The following happen to all be true. But why is this closed under scalar multiplication? α[[a],[3a+b],[4a+4b],[4a+3b]] =[[aα],[3aα+bα],[4aα+4bα],[4aα+3bα]] and sure, this is in R but this is NOT in the subset so is it not closed?
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I don't see the problem, clearly $$\alpha \begin{bmatrix} a \\ 3a+b \\ 4a+4b \\ 4a+3b \end{bmatrix} = \begin{bmatrix} \color{red}{\alpha a} \\ 3(\color{red}{\alpha a})+(\color{blue}{\alpha b}) \\ 4(\color{red}{\alpha a})+4(\color{blue}{\alpha b}) \\ 4(\color{red}{\alpha a})+3(\color{blue}{\alpha b}) \end{bmatrix} $$ is in the form of $$ \begin{bmatrix} \color{red}{u} \\ 3\color{red}{u}+\color{blue}{v} \\ 4\color{red}{u}+4\color{blue}{v} \\ 4\color{red}{u}+3\color{blue}{v} \end{bmatrix}.$$