Let's use the **rank-nullity theorem**, which states: \[ \text{dim}(U) = \text{rank}(T) + \text{nullity}(T) \] We'll fill in the missing entries step by step. --- ### **Step 1: Fill in the missing values for \(\mathbb{R}^7\)** - Given: \(\text{nullity}(T) = 7\) - The dimension of \(U\) (i.e., \(\text{dim}(U)\)) is the sum of rank and nullity: \[ \text{dim}(U) = \text{rank}(T) + \text{nullity}(T) \] Let \(\text{rank}(T) = r\). \[ \text{dim}(U) = r + 7 \] But the space is \(\mathbb{R}^7\), so \(\text{dim}(U) = 7\): \[ 7 = r + 7 \implies r = 0 \] So, \(\text{rank}(T) = 0\). --- ### **Step 2: Fill in the missing values for \(\mathbb{R}^n\)** - Given: \(\text{nullity}(T) = 6\) - Let \(\text{rank}(T) = r\). - The dimension of \(U\) is \(n\): \[ n = r + 6 \implies r = n - 6 \] Thus, \(\text{rank}(T) = n - 6\). --- ## **Final Table:** \[ \begin{array}{|c|c|c|c|} \hline U & \mathbb{R}^5 & \mathbb{R}^7 & \mathbb{R}^n \\ \hline \dim(U) & 5 & 7 & n \\ \hline \text{rank}(T) & 4 & 0 & n-6 \\ \hline \text{nullity}(T) & 1 & 7 & 6 \\ \hline \end{array} \] --- ### **Summary:** - For \(\mathbb{R}^7\): \(\text{dim}(U)=7\), \(\text{rank}(T)=0\), \(\text{nullity}(T)=7\) - For \(\mathbb{R}^n\): \(\text{dim}(U)=n\), \(\text{rank}(T)=n-6\), \(\text{nullity}(T)=6\)