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Let $T$ is a linear transformation on $\mathbb{R}^2$. $x,y$ linearly indipendent vector in $\mathbb{R}^2$, $T(y)=\alpha x$ and $T(x)=0$, Then with respect to some basis in $\mathbb{R}^2$, $T$ is of the form...

  1. $$\pmatrix{a&0\\0&a},\;a>0$$

  2. $$\pmatrix{a&0\\0&b},\;a,b>0,a\ne b$$

  3. $$\pmatrix{0&1\\0&0}$$

  4. $$\pmatrix{0&0\\0&0}$$

I have calculated the answer to be 3. Can anyone verify this?

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Please make sure that I used the matrices that you intended. –  Brian M. Scott Jun 11 '12 at 6:10
    
thank you Brian, its fine. –  El Angel Exterminador Jun 11 '12 at 6:11
    
Your answer is correct. –  Brandon Carter Jun 11 '12 at 6:11
    
The correct answer is indeed (3): it’s the only one that has a range of the right dimension, namely, $1$. –  Brian M. Scott Jun 11 '12 at 6:11
    
thank you....... –  El Angel Exterminador Jun 11 '12 at 6:12

2 Answers 2

up vote 1 down vote accepted

We can reason as follows:

If $x,y$ are linearly independent then $y,x\neq 0$. Since $Tx=0$ we know that the transformation has a non-trivial kernel. The first two matrices do not have this property, so they are out.

We also know that $Ty\neq 0$, therefore the transformation is not properly $0$, so the last one is out.

$\implies$ It is the third matrix.

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Well, first I would try the basis $B_1=\{x,y\}$. With the data given, the matrix representation of $T$ with the basis $B_1$ for domain and range will be: $\begin{bmatrix} 0 & \alpha \\ 0 & 0 \end{bmatrix}$.

Given the range of options you have above, I would assume that $\alpha \neq 0$, and look at the basis $B_2=\{\alpha x,y\}$. With the basis $B_2$ for domain and range, the matrix representation of $T$ will be: $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$.

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