find the coordinates of the vector $(a,b,c)?$

Question

Suppose $a_1 = (1,0,-i) $ , $a_2 =(1+i , 1-i,1)$ , $a_3 =(i,i,i)$ in $\mathbb{C}^3$ . What are the coordinates of the vector $(a,b,c)$ in this basis ?

My attempt : Here obviously $a_1,a_2$ and $a_3$ will form basis since det $A \neq 0$

take $A=\begin{bmatrix} 1 & 0 & -i \\ 1+i & 1-i & 1 \\ i & i & i \end{bmatrix}$

and we have $$A^{-1}=\begin{bmatrix} \frac{1-2i}{5} & \frac{1-2i}{5} & \frac{3-i}{5}\\ \frac{1-2i}{5}& \frac{1+3i}{5} & \frac{-2-i}{5} \\ \frac{-2+4i}{5} & \frac{-2-i}{5} & \frac{-1-3i}{5} \end{bmatrix}$$

The coordinates of the vector $(a,b,c)$ in this basis =$A^{-1} \begin{bmatrix} a \\ b\\c\end{bmatrix}=\begin{bmatrix} \frac{1-2i}{5}a + \frac{1-2i}{5}b + \frac{3-i}{5}c\\ \frac{1-2i}{5}a+\frac{1+3i}{5}b + \frac{-2-i}{5} c \\ \frac{-2+4i}{5}a + \frac{-2-i}{5} b + \frac{-1-3i}{5}c \end{bmatrix}$

Is my solution is correct or not ?

Answer 1

No, it isn't. Take the vector $(1,0,-i)$ for example. Then you get

$$A^{-1}\begin{bmatrix} 1\\ 0\\ -i \end{bmatrix}=\begin{bmatrix} -i\\ 0\\ -1+i \end{bmatrix}$$

But $(1,0,-i)$ is the first element of your basis so you should be getting $(1,0,0)$.

To see where you went wrong, look at how you constructed your change of basis matrix $A$.

If $A^{-1}$ takes a vector written in the canonical basis and gives you the vector written in the $\{a_1,a_2,a_3\}$ basis, $A$ should do the opposite. Then what should you get for $$A \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}?$$ Notice that this tells you what $A$'s first column should be.

Answer 2

From Paulo answer

i notice that to find the coordinate

$$A^{-1}\begin{bmatrix} 1\\ 0\\ -i \end{bmatrix}=\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$$

must satisfied

Therefore the coordinates of the vector $(a,b,c)$ in this basis =$(A^{-1})^T \begin{bmatrix} a \\ b\\c\end{bmatrix}=\begin{bmatrix} \frac{1-2i}{5}a + \frac{1-2i}{5}b + \frac{-2+4i}{5}c\\ \frac{1-2i}{5}a+\frac{1+3i}{5}b + \frac{-2-i}{5} c \\ \frac{3-i}{5}a + \frac{-2-i}{5} b + \frac{-1-3i}{5}c \end{bmatrix}$