If we assume that $\mathbf{U}$ is a matrix $n\times2$, with rank $2$ (two independent columns) and $\mathbf{A}$ is a positive definite matrix of order $n$, what would be the best way to see that the product $\mathbf{U}^{T}\mathbf{AU}$ is nonsingular? I know that If I can show that $\mathbf{U}^{T}\mathbf{A}^{1/2}$ is of rank $2$ the result would follow but I don't know how to proceed other than maintaining that the rank of the product should be less than $2$. But this does not exclude the case of the product having rank $1$ and thus $\mathbf{U}^{T}\mathbf{AU}$ would be singular.
I would be grateful if someone could provide an argument in that direction. Thank you.
If $V=U^TA^{1/2}$ had rank $<2$ then so would $U^T=VA^{−1/2}$, since the rank of a matrix can only drop after multiplication with another matrix.