Linear subspaces, $\text{Im} (T)$ and $\text{ker}(T)$ in $\mathbb{R}^3$ Can someone help me with this Linear Subspace problem?

If $U$ and $W$ are linear subspaces of $\Bbb R^3$ defined by:
$$U = \{(x,y,z)\in \Bbb R^3 \mid x+y+z =0\} \\
W = \operatorname{span}[(1,0,1),(0,-1,1)]$$
  Determine a linear operator $T$ of $\Bbb R^3$ such that $\operatorname{Im}(T) = U$ and $\operatorname{Ker}(T) = U\cap W$.

 A: Here’s another way to approach the problem:
You’ve already worked out that $v_1=(0,-1,1)$ spans $U\cap W$, so you know that you must have $T(v_1)=0$. Extend this to a complete basis $(v_1,v_2,v_3)$ of $\mathbb R^3$ and choose a basis $(u_1, u_2)$ for $U$. Define $T(v_2)=u_1$ and $T(v_3)=u_2$. Since $(v_1,v_2,v_3)$ is a basis of $\mathbb R^3$, this specifies $T$ completely. It’s clear from the construction that $\ker(T)=U\cap W$ and $\operatorname{im}(T)=U$. The solution is obviously not unique, since there are many choices for these bases.  
As a concrete example, take $u_1=v_1=(0,-1,1)$ and $u_2=v_2=(1,-1,0)$. You can easily verify that this is indeed a basis for $U$. For $v_3$, we know from the definition of $U$ that $(1,1,1)$ annihilates this space, so it will do for a third basis vector. We can immediately write down the matrix of $T$ relative to this basis: $$\pmatrix{0&1&0\\0&0&1\\0&0&0}.$$ A simple change of basis operation gives us the matrix of $T$ relative to the standard basis: $$\pmatrix{0&1&1\\-1&-1&1\\1&0&1}\pmatrix{0&1&0\\0&0&1\\0&0&0}\pmatrix{0&1&1\\-1&-1&1\\1&0&1}^{-1}=\pmatrix{\frac13&\frac13&\frac13\\-1&0&0\\\frac23&-\frac13&-\frac13}.$$ In fact, this construction can be used to find a matrix for $T$ relative to the standard basis in the general case, above: Extend $(u_1,u_2)$ to a basis $(u_1,u_2,u_3)$ of $\mathbb R^3$. The matrix of $T$ relative to the standard basis will then be $$\pmatrix{\mid&\mid&\mid\\u_1&u_2&u_3\\\mid&\mid&\mid}\pmatrix{0&1&0\\0&0&1\\0&0&0}\pmatrix{\mid&\mid&\mid\\v_1&v_2&v_3\\\mid&\mid&\mid}^{-1}=\pmatrix{\mid&\mid&\mid\\0&u_1&u_2\\\mid&\mid&\mid}\pmatrix{\mid&\mid&\mid\\v_1&v_2&v_3\\\mid&\mid&\mid}^{-1}.$$ Note that this matrix is in fact independent of the choice of $u_3$, which makes sense since it’s not part of the image.
