I am looking at this answer. After saying a matrix is idempotent it say it can diagonalize in this way:
$$V'QV = \Delta = \textrm{diag}(\underbrace{1, \ldots, 1}_{n-p \textrm{ times}}, \underbrace{0, \ldots, 0}_{p \textrm{ times}})$$
Then he defines $K = V' \hat{\epsilon}$ (after having deduced $\hat{\epsilon} = Q\epsilon$) and says
Since $\hat{\epsilon} \sim N(0, \sigma^2 Q)$, we have $K \sim N(0, \sigma^2 \Delta)$ and therefore $K_{n-p+1}=\ldots=K_n=0$.
Can someone explain this passage?