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?
Best Answer
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. $\square$