My book first defines a hyper plane in $R^n $ as set $H= \{p^tx=\alpha \} $ where $p $ is a nonzero vector and $\alpha $ is scalar, or equivalently as the set $H=\{x:p^t(x-\bar{x })=0$.

Next it states that the set defined as $\{(x,y): y=f(\bar {x } )+ \xi^t(x-\bar {x } )$ is a hyperplane.

But how can this later set be written in either of the first two forms?

Thanks in advance!

These two definitions of a hyperplane are equivalent?


The equation $y=f(\bar {x } )+ \xi^t(x-\bar {x } )$ can be rewritten $y=\xi^t(x-\bar {x } )+f(\bar {x } )-y=0$, which has the form $$(\xi,-1)^t\bigl((x,y)-(\bar x,f(\bar x)\bigr)=0.$$ (You will need to read $(a,b)$ above as a column vector with $a$ above $b$.)


$\newcommand{\R}{\mathbf{R}}$Your book's assertion is correct. To see why, note that the pair $X := (x, y)$, which may be viewed as an element of $\R^{n+1}$ in the second instance, plays the role of $x$ in the definition. Setting $$ \Xi = (\xi, -1),\quad \bar{X} = \bigl(\bar{x}, f(\bar{x})\bigr), $$ the condition $$ y = f(\bar{x}) + \xi^{t}(x - \bar{x}) $$ can be written $$ \Xi^{t}(X - \bar{X}) = 0. $$

Your two conditions are not equivalent, since not every non-zero vector in $\R^{n+1}$ has a non-zero final component. That is, the "graph-like" definition cannot accommodate hyperplanes whose normal vector has final component equal to zero. (Analogously, every line in the plane is the solution set of some equation $ax + by = c$, and every equation $y = b + mx$ defines a line, but not every line can be written in slope-intercept form.)


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