Finding base B'

Question

I have B = {(0,2,1),(-2,2,1),(-1,2,1)} how can I find B' so $ x + [x]_B + [x]_{B'} = 0 $ (equlas zero vector). For every vector $ x \in \mathbb{R}^{3} $.

Answer 1

The easiest approach is as follows: let $M_B$ denote the matrix whose columns are given by the vectors in $B$. Then $$ [x]_B = M_B^{-1}x $$ where $x$ denotes the column-vector for $x$ with respect to the standard basis. What we're looking for, then, is a matrix $M_{B'}$ such that $$ x + M_B^{-1}x + M_{B'}^{-1}x = 0 \quad (\text{for all }x)\implies\\ (I + M_B^{-1} + M_{B'}^{-1})x = 0 \quad (\text{for all }x)\implies\\ I + M_B^{-1} + M_{B'}^{-1} = 0 $$ Solve the above for $M_{B'}$.