Exercise 9.2

Question

Exercise 2: 
    Prove that $BA$ is linear if $A$ and $B$ are linear transformations. Prove also that $A^{-1}$ is linear and invertible.

analambanomenos · Accepted Answer

Let $A\in L(X,Y)$ and $B\in L(Y,Z)$. Let $\v x,\v y\in X$ and let $c$ be a scalar. Then
    \begin{align*}
        (BA)(\v x+\v y) &= B\big(A(\v x+\v y)\big) \
        &= B\big(A(\v x)+A(\v y)\big) \
        &= B\big(A(\v x)\big) + B\big(A(\v y)\big) \
        &= (BA)(\v x)+(BA)(\v y) \
        (BA)(c\v x) &= B\big(A(c\v x)\big) \
        &= B\big(cA(\v x)\big) \
        &= cB\big(A(\v x)\big) \
        &= c(BA)(\v x)
    \end{align*}
    which shows that $BA$ is linear.

Let $A\in L(X)$ be invertible. Let $\v x,\v y\in X$ and let $c$ be a scalar. Then there are $\v x',\v y'\in X$ such that
    $$
        A(\v x')=\v x, \quad A(\v y')=\v y, \quad A^{-1}(\v x)=\v x', \quad A^{-1}(\v y)=\v y'
    $$
    Then
    \begin{align*}
        A^{-1}(\v x+\v y) &= A^{-1}\big(A(\v x')+A(\v y')\big) \
        &= A^{-1}\big(A(\v x'+\v y')\big) \
        &= \v x'+\v y' \
        &= A^{-1}(\v x)+A^{-1}(\v y) \
        A^{-1}(c\v x) &= A^{-1}\big(cA(\v x')\big) \
        &= A^{-1}\big(A(c\v x')\big) \
        &= c\v x' \
        &= cA^{-1}(\v x)
    \end{align*}
    which shows that $A^{-1}$ is linear. Since $A^{-1}$ is 1-1 and maps $X$ onto itself, it is also invertible.

Exercise 9.2

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Exercise 9.2

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