Mathematics Asked by Kurt Muster on November 12, 2021
How can I generally solve equations of the form $mathbf{A} mathbf{w} =
begin{pmatrix} x \ y \ z end{pmatrix}
times mathbf{w}$ for the matrix $mathbf{A},$ where $mathbf{w}$ can be any vector? I recognize that you could just set $mathbf{w}$ to a vector with simple values, such as $begin{pmatrix} 1 \ 2 \ 1 end{pmatrix}$, but doing so still isn’t helpful. Also, $x,$ $y,$ and $z$ are entirely independent variables.
OK, let's put it other way as $mathbf{w}times mathbf{v}=-mathbf{A}mathbf{w}$. We can write the the cross product as vector-matrix multiplication: $$mathbf{w}timesmathbf{v} =[mathbf{w}]_times mathbf{v}=begin{bmatrix},0&!-w_{3}&,,w_{2}\,,w_{3}&0&!-w_{1}\-w_{2}&,,w_{1}&,0end{bmatrix}mathbf{v}.$$ So you can write your equation as a system of linear equations $$[mathbf{w}]_times mathbf{v}=-mathbf{A}mathbf{w}.$$ Matrix $[mathbf{w}]_times$ has rank $2$ and its nullspace is spanned by $[w_1,,w_2,,w_3]^top$.
Now depending on whether you assume $w_2neq 0$ or $w_3neq 0$, you can transform this system and find a particular solution. However, this solution can be found only if $langlemathbf{w},mathbf{Aw}rangle=0$. In particular, this implies that $mathbf{A}^top=-mathbf{A}$.
Answered by orthxx on November 12, 2021
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