Question

rank nullity theorem solution​

Answer 1

Answer:

Step-by-step explanation:

Proof of Rank-Nullity Theorem

THEOREM 15.5.1

Let $ T: V {\longrightarrow}W$ be a linear transformation and $ \{ u_1, u_2, \ldots, u_n \}$ be a basis of $ V$ . Then

$ {\cal R}(T) = L( T(u_1), T(u_2), \ldots, T(u_n) ).$

$ T$ is one-one $ \Longleftrightarrow \; \; {\cal N}(T) = \{{\mathbf 0}\}$ is the zero subspace of $ V \Longleftrightarrow \; \;$ $ \{ T(u_i): 1 \leq i \leq n \}$ is a basis of $ {\cal R}(T).$

If $ V$ is finite dimensional vector space then $ \dim ({\cal R}(T)) \leq

\dim (V).$ The equality holds if and only if $ {\cal N}(T) = \{{\mathbf 0}\}.$

Proof. Part $ 1)$ can be easily proved. For $ 2),$ let $ T$ be one-one. Suppose $ u \in {\cal N}(T).$ This means that $ T(u) = {\mathbf 0}=

T({\mathbf 0}).$ But then $ T$ is one-one implies that $ u = {\mathbf 0}.$ If $ {\cal N}(T) = \{{\mathbf 0}\}$ then $ T(u) = T(v) \; \Longleftrightarrow T(u - v) = {\mathbf 0}$ implies that $ u = v.$ Hence, $ T$ is one-one.

The other parts can be similarly proved. Part $ 3)$ follows from the previous two parts. height6pt width 6pt depth 0pt

The proof of the next theorem is immediate from the fact that $ T({\mathbf 0}) = {\mathbf 0}$ and the definition of linear independence/dependence.