Exercise 9.4

Question

Exercise 4: 
    Prove that null spaces and ranges of linear transformations are vector spaces.

analambanomenos · Accepted Answer

Let $\v x,\v y\in\Null(A)$ and let $c$ be a scalar. Then
    \begin{align*}    
        A(\v x+\v y) &= A(\v x)+A(\v y) = \v0+\v0 = \v0 	ext{ so that }\v x+\v y\in\Null(A) \
        A(c\v x) &= cA(\v x) = c\v0 = \v0 	ext{ so that }c\v x\in\Null(A)
    \end{align*}    
    which shows that $\Null(A)$ is a vector space.

Let $\v x,\v y\in\Range(A)$, so that $\v x=A(\v x')$ and $\v y=A(\v y')$, and let $c$ be a scalar. Then 
    \begin{align*}    
        \v x+\v y &= A(\v x')+A(\v y') = A(\v x'+\v y') 	ext{ so that }\v x+\v y\in\Range(A) \
        c\v x &= cA(\v x') = A(c\v x') 	ext{ so that }c\v x\in\Range(A)
    \end{align*}
    which shows that $\Range(A)$ is a vector space.

Exercise 9.4

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Exercise 9.4

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