $T^0$ is the identity operator "$I$" or "$\operatorname{id}$" by the same convention that $x^0 = 1$ in other contexts.
