Linear independence of projection matrices Probably it is a rather simple question ( I'm not sure)  but I would like to know answer supported by a proof.
Can it be proved that   

IF
  unit vectors $v_1,v_2,.. v_n$ are linearly independent
  THEN
  their projection matrices
     ${v_1}{v_1}^T,{v_2}{v_2}^T,.. {v_n}{v_n}^T$  are also linearly
     independent?

...and second case..

IF
  unit vectors $v_1,v_2,.. v_n$ are linearly dependent
  THEN
  their projection matrices
     ${v_1}{v_1}^T,{v_2}{v_2}^T,.. {v_n}{v_n}^T$  are also linearly
     dependent?

 A: You've already found a counterexample to the second statement, so I'll just prove the first.

Proof that linear independence of vectors implies linear independence of projections:
Let $v_1,v_2, \dots, v_n$ be linearly independent. Suppose that $P_1, P_2 , \dots , P_n$ are not linearly independent.
Without loss of generality, say that $P_n$ can be expressed as a combination of the other projections (we can reorder the projections so that this happens):
$$P_n = \sum_{i=1}^{n-1} \lambda_i \cdot P_i$$
We would hence expect that a vector projected onto the axis of $v_n$ can be created instead as a combination of the other projections of that vector. Pick an $\mathbf x \in V$ and let $P_i(\mathbf x) = \alpha_i \cdot v_i$. Then:
$$\begin{align}P_n(\mathbf x) &= \sum_{i=1}^{n-1} \lambda_i \cdot P_i(\mathbf x) \\
\alpha_n \cdot v_n &= \sum_{i=1}^{n-1}\lambda_i \cdot (\alpha_i \cdot v_i) \\
v_n &= \frac 1 {\alpha_n} \sum_{i=1}^{n-1}\lambda_i \cdot (\alpha_i \cdot v_i)
\end{align}$$
Provided that $\alpha_n \neq 0$, we have shown that $v_n$ is a linear combination of the other $v_i$. All that is left to do is pick an $\mathbf x$ so that $\alpha_n \neq 0$. This is easy-- pick $\mathbf x = v_n$. This makes $\alpha_n = v_n^T v_n = ||v_n||_2^2$ which is non-zero (because otherwise $v_n$ is the zero vector) and so
$$v_n = \frac 1 {||v_n||_2^2} \sum_{i=1}^{n-1}\lambda_i \cdot (\alpha_i \cdot v_i)$$
This reveals that $v_n$ is a linear combination of the other $v_i$; a contradiction. $\square$
A: The first statement is easy to prove, the second is easy to disprove.
For the first, I'll show that if the projections $v_iv_i^T$ are linearly dependent, then so are the vectors$~v_i$. So by assumption we have scalars $\lambda_1,\ldots,\lambda_n$, not all zero, such that $\sum_i\lambda_i v_iv_i^T=0$. Let $k$ be an index with $\lambda_k\neq0$; we can assume $v_k\neq0$ for otherwise the linear dependency is obvious. Then let $w$ be a vector with $v_k^T(w)\neq 0$ (if this is over the real numbers one can take $w=v_k$, but in any case $v_k^T\neq0$ ensures that such $w$ exists). Then $0=0(w)=\sum_i\lambda_i v_iv_i^T(w)=\sum_i(\lambda_i v_i^T(w))v_i$, which is a relation with $\lambda_kv_k^T(w)$ a nonzero coefficient, so it establishes linear dependency.
To disprove the second claim, you can take $v,w,v+w$ for any pair of orthogonal unit vectors $v,w$, and compute their projections explicitly to see that they are linearly independent.
