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Let:

\begin{equation} \mathbf{A} = \begin{bmatrix} 0.8 & 0.4 & 0.4 & 0.8\\ 0.4 & 0.8 & -0.4 & -0.8 \end{bmatrix} \end{equation}

We can find a null space basis of $\mathbf{A}$ whose vector's components absolute values are (at best) equals by combining its columns and trying to produce zero. If it is not possible that all components are equals, a maximum number of them should be.

\begin{equation} \mathbf{x} = \begin{bmatrix} 1 & 1 \\ -1 & -1 \\ 1 & -1 \\ -1 & 0 \\ \end{bmatrix} \end{equation}

This is done by hand on a simple matrix. Is there a way to calculate (if it exists) in a systematic way a null space basis for which the absolute values of the vector's components are equals, for an arbitraty matrix?

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    $\begingroup$ I don't understand what you mean by "whose vector's components are equals". When we are asked for the null space of a matrix we are given only the matrix, not a vector. The vectors in the null space may or may not have "all components equal". $\endgroup$ Sep 15, 2022 at 12:45
  • $\begingroup$ I edited my question (the absolute values of the components should be equal). I mean that the components of the vectors of the basis have equal absolute values. Whether it is possible to find it or not is the entire question, but if not, the maximum number of components of each vector should be. $\endgroup$
    – Gab
    Sep 15, 2022 at 13:03

1 Answer 1

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The null space of $\begin{bmatrix}0.8 & 0.4 & 0.4 & 0.80 \\ 0.4 & -0.4 & 0.8 & -0.8 \end{bmatrix}$ is the space of all vectors $\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$ such that $\begin{bmatrix}0.8 & 0.4 & 0.4 & 0.8 \\ 0.4 & -0.4 & 0.8 & -0.8 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}= \begin{bmatrix}0.8a+ 0.4b+ 0.4c+ 0.8d \\ 0.4a- 0.4b+ 0.8c- 0.8d\end{bmatrix}= \begin{bmatrix}0 \\ 0\end{bmatrix}$.

So we have 0.8a+ 0.4b+ 0.4c+ 0.8d= 0 and 0.4a- 0.4b+ 0.8c- 0.8d= 0. if we add the two equations we get 1.2a+ 1.2c= 0 so c= -a. If we subtract the second equation from the first we get 0.4a+ 0.8b- 0.4c+ 1.6d= 0.8a+ 0.8b+ 1.6d= 0 so b= -2d- a. We can write $\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}= \begin{bmatrix}a \\ -2d- a \\ -a \\ d \end{bmatrix}= a\begin{bmatrix} 1 \\ -1 \\ -1 \\ 0 \end{bmatrix}+ d\begin{bmatrix}0 \\ -2 \\ 0 \\ 1 \end{bmatrix}$.

The null space is the two dimensional space with basis $\begin{bmatrix} 1 \\ -1 \\ -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix}0 \\ -2 \\ 0 \\ 1 \end{bmatrix}$.

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  • $\begingroup$ Thank you for your answer. This does not answer the question though, as I'm asking for a generic method that would constrain the null space basis in a certain way: equal absolute values of all the components of the basis's vectors, and if not possible, a maximum number of component's absolute values should be equal, as in the basis found in the question formulation $\endgroup$
    – Gab
    Sep 16, 2022 at 7:44

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