To solve the problem using the rank-nullity theorem, let's first recall the theorem:

Rank-Nullity Theorem: For a linear transformation $T: U \to V$, the theorem states: $\text{dim}(U) = \text{rank}(T) + \text{nullity}(T)$

Let's apply this theorem to each case in the table:

Case 1: $U = \mathbb{R}^4$

• $\text{dim}(U) = 4$
• $\text{rank}(T) = 3$
• $\text{nullity}(T) = \text{Ex. 5}$

Using the rank-nullity theorem: $4 = 3 + \text{nullity}(T)$ $\text{nullity}(T) = 4 - 3 = 1$

So, $\text{nullity}(T) = 1$.

Case 2: $U = \mathbb{R}^9$

• $\text{dim}(U) = 9$
• $\text{rank}(T) = \text{Ex. 5}$
• $\text{nullity}(T) = 6$

Using the rank-nullity theorem: $9 = \text{rank}(T) + 6$ $\text{rank}(T) = 9 - 6 = 3$

So, $\text{rank}(T) = 3$.

Case 3: $U = \mathbb{R}^n+2$

• $\text{dim}(U) = n+2$
• $\text{rank}(T) = n+2$
• $\text{nullity}(T) = \text{Ex. 5}$

Using the rank-nullity theorem: $n+2 = (n+2) + \text{nullity}(T)$ $\text{nullity}(T) = 0$

So, $\text{nullity}(T) = 0$.

Summary of the Table

$U$	$\mathbb{R}^4$	$\mathbb{R}^9$	$\mathbb{R}^{n+2}$
$\text{dim}(U)$	4	9	$n+2$
$\text{rank}(T)$	3	3	$n+2$
$\text{nullity}(T)$	1	6	0

Would you like more details or have any other questions?

Here are 5 questions that expand on this topic:

1. How is the rank-nullity theorem useful in understanding the structure of linear transformations?
2. What happens if the nullity of a transformation is zero? What does that tell us about the transformation?
3. Can the rank of a linear transformation be greater than the dimension of the codomain $V$?
4. How would you interpret a linear transformation with high nullity in terms of its impact on the space $U$?
5. What role does the dimension of the codomain $V$ play in determining the properties of $T$?

Tip: Remember that the rank of a transformation gives us the dimension of the image (range) of the transformation, while the nullity tells us the dimension of the kernel (the set of vectors that map to zero).