$T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined as $T(x,y,z) = (2x,z,y)$ is a linear transformation.
I need to prove that the following matrices cannot represent $T$ in ANY basis:
$$\begin{bmatrix} 2 & 0 & 1 \\ 0 & 1 & 0\\ 2 & 0 & 1 \end{bmatrix}$$
$$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1\end{bmatrix}$$
My attempt for the first matrix was to assume (negatively) existence of a basis in which the given matrix is the representation, and left multiplying it by $(x,y,z)^{T}$ to get $(2x+z,y,2x+z)^{T}$.
Can I conclude that given it is not in the form $(2x,z,y)^{T}$, there is NO such basis?
I want to know if this is correct and, in addition (Or alternatively), learn other ways to solve this exercise.
Thank you.
In the canonical basis: $[T]_e= \left( \begin{matrix} 2 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right)$
Any other representation of $T$ would be similar to $[T]_e$.
Similar matrices have the same trace, and this rules out the first option.
Similar matrices have the same determinant, and $|[T]_e|=-2$. This rules out the second option (Has determinant $2$).
About the reasoning you tried and asked about: It is not correct. The form you were talking about is just the form of the coordinates (The numbers you multiply some set of basis vectors by) and would certainly not retain It's form, generally, in different basis.