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7983Geometrical Interpretation of Hyperplanes

As we have seen in the above proposition, a hyperplane solving a linear equation with $n=2$ unknowns is a straight line. But what about the general case $n\ge 1$?

If $A=(\alpha_1,\ldots,\alpha_n)\in\mathbb R^n$ is not the origin, and $\beta\in\mathbb R$, the linear equation might have the form

$$\alpha_1x_1+\alpha_2x_2+\ldots+\alpha_nx_n=\beta.\quad ( * ) $$

We have seen how to identify points with vectors in number spaces. By interpreting the points $P\in \mathbb R^n$ solving the equation $( * )$ as vectors $\overrightarrow{OP}$ we can easily interpret hyperplanes geometrically. In order to do so, we first have to reformulate the equation $( * )$ a little bit. We can re-index the $\alpha_i$ to achieve that $\alpha_1\neq 0$. In this case, we can solve the linear equation for $x_1$ and get

$$x_1=\frac{\beta}{\alpha_1}-\frac{\alpha_2}{\alpha_1}x_2-\ldots-\frac{\alpha_n}{\alpha_1}x_n.\quad( * * )$$

Therefore, the value of the unknown $x_1$ depends on the specific choices of the values of the unknowns $x_2, \ldots,x_n$. In other words, we can represent $x_1$ by choosing specific values of the other variables. Any choice of these $n-1$ variables will give us one value of $x_1$. Thus, the set of all solutions of an equation of $n$ unknowns can be completely described by all possible choices of $n-1$ variables. In order to illustrate this, we write any point $P$ solving the equation $( * )$ as a vector $\overrightarrow{OP}$ with the coordinates $x_1,\ldots,x_n$, in which we replace $x_1$ by the expression $( * * )$:

$$\overrightarrow{OP}=\pmatrix{\frac{\beta}{\alpha_1}-\frac{\alpha_2}{\alpha_1}x_2-\ldots-\frac{\alpha_n}{\alpha_1}x_n\\x_2\\\vdots\\x_n}.$$

While the vector $\overrightarrow{OP}$ depends on the values of the $n-1$ variables $x_2,\ldots,x_n$, we can use vector addition and scalar multiplication to write it as a sum of vectors depending on only one of these variables each:

$$\pmatrix{\frac{\beta}{\alpha_1}-\frac{\alpha_2}{\alpha_1}x_2-\ldots-\frac{\alpha_n}{\alpha_1}x_n\\x_2\\\vdots\\x_n}=\pmatrix{\frac{\beta}{\alpha_1}\\0\\\vdots\\0}+x_2 \pmatrix{-\frac{\alpha_2}{\alpha_1}\\1\\\vdots\\0}+\cdots+x_n\pmatrix{-\frac{\alpha_n}{\alpha_1}\\0\\\vdots\\1}.$$

In case the equation $( * )$ is homogeneous, this representation simplifies to

$$\pmatrix{-\frac{\alpha_2}{\alpha_1}x_2-\ldots-\frac{\alpha_n}{\alpha_1}x_n\\x_2\\\vdots\\x_n}=x_2 \pmatrix{-\frac{\alpha_2}{\alpha_1}\\1\\\vdots\\0}+\cdots+x_n\pmatrix{-\frac{\alpha_n}{\alpha_1}\\0\\\vdots\\1}.$$

Obviously, the $n-1$ vectors $$\pmatrix{-\frac{\alpha_2}{\alpha_1}\\1\\\vdots\\0},\cdots,\pmatrix{-\frac{\alpha_n}{\alpha_1}\\0\\\vdots\\1}$$ are not multiplies of each other, thus they form an $n-1$ dimensional basis of an $n-1$ dimensional number space $H\subseteq \mathbb R^{n}$. This is the reason, why a hyperplane $H$ being a solution of a linear equation with $n$ unknowns has always a dimension which is one less than the number of dimensions of the space $\mathbb R^n$ it is part of.

Now, it is easy to use all these observations to conclude, how to interpret hyperplanes for different values of $n$.

Case $n=1$

A linear equation in the one-dimensional number line $\mathbb R^1$ has the form $\alpha_1x_1=\beta,$ and the hyperplane solving it is always a zero-dimensional point $\pmatrix{\frac{\beta}{\alpha_1}}.$

Case $n=2$

A linear equation in the two-dimensional number plane $\mathbb R^2$ has the form $\alpha_1x_1+\alpha_2x_2=\beta,$ and the hyperplane solving it is always a one-dimensional straight line
$$\pmatrix{\frac{\beta}{\alpha_1}\\0}+x_2 \pmatrix{-\frac{\alpha_2}{\alpha_1}\\1}.$$

Case $n=3$

A linear equation in the three-dimensional number space $\mathbb R^3$ has the form $\alpha_1x_1+\alpha_2x_2+\alpha_3x_3=\beta,$ and the hyperplane solving it is always a two-dimensional plane

$$\pmatrix{\frac{\beta}{\alpha_1}\\0\\0}+x_2 \pmatrix{-\frac{\alpha_2}{\alpha_1}\\1\\0}+x_3\pmatrix{-\frac{\alpha_3}{\alpha_1}\\0\\1}.$$

Case $n\ge 4$

A linear equation in the $n$-dimensional number space $\mathbb R^n$ has the form $\alpha_1x_1+\ldots+\alpha_n x_n=\beta,$ and the hyperplane solving it is always an $n-1$-dimensional hyperplane

$$\pmatrix{\frac{\beta}{\alpha_1}\\0\\\vdots\\0}+x_2 \pmatrix{-\frac{\alpha_2}{\alpha_1}\\1\\\vdots\\0}+\cdots+x_n\pmatrix{-\frac{\alpha_n}{\alpha_1}\\0\\\vdots\\1}.$$

| | | | Contributors: bookofproofs | References: [7937]


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Bibliography (further reading)

[7937] Knabner, P; Barth, W.: “Lineare Algebra – Grundlagen und Anwendungen”, Springer Spektrum, 2013

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