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Could somebody please check my solution?

Let $K$ be a field. Given is the $K$-linear transformation

$f:Mat (3 \times 2, K) \rightarrow K^3$

$\begin {pmatrix} a_{1 1} & a_{2 1}\\ a_{2 1} & a_{2 2} \\ a_{3 1} & a_{3 2} \end{pmatrix} \longmapsto \begin {pmatrix} a_{1 1} + a_{2 2}\\ a_{2 1} + a_{3 2} \\ a_{3 1} + a_{1 2} \end{pmatrix} $

For $i \in \{1,2,3\}$ and $j \in \{1,2\}$ let $E_{i,j} \in Mat (3 \times 2, K)$ the Matrix that has a $1$ in row $i$ and column $j$ and other entries are $0$. Then is $B:=\{E_{ij } | i \in \{1,2,3\}, j \in \{1,2\}\} $ a Bais of $Mat (3 \times 2, K)$. Let furthermore $C=\{e_q,e_2,e_3\}$ the standard basis of $K^3$.

(i) Determine the Transformation matrix $_BM_C(f)$ of $f$ relating to bases $B$, $C$.

(ii) Determine the Dimension of $Im(f)$ and $Ker(f)$

To (i): My solution is, that $_BM_C(f) = \begin {pmatrix} b_{1 1} + b_{2 2}\\ b_{2 1} + b_{3 2} \\ b_{3 1} + b_{1 2} \end{pmatrix}$

To (ii): The whole dimension is $3$ and so my idea was to determine the kernel, so $\begin {pmatrix} a_{1 1} + a_{2 2}\\ a_{2 1} + a_{3 2} \\ a_{3 1} + a_{1 2} \end{pmatrix} = \begin {pmatrix} 0 \\0 \\ 0 \end{pmatrix} $ So contains the Kernel only the null vector?

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Ad (i): $\dim \operatorname{Mat}(3 \times 2,K) = 3 \cdot 2 = 6$, whilst $\dim K^3 = 3$. Hence, $_B M_C(f)$ must be a $3 \times 6$ matrix. Ad (ii): What is $f(A)$ for $A =\begin{pmatrix} 1 & 2 \\ 3 & -1 \\ -2 & -3 \end{pmatrix}$? –  Branimir Ćaćić Dec 16 '13 at 7:08
    
to (ii) it's $\begin {pmatrix} 0 \\0 \\ 0 \end{pmatrix}$. How can i determine $_BM_C$? –  fear.xD Dec 16 '13 at 7:57

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