If $P_{i}=\frac{\alpha_{i}\otimes\beta_{i}}{(\alpha_{i},\;\beta_{i})}=\frac{\alpha_{i}\beta_{i}^{T}}{\alpha_{i}^{T}\beta_{i}}$, where $P_{i}$ are rank-1 projection matrices and $(\alpha_{i},\;\beta_{i})\neq0$, then how we can find $\text{Tr}(P_{1}P_{2})$?
As I understand $$ \text{Tr}(P_{1}P_{2})=\text{Tr}\left(\frac{\alpha_{1}\beta_{1}^{T}}{\alpha_{1}^{T}\beta_{1}}.\frac{\alpha_{2}\beta_{2}^{T}}{\alpha_{2}^{T}\beta_{2}}\right), $$ but the answer should be $$ \text{Tr}(P_{1}P_{2})=\frac{(\alpha_{1},\;\beta_{2})(\alpha_{2},\;\beta_{1})}{(\alpha_{1},\;\beta_{1})(\alpha_{2},\;\beta_{2})}. $$
Using the cyclic property of the trace, that is $\text{Tr}(AB) = \text{Tr}(BA)$ (choose $A = \alpha_{1}\beta_{1}^{T}\alpha_{2}$ and $B = \beta_2^T$. (The denominator is just a scalar $k =\alpha_{1}^{T}\beta_{1}\alpha_{2}^{T}\beta_{2} $ so it doesn't affect the trace $$\text{Tr}(P_{1}P_{2})=\text{Tr}\left(\frac{\beta_{2}^{T}\alpha_{1}}{\alpha_{1}^{T}\beta_{1}}.\frac{\beta_{1}^{T}\alpha_{2}}{\alpha_{2}^{T}\beta_{2}}\right)$$ Notice each term in the numerator and denominator is a scalar, so the whole quantity inside the trace is a scalar and we know that $\text{Tr}(x) = x$ for a scalar $x$, hence we can omit the trace, $$\text{Tr}(P_{1}P_{2})=\frac{\beta_{2}^{T}\alpha_{1}}{\alpha_{1}^{T}\beta_{1}}.\frac{\beta_{1}^{T}\alpha_{2}}{\alpha_{2}^{T}\beta_{2}}$$ and that's about it.