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$$A= \begin{pmatrix} k & -2 \\ 1-k & k \end{pmatrix}\text{, where k is a constant}$$

$$\text {A transformation } T : \mathbb{R}^2 \rightarrow \mathbb{R}^2 \text{ is represented by the matrix A.}$$

$$\text {Find the value of k for which the line } y = 2x \text{ is mapped onto itself under T.}$$

Working: $$\begin{pmatrix} k & -2 \\ 1-k & k \end{pmatrix}\cdot\begin{pmatrix} x \\ 2x \end{pmatrix}=\begin{pmatrix} x \\ 2x \end{pmatrix}$$ $$\begin{pmatrix} x(k-4) \\ x(1+k) \end{pmatrix}=\begin{pmatrix} x \\ 2x \end{pmatrix}$$ $$x(k-4)=x$$ $$x(1+k)=2x$$

Leaving me with $k=5$ and $k=1$, However the answer is $k = 9$ why?

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up vote 3 down vote accepted

The line $y = 2x$ is mapped to itself if we have that for each point $(x,y)$ on the line $T(x,y)$ also lies on the line. We needn't have $T(x,2x) = (x,2x)$ for all $x$, it suffices to have $T(x,2x) = \bigl(\phi(x), 2\phi(x)\bigr)$ for all $x$ and some function $\phi$.

Having this in mind, we look at $T(x,2x) = \bigl(x(k-4), x(k+1)\bigr)$. We must have $2(k-4) = k+1$ for $(x(k-1), x(k+1))$ to lie on our line. This gives $k=9$.

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+1. In fact, Construct the matrix A and multiply it with $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$, you get $\begin{pmatrix} 5 \\ 10 \end{pmatrix}$ – Inquest May 23 '12 at 11:00

On the same line (!) of thought: the line $\,l:y=2x\,$ is the same as the vector space $\,\operatorname{Span}\{(1,2)\}\leq\mathbb{R}^2$ , or if you prefer: $\,l:\{(r,2r)\,/\,r\in\mathbb{R}\}\,$ , and then what we really want to happen is $$\begin{pmatrix}k&-2\\1-k&k\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}r\\2r\end{pmatrix}\Longrightarrow \begin{array}\\k =\,\,\,r+4\\k=2r-1\end{array}$$so $\,r+4=2r-1\Longrightarrow r=5\,$ and thus $\,k=9\,$

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