Question

Prove the Rank-Nullity Theorem (Theorem 3.35) in the case that kerT={0}

Answer 1

The Rank-Nullity Theorem states that for a linear transformation T:V → W, where V and W are vector spaces, the rank of T plus the nullity of T equals the dimension of V. If ker(T) = {0}, it means that the only vector in V that is mapped to 0 is the zero vector 0∈V. In this case, the nullity of T is 0 because there are no other vectors in V that are mapped to 0. Therefore, in the case that ker(T) = {0}, the Rank-Nullity Theorem simplifies to: rank(T) = dim(V).

Step-by-step explanation:

The Rank-Nullity Theorem states that for a linear transformationWe'll use> $T:V \rightarrow W$,

where $V$ and $W$ are vector spaces, the rank of $T$ (denoted by $\text{rank}(T)$) plus the nullity of $T$ (denoted by $\text{nullity}(T)$) equals the dimension of $V$:

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

If $\ker(T) = \{0\}$, it means that the only vector in $V$ that is mapped to $0$ is the zero vector $0 \in V$. In this case, the nullity of $T$ is $0$ because there are no other vectors in $V$ that are mapped to $0$.

Therefore, in the case that $\ker(T) = \{0\}$, the Rank-Nullity Theorem simplifies to:

$$\text{rank}(T) = \dim(V)$$