Hey I want to check my solutions to this exercise:
a) Let $B\in M_{3,2}(\mathbb{R})$ be a matrix. Show that there is at most one matrix $A \in M_{2,3}(\mathbb{R})$ such that $BA=\begin{pmatrix} 1 &0 &0 \\ 0& 1&0 \\ 0& 0 &0 \end{pmatrix}$.
b) Show that if such an $A$ exists, then $AB= \begin{pmatrix} 1 & 0\\ 0& 1 \end{pmatrix}$.
[In order for $AB$ to be defined, the number of columns in $A$ must be equal to the number of rows in $B$. Similarly, for $BA$ to be defined, the number of columns in $B$ must be equal to the number of rows in $A$.
Thus, the dimensions of $A$ must be $2 \times 3$ in order for $AB$ to be defined, and the dimensions of $B$ must be $3 \times 2$ in order for $BA$ to be defined.]
a) To show that $A$ is unique, suppose that there exist two matrices $A_1$ and $A_2$ such that $B\cdot A_1 = B \cdot A_2 = \begin{pmatrix} 1 &0 &0 \\ 0& 1&0 \\ 0& 0 &0 \end{pmatrix}$. Then we have $B\cdot A_1 = B\cdot A_2$, and multiplying both sides on the left by $B^{-1}$ (if $B$ is invertible) or by the pseudoinverse of $B$ (if $B$ is not invertible) yields $A_1 = A_2$. Therefore, $A$ is unique.
b) Assuming that such an $A$ exists, we can multiply both sides of the equation $B\cdot A = \begin{pmatrix} 1 &0 &0 \\ 0& 1&0 \\ 0& 0 &0 \end{pmatrix}$ on the right by $B$ to obtain $AB = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.
[To show that $BA = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$, we first note that the first two rows of $B\cdot A$ are linearly independent, which implies that the rank of $B\cdot A$ is at least $2$. However, the rank of $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ is also $2$, which implies that $BA$ must have rank $2$. This in turn implies that the third column of $BA$ must be zero, since the third column of $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ is zero.
Therefore, $BA = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$, as desired. ]
Am I right or have I done any mistake?