Injective, surjective map!

Hello everyone!

May anyone can help in this case.

I have a map F: $\displaystyle \left F( \begin{bmatrix}

d_1_1 & d_1_2 & \ d_1_3 \\

d_2_1 & d_2_2 & d_2_3 \\

d_3_1 & d_3_2 & d_3_3

\end{bmatrix}) \right = \begin{bmatrix}

d_1_1 & d_1_2-d_2_1 & d_3_1 \\

0 & 2d_2_2 & d_2_1-d_1_2 \\

0 & 0 & 3d_3_3

\end{bmatrix}$

in a different note:

$\displaystyle \left F( \begin{bmatrix}

d_1_1 & d_1_2 & \ d_1_3 \\

d_2_1 & d_2_2 & d_2_3 \\

d_3_1 & d_3_2 & d_3_3

\end{bmatrix}) \right = \begin{bmatrix}

g_1_1 & g_1_2 & g_1_3 \\

0 & g_2_2 & g_2_3 \\

0 & 0 & g_3_3

\end{bmatrix}$

eg. $\displaystyle \left F( \begin{bmatrix}

0 & 3 & \ 6 \\

1 & 4 & 7 \\

2 & 5 & 8

\end{bmatrix}) \right = \begin{bmatrix}

0 & 2 & 2 \\

0 & 8 & -2 \\

0 & 0 & 24

\end{bmatrix}$.

I must to find out if this map is linear, surjective, injective?

The first point I already done, so I know the map is linear. I have more problems with second and third point. My thinking:

1) the map is not injective because:

$\displaystyle for \ d_2_1 \ne d_3_1 \ is \ F(d_2_1)=F(d_3_1) \ or \ g_2_1=g_3_1=0$

2) but the map is surjective because:

for every g in G exits d in D that is F(d)=g

Are those two arguments correct?

Thanks in advance!

Enrico