<s>[INST] Subspaces of the projection - matrix 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?
 [/INST] 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).
</s>