Exercise 2.7.5

Question

Prove that if $A: X \rightarrow Y$ and $V$ is a subspace of $X$ then dim $AV \leq$ dim $V$. Deduce from here that rank$(AB) \leq$ rank $B$.

Jing Huang · Accepted Answer

Suppose dim $V = k$ and let $\mathbf{v}_1, \mathbf{v}_2, ... , \mathbf{v}_k$ be a basis of $V$. $AV$ is defined by $A\mathbf{v}_1, A\mathbf{v}_2, ... , A\mathbf{v}_k$.
dim $AV$ = rank $[A\mathbf{v}_1, A\mathbf{v}_2, ... , A\mathbf{v}_k] \leq k = $ dim $V$.   \par
Similarly, assume rank $B = k$ and $\mathbf{b}_1, \mathbf{b}_2, ... , \mathbf{b}_k$ are linearly independent column vectors in $B$. Then rank $AB$ = rank $[A\mathbf{b}_1, A\mathbf{b}_2, ... , A\mathbf{b}_k] \leq k = $ rank $B$.

Exercise 2.7.5

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Exercise 2.7.5

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