There's matrix $M =\left[\begin{array}{cc} I &B\\ B^T& I\end{array}\right]$, where $B$ is $m × n$ (note that the upper left corner is the $m × m$ identity, while the lower right corner is the $n × n$ identity).
(a) Show that if $I − (B^T)B$ is invertible then $M$ is also invertible.
(b) Show that if $I − (B^T)B$ is singular then $M$ is also singular.
What I tried: I tried calculating $\det(M) = \det (I) \cdot \det (I-(B^T)B)$, and since $I-(B^T)B$ is invertible, then $\det(M)$ is not equal to zero and then $M$ is invertible. I am not sure if this is correct and I am looking for other methods than the determinant method.
Note that $$ \pmatrix{I & 0\\-B^T & I}\pmatrix{I & B\\B^T & I} = \pmatrix{I & B\\0 & I - B^TB}, $$ which means that $$ \det \pmatrix{I & 0\\-B^T & I}\det \pmatrix{I & B\\B^T & I} = \det\pmatrix{I & B\\0 & I - B^TB} \implies\\ 1 \cdot \det\pmatrix{I & B\\B^T & I} = \det(I - B^TB). $$