Chapter 6 - Linear Transformations - 6.3 The Kernel and Range of a Linear Transformation - True-False Review - Page 405: e

True

Given: $T\begin{bmatrix} 1 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix} \in Ker T\\ T \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} \in Ker T \\ \rightarrow \dim Ker T\geq2$ According to Rank-Nullity Theorem: $\dim RngT=\dim V-\dim Ker T\\ =6-\dim Ker T$ But $\dim Ker T\geq2\\ \rightarrow 6-\dim Ker T \leq 2\\ \rightarrow \dim RngT \leq 6-2=4$ Hence, $Rng(T )$ is at most four-dimensional.