I have some problems with the following exercise:
Let $v_1 = [1,1,0]^T,v_2 = [1,0,1]^T, v_3 = [0,0,1]^T$. Find the dual basis $\bar{v}^1,\bar{v}^2,\bar{v}^3$ such that $\bar{v}^{i}(v_{j})= \delta^{i}_j$. Calculate $v_i \bar{v}^i$. Along which subspace are these projections and on which subspace are they projected into?
I have found the dual basis: $\bar{v}^1 = [\frac{1}{2},\frac{1}{2}, -\frac{1}{2}],\bar{v}^2 = [\frac{1}{2},-\frac{1}{2},\frac{1}{2}], \bar{v}^3 = [-\frac{1}{2},\frac{1}{2}, \frac{1}{2}]$ and the projection matrices: \begin{align*} v_1 \bar{v}^1 = \left[ \begin{array}{ccc} \frac{1}{2}& \frac{1}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& -\frac{1}{2}\\ 0& 0& 0 \end{array} \right],\quad v_2 \bar{v}^2 = \left[ \begin{array}{ccc} \frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\\ 0& 0& 0\\ \frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\\ \end{array} \right], \quad v_3 \bar{v}^3 = \left[ \begin{array}{ccc} 0& 0& 0\\ -\frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ \end{array} \right]. \end{align*} I don't know how from the projection operators one can determine along which subspace are these projections and on which subspace are they projected into?
In general if you have a rank one matrix of the form $a\otimes b$, its annihilator is the subspace of vectors orthogonal to $b$ and its range is the subspace spanned by $a$. This is because $(a\otimes b)(x)=a(b\cdot x)$ by definition.
Another way: From your matrices you can tell that any vector orthogonal to $(1,1,-1)$ transforms to the origin, and the image of any vector is parallel to $(1,1,0)$ (being a linear combination of the columns).