Exercise 2.7.4

Question

Prove that if $A: X \rightarrow Y$ and $V$ is a subspace of $X$ then dim $AV \leq $ rank $A$. ($AV$ here means the subspace $V$ transformed by the transformation $A$, i.e., any vector in $AV$ can be represented as $A \mathbf{v}, \mathbf{v} \in V$). Deduce from here that rank$(AB) \leq$ rank$A$.

Jing Huang · Accepted Answer

\begin{proof}
dim$AV \leq$ dim$AX \leq$ dim Ran$A$ $=$ rank$A$ 
ewline
Suppose that the column vectors of $B$  compose a basis of space $V$. Then rank$(AB) \leq$ rank$A$.
\end{proof}

Exercise 2.7.4

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Exercise 2.7.4

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