Given the linear transformation $ T: \mathbf{R}^{3} \to \mathbf{R}^{3} $ defined by $$ \forall (x,y,z) \in \mathbf{R}^{3}: \qquad T(x,y,z) = (−2 x − 36 z,−3 y,−36 x − 23 z), $$ find an orthonormal basis $ B $ for $ \mathbf{R}^{3} $ such that $ [T]_{B} $ is diagonal.
I’m really very lost here. It seems like there’re multiple ways to get an orthonormal basis, so I’m not sure how to work with the fact that $ T $ is a linear transformation. All I know is that a diagonal matrix has non-zero entries only down the diagonal and that a matrix $ A $ is orthonormal if $ A A^{\intercal} = \mathsf{I} $.
I tried writing the transformation $ T $ as a $ (3 \times 3) $-matrix and getting it in reduced row echelon form, but that was just a $ (3 \times 3) $-identity matrix, so I’m not sure what to do with this information. Thank you so much for any help!