Exercise 9.5

I’m going to show this by induction on $n$. For the case $n=1$, let $A{ě}_1=k$, and let $\check{y}=k{ě}_1$. If $\check{x}\in 1$, then $\check{x}=x{ě}_1$ for some scalar $x$, so $A\check{x}=\mathit{xA}{ě}_1=\mathit{ck}=\check{x}\cdot \check{y}$.

Suppose the assertion is true for the case $n$ and let $A\in L(n+1,1)$. Restricting $A$ to the subspace $n$ (spanned by ${ě}_1,\dots ,{ě}_n$) yields an element of $L(n,1)$, so by the induction assumption there is a ${\check{y}}^{\prime }\in n$ such that $A{\check{x}}^{\prime }={\check{x}}^{\prime }\cdot {\check{y}}^{\prime }$ for all ${\check{x}}^{\prime }\in n$. If $A{ě}_{n_1}=k$, let $\check{y}={\check{y}}^{\prime }+k{ě}_{n+1}$. If $\check{x}\in n+1$, then $\check{x}={\check{x}}^{\prime }+x{ě}_{n+1}$ for some ${\check{x}}^{\prime }\in n$ and some scalar $x$. Then, since the scalar product of ${ě}_{n+1}$ with every element of $n$ is 0, we have

If $\left|\check{x}\right|\le 1$, then $\left|A\check{x}\right|=\left|\check{x}\cdot \check{y}\right|\le \left|\check{x}\right|\left|\check{y}\right|\le \left|\check{y}\right|$, so that $A\le \left|\check{y}\right|$. Let $\widetilde{\stackrel{ˇ}{y}}=\check{y}∕\left|\check{y}\right|$. Then $\left|\widetilde{\stackrel{ˇ}{y}}\right|=1$ and

Hence $A=\left|\check{y}\right|$.