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$$\begin{bmatrix}a&b\\\\c&d\end{bmatrix} \longrightarrow \begin{bmatrix}a\\\\a\\\\3a+b\end{bmatrix}$$

As the kernel maps everything to zero then $a=0$ and $3a+b=0$ so the basis is of the form

$$\left\{\begin{bmatrix}0&0\\\\1&0\end{bmatrix},\begin{bmatrix}0&0\\\\0&1\end{bmatrix}\right\}$$

However i'm not sure if this is correct so far

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Looks good. Have you tried to find the image yet? –  icurays1 Dec 4 '12 at 18:05
    
Do you mean $3a + b = 0$? (it doesn't change the solution, of course). –  Christopher A. Wong Dec 4 '12 at 18:06
    
@ChristopherA.Wong yes I have changed it now –  Adam Dec 4 '12 at 18:32
    
@icurays1 No I'm not sure where to start for the image –  Adam Dec 4 '12 at 18:37

1 Answer 1

To find the image of this transformation, I would first break up my image vector into a sum:

$$ \left[\begin{array}{c} a\\a\\3a+b \end{array}\right]= \left[\begin{array}{c} a\\a\\3a \end{array}\right]+ \left[\begin{array}{c} 0\\0\\b \end{array}\right]=a\left[\begin{array}{c} 1\\1\\3 \end{array}\right]+ b\left[\begin{array}{c} 0\\0\\1 \end{array}\right] $$

You should be able to take it from there (hint: what do you need to show to prove something is a basis?)

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so from here I just need to show they are linearly independant and span –  Adam Dec 4 '12 at 19:07
    
Yep, you've got it! –  icurays1 Dec 4 '12 at 19:07

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