I'd like to prove that:
$$ s \, \overrightarrow{\boldsymbol{Q}}(s) - \boldsymbol{A} \, \overrightarrow{\boldsymbol{Q}}(s) = \overrightarrow{\boldsymbol{Q}}(s) \, \left( s \, \boldsymbol{I} - \boldsymbol{A} \right)$$
where $s$ is a scalar, $Q$ is a column vector of dimension $n$, and $A$ is a square matrix $n \times n$.
My problem is to understand where identity matrix comes from. I tried to do a proof, but it is wrong: $Q^{-1}Q=I$ does not have meaning because $Q$ is a column vector, thus $Q^-1$ is absurd.
$$ s \, \overrightarrow{\boldsymbol{Q}}(s) - \boldsymbol{A} \, \overrightarrow{\boldsymbol{Q}}(s) = $$
$$ =\overrightarrow{\boldsymbol{Q}}(s) \, \overrightarrow{\boldsymbol{Q}}^{-1}(s) \left(s \, \overrightarrow{\boldsymbol{Q}}(s) - \boldsymbol{A} \, \overrightarrow{\boldsymbol{Q}}(s) \right) = $$
$$=\overrightarrow{\boldsymbol{Q}}(s) \left( s \, \overrightarrow{\boldsymbol{Q}}(s) \, \overrightarrow{\boldsymbol{Q}}^{-1}(s) - \boldsymbol{A} \, \overrightarrow{\boldsymbol{Q}}(s) \, \overrightarrow{\boldsymbol{Q}}^{-1}(s) \right) = $$
$$=\overrightarrow{\boldsymbol{Q}}(s) \, \left( s \, \boldsymbol{I} - \boldsymbol{A} \, \boldsymbol{I} \right) = $$
$$=\overrightarrow{\boldsymbol{Q}}(s) \, \left( s \, \boldsymbol{I} - \boldsymbol{A} \right) $$
Thank you for your time.
If we adopt the usual row- columns rule for the multiplication of a matrix and a vector, the correct result is: $$ s\overrightarrow{\boldsymbol{Q}}(s) - \boldsymbol{A} \, \overrightarrow{\boldsymbol{Q}}(s) = \, \left( s \, \boldsymbol{I} - \boldsymbol{A} \right)\overrightarrow{\boldsymbol{Q}}(s) $$
and the identity matrix comes from the fact that $$ s\overrightarrow{\boldsymbol{Q}}(s)=s\,({\bf I} \overrightarrow{\bf Q}(s))= (s \,{\bf I} \,)\overrightarrow{\boldsymbol{Q}}(s) $$