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A linear operator A in the three-dimensional space is determined by the transformation of the coordinates of a vector $\begin{pmatrix} x\\ y\\ z\\ \end{pmatrix}$ whose coordinates are given in the standard basis, this vector after transformation becomes $\begin{pmatrix} x-2y\\ y+z\\ y-x\\ \end{pmatrix}$. The new basis in the space is chosen as $\begin{pmatrix} 1\\ 1\\ 1\\ \end{pmatrix}$, $\begin{pmatrix} 1\\ 1\\ 0\\ \end{pmatrix}$, $\begin{pmatrix} 1\\ 0\\ 0\\ \end{pmatrix}$. Find the matrix representation of this operator relative to the given basis.

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You need to check again what a matrix $M:V_1\to V_2$ represents. In particular for an endomorphism like in your case (meaning both vector spaces are the same $V_1=V_2$), each column $j$ of the matrix $M$ gives the image of the basis vector $b_i$ by that matrix, i.e., $Mb_i$.

Thus, in the standard basis $$M=\begin{pmatrix}1 & -2& 0\\0 & 1& 1\\-1 & 1& 0\end{pmatrix}$$

Let's call $v_1=(1,1,1)^T,\,v_2=(1,1,0)^T,\,v_3=(1,0,0)^T$. We have $$Mv_1=(-1,2,0)^T=2v_2-3v_3\\Mv_2=(-1,1,0)=v_2-2v_3\\Mv_3=(1,0,-1)=-v_1+v_2+v_3$$ Therefore, in the new basis $$M'=\begin{pmatrix}0 & 0& -1\\2 & 1& 1\\-3 & -2& 1\end{pmatrix}$$ Which is the answer you are seeking.

Notice, if we call $C$ the matrix providing the change of basis from the new $\{v_i\}$ to the old one $\{e_i\}$, it is $$C=\begin{pmatrix}1 & 1& 1\\1 & 1& 0\\1 & 0& 0\end{pmatrix}$$ and it holds that $$M'=C^{-1}MC$$

You could have also proceeded by determining $C,\,C^{-1}$ and doing the above matrix multiplication. For a "small" matrix ($3\times 3$) like in the present case, however, it is simpler and more elegant to do as we did.

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  • $\begingroup$ If an answer was helpful in solving your question, it would be nice if you accept it. Thanks. $\endgroup$ – MASL Sep 18 '15 at 14:14
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If I understood right, $T(x,y, z)=(x-2y,y+z,y-x)$ thus $$T(1, 1, 1)=(1-2, 1+1, 1-1)=(-1, 2, 0)=0(1, 1, 1)+2(1,1,0)-3(1, 0, 0)$$ and $$T(1, 1, 0)=(-1,1,0)=0(1, 1, 1)+1(1,1,0)-2(1, 0, 0)$$ and $$T(1, 0, 0)=(1,0,-1)=-1(1, 1, 1)+1(1,1,0)+1(1, 0, 0)$$

\begin{bmatrix} 0 & 0 & -1 \\ 2 & 1 & 1 \\ -3 & -2 & 1 \end{bmatrix}

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