Rank of submatrices: clarify an apparent contradiction Consider the matrix
$$
A\equiv \begin{pmatrix}
B & C & D\end{pmatrix}^\top \begin{pmatrix}
B & C & D\end{pmatrix}= \begin{pmatrix}
B^\top B & B^\top C &  B^\top D \\
C^\top B & C^\top C &  C^\top D \\
D^\top B & D^\top C &  D^\top D \\
\end{pmatrix}
$$
Consider the following three observations. It seems to me that they lead to a contradiction. Hence, some (all) of them should be wrong. Could you point out the mistake?
(1) If $A$ is invertible, then it does not imply that every square submatrix of $A$ is invertible. For example, we could have that $A$ is invertible but the submatrix
$$
Q\equiv \begin{pmatrix}
B^\top B  &  B^\top D \\ 
D^\top B  &  D^\top D \\
\end{pmatrix}
$$
is not invertible.
(2) $A$ is invertible if and only if $\begin{pmatrix}
B & C & D\end{pmatrix}$ is full column rank.
(3) If $\begin{pmatrix}
B & D \end{pmatrix}$ is not full column rank, then $\begin{pmatrix}
B & C& D \end{pmatrix}$ is not full column rank.

The contradiction is the following:
If $\begin{pmatrix}
B & D \end{pmatrix}$ is not full column rank (which holds if and only if $Q$ is invertible), then by (3) $\begin{pmatrix}
B & C& D \end{pmatrix}$ is not full column rank. In turn, by (2), $A$ is not invertible. This contradicts (1).
 A: Note. The first version of this answer was written under the misunderstanding that  full column rank means
that the columns generate the space where they live.  The OP has later called my attention to the fact that some people take it to
mean that instead, the columns are linearly independent.   I must say I find the latter terminology a bit misleading, so
I would try to avoid it by  refering to it by what it means.  Nevertheless, should  you like to further investigate
the terminology issue, I invite you to
compare
https://en.m.wikipedia.org/wiki/Rank_(linear_algebra)
and
https://www.cds.caltech.edu/~murray/amwiki/index.php/FAQ:_What_does_it_mean_for_a_non-square_matrix_to_be_full_rank%3f

Statement 1.  Every submatrix of an invertible matrix is invertible.
Verdict: False.
Counter example: Let  $A$ be the $3×3$ identity matrix, and consider the  $2×2$ submatrix
$$
  \pmatrix{0 & 1\cr 0 & 0}
  $$
obtained by removing  the first row and the last  column of $A$.  It is obviously not invertible.
Statement 1 (bis).  Every principal submatrix of a positive invertible matrix is invertible.
Verdict: True.
Proof: Let  $A$ be a positive invertible $n×n$ matrix.  A principal submatrix of $A$ is one that is obtained by removing some
rows, say the ones indexed by
$$
  k_1,k_2,…,k_p,
  $$
as well as some columns, but not just any columns, rather the columns indexed by the very same $k_1,k_2,…,k_p$.
Letting $R$ be the $n×(n-p)$ matrix whose columns  are the canonical basis vectors corresponding to the columns of $A$ which were not
removed, the corresponding submatrix may be expressed as $R^TAR$.
In order to prove that $R^TAR$ is invertible, it is enough to prove it is injective, so suppose $x$ is a ($n-p)$-column
matrix such that $R^TARx=0$.  Then
$$
  0 = ⟨R^TARx,x⟩ =
  ⟨A^{1/2}Rx,A^{1/2}Rx⟩ =
  \|A^{1/2}Rx\|^2,
  $$
so $A^{1/2}Rx=0$, and hence $ARx=0$, as well.  Since $A$ is injective, then $Rx=0$, but $R$ is also injective, so
$x=0$.  QED

Statement 2.  $A$ is invertible if and only if $\pmatrix{B & C & D}$ has linearly independent columns.
Verdict: True.
Proof: Denoting by the matrix $\pmatrix{B & C & D}$ simply by $E$, suppose that
$Ax=0$.  Then
$$
  0 = ⟨Ax,x⟩ = ⟨E^TEx,x⟩ =
  ⟨Ex,Ex⟩ = \|Ex\|^2,
  $$
so $Ex=0$, but since the columns of $E$ are linearly independent, one deduces that $x=0$.  QED

Statement 3.  If $\pmatrix{B & C}$ does  not have linearly independent columns then  $\pmatrix{B & C & D}$
does not have linearly independent columns.
Verdict: True.
Proof: Easy exercise.
