I have an equation which describes a line, plane or higher dimensional object in the form.


So for example:

$\begin{bmatrix}1 & 1\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} = 1$

This is a line through the points $(0,1)$ and $(1,0)$. How can i systematically write it to the form $P+\alpha Q=x$:

$\begin{bmatrix}1 \\ 0\end{bmatrix} + \alpha \begin{bmatrix}1 \\ -1\end{bmatrix} = \begin{bmatrix}x_1 \\ x_2\end{bmatrix}$

I am looking on how to do it with matrix operations only, so i can do it systematically, also with higher-dimensional equations.

  • 1
    $\begingroup$ If you are ok, you can set as solved. Thanks! $\endgroup$ – user Dec 6 '17 at 13:12
  • $\begingroup$ @gimusi, sorry, I didn't look at this for the last few days. Thanks for the help though. I added some questions to your answer since it is for me not fully clear yet, but i will accept the answer after that. $\endgroup$ – user3053216 Dec 6 '17 at 13:25

If $Ax=b$ we are dealing with vector spaces, thus let's consider a basis for [x]:$[v_1], [v_2],...,[v_m]$

you can find $\alpha_i \in \mathbb{R}$, $i=1,m$ such that:

$\begin{bmatrix}x_1 \\|\\ x_m\end{bmatrix}=\alpha_1\begin{bmatrix}v_{11} \\|\\ v_{1m}\end{bmatrix} +...+ \alpha_m\begin{bmatrix}v_{n1} \\|\\ v_{nm}\end{bmatrix}$

thus $P=0$.

Whereas $P+\alpha Q=x$ is about affine spaces.


  • $\begingroup$ Thanks for your answer. I have two questions: Should it not be the basis of A instead of [x]? And if not, how to find this basis. The second question is should the lost column not be vn1 - vnm? $\endgroup$ – user3053216 Dec 6 '17 at 13:23
  • 1
    $\begingroup$ I'm referring to [x] as the subspace that satisfy the system Ax=b (think for example to the Null Space related to Ax=0). How to find a basis its a pure question of Linear Algebra, I can't do a course of L.A. here but you can find all details in any good book of L.A. (I recomend G. Strang books for an elementary but strong indroduction to the subject). Yes it should be vni. $\endgroup$ – user Dec 6 '17 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.