gorgeousgen9487

2022-07-06

$A=\left(\begin{array}{cc}k& -2\\ 1-k& k\end{array}\right)\text{, where k is a constant}$

Valeria Wolfe

Expert

The line $y=2x$ is mapped to itself if we have that for each point $\left(x,y\right)$ on the line $T\left(x,y\right)$ also lies on the line. We needn't have $T\left(x,2x\right)=\left(x,2x\right)$ for all $x$, it suffices to have $T\left(x,2x\right)=\left(\varphi \left(x\right),2\varphi \left(x\right)\right)$ for all $x$ and some function $\varphi$.
Having this in mind, we look at $T\left(x,2x\right)=\left(x\left(k-4\right),x\left(k+1\right)\right)$. We must have $2\left(k-4\right)=k+1$ for $\left(x\left(k-1\right),x\left(k+1\right)\right)$ to lie on our line. This gives $k=9$

racodelitusmn

Expert

On the same line (!) of thought: the line $\phantom{\rule{thinmathspace}{0ex}}l:y=2x\phantom{\rule{thinmathspace}{0ex}}$ is the same as the vector space $\phantom{\rule{thinmathspace}{0ex}}\mathrm{Span}\left\{\left(1,2\right)\right\}\le {\mathbb{R}}^{2}$ , or if you prefer: $\phantom{\rule{thinmathspace}{0ex}}l:\left\{\left(r,2r\right)\phantom{\rule{thinmathspace}{0ex}}/\phantom{\rule{thinmathspace}{0ex}}r\in \mathbb{R}\right\}\phantom{\rule{thinmathspace}{0ex}}$, and then what we really want to happen is
$\left(\begin{array}{cc}k& -2\\ 1-k& k\end{array}\right)\left(\begin{array}{c}1\\ 2\end{array}\right)=\left(\begin{array}{c}r\\ 2r\end{array}\right)⟹\begin{array}{}k=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}r+4\\ k=2r-1\end{array}$
so $\phantom{\rule{thinmathspace}{0ex}}r+4=2r-1⟹r=5\phantom{\rule{thinmathspace}{0ex}}$ and thus $\phantom{\rule{thinmathspace}{0ex}}k=9\phantom{\rule{thinmathspace}{0ex}}$

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