Exercise 9.25

Question

Exercise 25: Suppose $A\in L(\R n,\R m)$, let $r$ be the rank of $A$.

(a) Define $S$ as in the proof of Theorem 9.32. Show that $SA$ is a projection in $\R n$ whose null space is $\Null(A)$ and whose range is $\Range(S)$.

(b) Use (a) to show that
    $$
        \dim\Null(A)+\dim\Range(A)=n.
    $$

analambanomenos · Accepted Answer

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(a) If $r$ equals the rank of $A$, and $\Range(A)$ is spanned by the independent set $\v y_1,\ldots,\v y_r$, then $\v y_i=A\v z_i$, $i=1,\ldots,r$ for some independent set $\v z_1,\ldots,\v z_r$
    in $\R n$, and $S$ is a map of $\Range(A)$ into $\R n$ defined as
    $$
        S(c_1\v y_1+\cdots+c_r\v y_r)=c_1\v z_1+\cdots+c_r\v z_r.
    $$
    Note that $S$ is a one-to-one map of $\Range(A)$ into $\R n$.
    Following the hint, note that
    $$
        AS(c_1\v y_1+\cdots+c_r\v y_r)=A(c_1\v z_1+\cdots+c_r\v z_r)=c_1\v y_1+c_r\v y_r,
    $$
    that is, $AS(\v y)=\v y$ for $\v y\in\Range(A)$. Hence, for $\v x\in\R n$,
    $$
        SASA(\v x)=S\Big(AS\big(A(\v x)\big)\Big)=S\big(A(\v x)\big)=SA(\v x)
    $$
    so $SASA$ is a projection on $\R n$.

If $SASA(\v x)=SA(\v x)=\v 0$, then $A(\v x)=\v 0$ since $S$ is one-to-one, so the null space of $SASA$ is $\Null(A)$. The range of $SASA$ is clearly a subset of $\Range(S)$, and if
    $\v z=c_1\v z_1+\cdots+c_r\v z_r\in\Range(S)$, then
    $$
        SASA(\v z)=SA(\v z)=S(c_1\v y_1+\cdots+c_r\v y_r)=\v z
    $$
    so that the range of SASA is $\Range(S)$.

(b) From the discussion in 9.31, you can conclude that if $P$ is a projection in $\R n$, then $n=\dim\Null(P)+\dim\Range(P)$. Hence from part (a), we have
    \begin{align*}
        n &= \dim\Null(SASA)+\dim\Range(SASA) \
        &= \dim\Null(A)+\dim\Range(S) \
        &= \dim\Null(A)+\dim\Range(A)
    \end{align*}
    where the last equality follows from the fact that $S$ is a one-to-one map on $\Range(A)$.

Exercise 9.25

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Exercise 9.25

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