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I was thinking about the following problem:

Let $x,y$ be linearly independent vectors in $\mathbb R^2$. Suppose,$T:\mathbb R^2 \rightarrow \mathbb R^2$ is a linear transformation such that $Ty=\alpha x$ and $Tx=0.$ Then with respect to some basis in $\mathbb R^2$, $T$ is of the form:
(a)\begin{pmatrix} a & 0\\ 0 & a \end{pmatrix}, where $a>0$

(b)\begin{pmatrix} a & 0\\ 0 & b \end{pmatrix}, where $a,b>0, a \neq b$
(c)\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} (d)\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}

I do not know how to proceed with the problem. If I take $T$ of the form \begin{pmatrix} a & 0\\ 0 & a \end{pmatrix}, where $a>0$
and then after satisfying the given conditions $Ty=\alpha x$ and $Tx=0,$ i see that $x,y$ are linearly independent.But i want a more direct way of solving it. Could someone point me in the right direction?Thanks in advance for your time.

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If you take as a basis $\{x,y\}$, which matrix is associated to $T$? (Remember that the columns of the matrix are the coordinates of $Tx$ and $Ty$ with respect to the basis $\{x,y\}$). Now, can you change a bit the basis $\{x,y\}$ so that the matrix of $T$ looks like one of the given ones? – wisefool Dec 18 '12 at 15:18
@wisefool: Perfect. – Sugata Adhya Dec 18 '12 at 16:04
up vote 0 down vote accepted

Clearly $\,\{x,y\}\,$ is a basis of $\,\Bbb R^2\,$ , so also $\,A:=\{x,\alpha^{-1}y\}\,$ is a basis (assuming $\,\alpha\neq 0\,$), so:

$$\begin{align*}Tx&=0=&0\cdot x+0\cdot \alpha^{-1}y\\T(\alpha^{-1}y)&=x=&1\cdot x+0\cdot\alpha^{-1}y\end{align*}\,\,\Longrightarrow\;\;\;[T]_A:=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$

Of course, the above is just the fleshing out of wisefool's comment.

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