Given a Linear Transformation, to find Nullity. (Linear algebra) $V$ is a vector space of polynomials of degree less than equal to n with real coefficients, a linear transformation $$T: V(\mathbb{R})\rightarrow\mathbb{R}^2$$ defined by $$T(P(x)=a_0+a_1x+...+a_nx^n)= (P(1), P(-1))$$ Then find the $\bf dim~N(T).$
My approach to solve this problem is as follows;
For N(T), we first take $T(X)=\bf 0$. So we get $((P(1), P(-1))= (0,0)$
$\implies P(1)=0~ \& ~P(-1)=0$. This means $x=-1,x=1$ are two roots of P(x). So we can write $P(x)$ as $(x^2-1)P(x)$. Any suggestion how to proceed further will be very helpful.
 A: You're completely right that we need to take $T(X) = \bf 0$, which means
$$(P(1),P(-1)) := (0,0)$$
Now we can actually just write up this condition:
$$P(1) = a_0 + a_1 + a_2 + \dots + a_n = \sum_{k=0}^n a_k := 0\\
P(-1) = a_0 - a_1 + a_2 - \dots + (-1)^n a_n = \sum_{k=0}^n (-1)^k a_k := 0$$
Now our question is, what's the dimensionality of the values $a_0,\dots,a_n$ where these conditions hold?
First off, if there were no conditions, then the dimensionality of $a_0,\dots,a_n$ is $n+1$, since all $n+1$ of them can be distinct real values.
But with the first condition, we can actually express $a_0$ from the others, namely:
$$a_0 = -\sum_{k=1}^n a_k$$
With $a_0$ known, we can also express $a_1$ from the others, using the second condition (but this one's a bit more tricky, I've started by expressing $a_2$):
$$a_2 = -a_0+a_1-\sum_{k=3}^n (-1)^k a_k \stackrel{\text{first cond.}}{=} \sum_{k=1}^n a_k + a_1 - \sum_{k=3}^n (-1)^k a_k = \\ = a_1 + a_2 + \sum_{k=3}^n a_k + a_1 - \sum_{k=3}^n (-1)^k a_k = 2a_1 + a_2 + \sum_{k=3}^n (a_k + (-1)^k a_k)$$
Simplifying by $a_2$, we get a formula for $a_1$:
$$0 = 2a_1 + \sum_{k=3}^n (a_k + (-1)^k a_k) \\
a_1 = -\frac{1}{2}\sum_{k=3}^n (a_k + (-1)^k a_k)$$
So since we can express $a_0$ and $a_1$ from the others, and there are $n+1$ of these values, that must mean that the total dimension of $(a_0,a_1,\dots,a_n)$ can't be greater than $(n+1) - 2=n-1$. We would also need to prove that their dimensionality can't be less than $n-1$, but given no more equations, we aren't able to express $(a_2,a_3,\dots,a_n)$ from each other.
There is one more caveat, if $n=0$, then the first and second equations are actually the same $a_0 = 0$, so the final answer is
$$N(T) = \begin{cases} 0 \quad \text{if }\  n=0 \\ n-1 \quad \text{otherwise}\end{cases}$$
