How to show that Im R is an A-invariant subspace?

Question

Suppose we have $A=\begin{pmatrix}4 & -4 & 2 \\ 3 & -3 & 2 \\ -3 & 2 & -3 \end{pmatrix}$, and $\operatorname{Im} R = \operatorname{basis}\{ \begin{pmatrix}1 & 4 & -4\end{pmatrix}^T,\begin{pmatrix}0 & 1 & -1\end{pmatrix}^T\}$. How can we proceed to show that $\operatorname{Im} R$ is $A$-invariant?

Answer 1

Take the matrix $B$ whose columns are the basis vectors of the image. Put the matrix into reduced column echelon form using the transpose of Gaussian elimination to get a matrix $B'$. Find the product $AB$, and put this in reduced column echelon form. The subspace is invariant if and only if the resulting matrix is $B'$.

Edit: This is almost true. However, it's possible the subspace is invariant but the dimension decreases. If that's the case, after you put it in echelon form you'll have a column of zeros. Take the column that is not zero and see if it is contained in the subspace by usual methods (e.g. determinant).