Let \( P_3(\mathbb{R}) \) be the vector space of all polynomials of degree at most three with real coefficients under usual polynomial addition and scalar multiplication. Let \( T: P_3(\mathbb{R}) \to \mathbb{R}^2 \) be the linear transformation defined as
\[ T(p) = \left(p(1), p'(1)\right) \]
for all \( p \in P_3(\mathbb{R}) \), where \( p' \) is the derivative of \( p \).
Then the nullity of \( T \) is equal to ______. (Answer in integer)
Show Hint
Use rank-nullity: find the rank of T first by checking whether it is onto, then nullity equals dim(domain) minus rank.
Step 1: Write a general element of $P_3(\mathbb{R})$.
Let $p(x) = a + bx + cx^2 + dx^3$, where $a,b,c,d$ are real numbers. This is the most general polynomial of degree at most $3$, so finding the kernel of $T$ means finding all $(a,b,c,d)$ with $T(p) = (0,0)$.
Step 2: Write down $p(1)$ and $p'(1)$ in terms of $a,b,c,d$.
$p(1) = a+b+c+d$. Also $p'(x) = b+2cx+3dx^2$, so $p'(1) = b+2c+3d$. The kernel condition $T(p)=(0,0)$ becomes the two linear equations:
\[ a+b+c+d = 0, \qquad b+2c+3d = 0 \]
Step 3: Count the free parameters in the solution set.
These are $2$ independent linear equations (the coefficient rows $(1,1,1,1)$ and $(0,1,2,3)$ are not multiples of each other) in $4$ unknowns $a,b,c,d$. So the solution set is a subspace of dimension $4-2=2$: fixing any $2$ of the unknowns freely, say $c$ and $d$, determines $a$ and $b$ uniquely from the two equations above.
Step 4: State the kernel dimension as the nullity.
The nullity of $T$ is, by definition, the dimension of its kernel. Since the kernel is a $2$-dimensional subspace of $P_3(\mathbb{R})$,
\[ \operatorname{nullity}(T) = 2 \]
This matches the count obtained through the rank-nullity theorem, confirming the answer a second, independent way.
Final Answer:
The nullity of $T$ is $2$.
\[ \boxed{2} \]
