Exercise 1.18

Question

Exercise 18: If $k \geq 2$ and $x \in R^k$, prove that there exists $y \in R^k$ such that $y 
eq 0$ but $x \cdot y = 0$ ($\langle x , y angle = 0$).  Is this also true if $k = 1$?

ghostofgarborg · Accepted Answer

(	extit{Matt ``frito'' Lundy}) \
    This solution uses some linear algebra.  
    If $x = 0$ then any non-zero $y$ will suffice, so assume that $x 
eq 0$.
    Also choose any $z \in R^k$ such that $x$ and $z$ are linearly indepedent (this is possible because $k \geq 2$). Then let
    $$ y = z - \frac{\langle z, x angle}{\langle x , x angle} x.$$
    Notice that $y 
eq 0$ because $x$ and $z$ are linearly independent and $y$ is a linear combination of $x$ and $z$ with not all coefficients equal to zero.  And so we have
    \begin{align*}
        \langle x ,y angle &= \langle x , z - \frac{\langle z, x angle}{\langle x , x angle} x angle\
        &= \langle x , z angle -\frac{\langle z ,x angle \langle x , x angle}{\langle x, x angle} \
        &= \langle x , z angle - \langle x , z angle \
        &= 0
    \end{align*}
    There is trouble when $k = 1$ because any two vectors in $R^1$ are linearly dependent.

Amit Awasthi · Answer

\begin{proof}
      Let $\mathbf{x}$ be given. We will construct such a $\mathbf{y}$. By Definition 1.36,
      \begin{equation}\label{18}
        \mathbf{x}\cdot\mathbf{y} = \sum_{i=1}^k x_iy_i.
      \end{equation}
      Now there are two cases for $\mathbf{x}$. If $2|k$, then we can let
      \begin{equation*}
        \mathbf{y} = (x_2,-x_1,\ldots,x_k,-x_{k-1}).
      \end{equation*}
      By construction we see that $\eqref{18} = 0$. Now consider the case where $2
mid k$. Then, we can construct $\mathbf{y}$ as above, except for one index $y_j$ which will be 0. Then,
      \begin{equation*}
        \mathbf{y} = (x_2,\ldots,-x_{j-2},0,x_{j+2},\ldots,-x_{k-1}).
      \end{equation*}
      By construction we see that $\eqref{18} = 0$. This construction, however, does not work for $k = 1$, since if $\mathbf{x} 
e 0$, there is no number it could be multiplied by to get 0.
\end{proof}

Exercise 1.18

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Exercise 1.18

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