To solve the problem using the rank-nullity theorem, we need to recall the relationship:

$\text{dim}(U) = \text{rank}(T) + \text{nullity}(T)$

This theorem states that for a linear transformation $T: U \to V$, the dimension of the domain $U$ is equal to the sum of the rank of the transformation and the nullity of the transformation.

Given the table:

$U$	$P_6$	$P_7$	$P_n$
$\text{dim}(U)$	6	?	?
$\text{rank}(T)$	Ex: 5	8	Ex: n+2
$\text{nullity}(T)$	1	Ex: 5	7

We'll solve each case using the formula:

1. For $P_6$:

   • Given: $\text{dim}(U) = 6$, $\text{nullity}(T) = 1$.
   • Using the formula: $\text{dim}(U) = \text{rank}(T) + \text{nullity}(T)$, we get: $6 = \text{rank}(T) + 1$ Therefore, $\text{rank}(T) = 5$.

2. For $P_7$:

   • Given: $\text{rank}(T) = 8$, and the missing values for $\text{dim}(U)$ and $\text{nullity}(T)$ are to be calculated.
   • Using the same formula: $\text{dim}(U) = 8 + \text{nullity}(T)$
   • Since the table suggests $\text{nullity}(T) = 5$, then: $\text{dim}(U) = 8 + 5 = 13$

3. For $P_n$:

   • Given: $\text{nullity}(T) = 7$, and $\text{rank}(T) = n + 2$.
   • Using the formula: $\text{dim}(U) = (n + 2) + 7 = n + 9$

Now, let's update the table:

$U$	$P_6$	$P_7$	$P_n$
$\text{dim}(U)$	6	13	$n + 9$
$\text{rank}(T)$	5	8	$n + 2$
$\text{nullity}(T)$	1	5	7

Would you like any further details or explanations?

Here are some related questions:

1. How does the rank-nullity theorem apply to different types of vector spaces?
2. What are some examples where the rank-nullity theorem is used in real-world applications?
3. How does changing the dimensions of the domain $U$ affect the rank and nullity of $T$?
4. Can the rank of a transformation exceed the dimension of the domain?
5. What happens to the nullity if the transformation is injective?

Tip: Always ensure that the sum of the rank and nullity matches the dimension of the domain when applying the rank-nullity theorem.