To determine Nullity of $T$ Let $V$ be vector space of polynomials of degree $\leq  n$ . And  $ T : V \rightarrow \mathbb R ^{m}$ be defined as $T (P (x)) =  (P (1) , P (2) ,..., P (m) )$
I have to determine nullity of $T$ .
Here for nullity I see I have to make two separate cases when $n> m$ and $n<m$ .Also for nullity I set $P (1) =0 , P (2) =0\cdots$ but i am not getting clear along these lines .Please help me to get through this .Thanks
 A: Revised answer
Theorem. Given pairwise distinct numbers $x_0,x_1,\dots,x_n$ in $\mathbb{R}$ and any choice of $\alpha_0, \alpha_1,\dots,\alpha_n\in\mathbb{R}$, there exists one and only one polynomial $p$ of degree at most $n$ such that
$$
p(x_0)=\alpha_0,p(x_1)=\alpha_1,\dots,p(x_n)=\alpha_n
$$
This solves the so-called “interpolation problem“. We don't need to determine it, but it can be done with the Lagrange method.
Proof.
Write the unknown polynomial as
$$
\beta_0+\beta_1x+\dots+\beta_nx^n.
$$
Then applying the requirements $p(x_k)=\alpha_k$ (for $k=0,1,\dots,n$), we get a linear system with coefficient matrix
$$
\begin{bmatrix}
1 & x_0 & x_0^2 & \dots & x_0^n \\
1 & x_1 & x_0^2 & \dots & x_1^n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & x_n & x_n^2 & \dots & x_n^n
\end{bmatrix}
$$
which is a Vandermonde matrix, having nonzero determinant. $\square$
Now we apply the theorem to your case. Let's write $T_m\colon V\to\mathbb{R}^m$ for the mapping
$$
T_m(p)=(p(1),p(2),\dots,p(m))
$$
and, for $r\ge s$, we consider the (linear) map $h_{r,s}\colon\mathbb{R}^r\to\mathbb{R}^s$ defined by
$$
h_{r,s}(\alpha_1,\alpha_2,\dots,\alpha_r)=
(\alpha_1,\dots,\alpha_s)
$$
In particular, $h_{r,r}$ is the identity map. Note also that $h_{r,s}$ is surjective.
The theorem above states that the map $T_{n+1}$ is bijective.
Now, suppose $m\le n+1$. Then it's easy to verify that
$$
T_m=h_{n+1,m}\circ T_{n+1}
$$
so $T_m$ is surjective as the composition of two surjective maps. Since the image has dimension $m$, the nullity of $T_m$ is
$$
\dim V-\dim\operatorname{Im}T_m=n+1-m.
$$
If $m>n+1$, we can see that $h_{m,n+1}\circ T_m=T_{n+1}$, which is injective, so also $T_m$ is injective and its nullity is $0$.
Thus we can write
$$
\dim\ker T_m=\max\{n+1-m,0\}.
$$

Original answer
It's a known fact that, given $n+1$ numbers (distinct or not) $\alpha_0,\alpha_1,\dots,\alpha_n$, there exists exactly one polynomial $p$ of degree at most $m$ such that
$$
p(0)=\alpha_0,p(1)=\alpha_1,\dots,p(n)=\alpha_n
$$
This is the “interpolation problem”. Instead of $0,1,\dots,n$ one can use any set of pairwise distinct numbers $x_0,x_1,\dots,x_n$.
As a consequence, if $m\le n+1$, the map you're given is surjective. Since the dimension of $V$ is $n+1$, the nullity is $(n+1)-m$.
If $m>n+1$, then the above result implies that the map is injective.
[For details on the interpolation problem, see the “revised answer” part.]
A: Divide the two cases as:
Case 1: $m>n$
Here we have a homogeneous system with more equations than unknowns, hence we get a unique solution, i.e if $P\in\ker T$ then $P=0$, i.e $\dim\ker T=0$
Case 2:$m\leq n$
Here we have less equations than number of unknowns. If $P\in\ker T$ From $m$ equations (since $P(1)=0,P(2)=0,...,P(m)=0$) we get $m$ relations among the coefficients of the $n$ degree polynomial $P$. Hence $\dim\ker T =n-m+1$
A: Ince $\dim V=n+1$ you get from rank-nullity that $\def\rk{\operatorname{rk}}\dim(\ker T)=n+1-\rk(T)$, where $\rk(T)\leq\dim(\Bbb R^m)=m$. This, together with non-negativity of dimensions, gives the lower bound
$$
  \dim(\ker T)\geq\max(n+1-m,0).
$$
I turns out this bound is always sharp (the "$\geq$" may be replaced by "$=$"). To prove that, consider two cases (overlapping for $m=n+1$):


*

*$m\geq n+1$, so that the right hand side is $0$. We must show that $T$ is injective: it cannot produce $0$ for a nonzero polynomial $P$. This is so because $T(P)=(0,\ldots,0)$ would mean that $1,2,\ldots,m$ are roots of $P$, which is too many roots for a nonzero polynomial of degree${}<n+1\leq m$ to have.

*$m\leq n+1$. Now equality is obtained if and only if $\operatorname{rk}(T)=m$, in other words if $T$ is surjective. This is indeed the case, because one can always produce a polynomial of degree${}<m$ (so a fortiori of degree${}<n+1$) with $m$ prescribed values in $m$ distinct points, by Lagrange interpolation.
