1. The problem states: Let $T: W \to V$ be a linear transformation where $W$ and $V$ are vector spaces. If $W$ is finite dimensional, prove that the nullity of $T$ plus the rank of $T$ equals the dimension of $W$. 2. Recall the definitions: - The nullity of $T$, denoted $\mathrm{nullity}(T)$, is the dimension of the kernel of $T$, i.e., $\dim(\ker(T))$. - The rank of $T$, denoted $\mathrm{rank}(T)$, is the dimension of the image of $T$, i.e., $\dim(\mathrm{im}(T))$. 3. The Rank-Nullity Theorem states that for a linear transformation $T: W \to V$ where $W$ is finite dimensional, $$\dim(W) = \dim(\ker(T)) + \dim(\mathrm{im}(T))$$ This means: $$\dim(W) = \mathrm{nullity}(T) + \mathrm{rank}(T)$$ 4. This theorem can be proved by choosing a basis for $\ker(T)$, extending it to a basis of $W$, and showing that the images of the additional basis vectors form a basis for $\mathrm{im}(T)$. 5. Hence the addition of nullity and rank of $T$ is equal to the dimension of $W$ as required. **Final answer:** $$\boxed{\mathrm{nullity}(T) + \mathrm{rank}(T) = \dim(W)}$$