TITLE: I have the following matrix B and I need to find matrix A so that kerA=ImB and A has as few rows as possible
QUESTION [1 upvotes]: Every thing is considered in Z_3 (modulo 3).
B =\begin{bmatrix}1&1&1\\0&1&2\\2&1&0\\0&2&2\end{bmatrix}
I think that the correct path is to show that ImB ⊆ KerA which is when AB=0 and KerA⊆ImB at the same time implies KerA=ImB. But I don't know when that is true. I also need to prove that A cannot have fewer rows.

REPLY [0 votes]: The columns of $B $ are linearly independent so the dimension of the image of $B $ is $3$. $A $ has to have $4$ columns for multiplication to work. Hence by a theorem we have that the image of $A $ has dimension $1$. So $A $ can be taken as a $1\times 4$ matrix. This row vector can be found by finding the orthogonal complement of image of $B $ in $\mathbb {R}^4$.
One way to do this is to solve for $A=(x_1,x_2,x_3,x_4) $ so that $AB=0$. Equivalently $B^TA^T=0$.
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