Proving a determinant of a particular type is always null

Question

Let $n\ge 1$ and $A,B\in\mathrm M_n(\mathbb R)$.

Let's assume that

$$\forall Q\in\mathrm M_n(\mathbb R), \quad \det\begin{pmatrix} I_n & A \\ Q & B\end{pmatrix}=0$$

where $I_n$ is the identity matrix of $\mathrm M_n(\mathbb R)$.

Can we prove that $\mathrm{rank} \begin{pmatrix}A\\ B\end{pmatrix}<n$?

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This fact seems quite obvious, but I can't find any straightforward argument to prove it.

Some ideas.

With $Q=0$, we deal with a block-triangular matrix, so we have $\det B=0$.

Moreover, with $Q=\lambda I_n$, $\lambda\in\mathbb R$, since it commutes with $B$, we have

$$\forall \lambda\in\mathbb R,\quad \det(B-\lambda A)=0,$$

so if $\det(A)\ne 0$, we have

$$\forall \lambda\in\mathbb R,\quad\det((BA-\lambda I_n)A^{-1})=\det(BA-\lambda I_n)\det(A)^{-1}=0,$$

which means that every $\lambda\in\mathbb R$ is an eigenvalue of $BA$ (since for all $\lambda\in\mathbb R$, $\det(BA-\lambda I_n)=0$), which is absurd.

So $\det(A)=0$ also.

Answer 1

It is true over any field, not just $\mathbb R$. Note that $\det\pmatrix{I&A\\ Q&B}=\det(B-QA)$. View $A$ and $B$ as two linear maps from a vector space $V$ to another vector space $W$ of the same dimension. View $Q$ as a linear operator on $W$. By changing the bases of $V$ and $W$ separately, we may assume that $A=I_r\oplus0$ where $r=\operatorname{rank}(A)$. Partition $B$ and $Q$ accordingly as $[B_1|B_2]$ and $[Q_1|Q_2]$, where $B_1$ and $Q_1$ each has $r$ columns. Then $B-QA=[B_1-Q_1|B_2]$ is singular for every $Q_1$. Hence $B_2$ has deficient column rank and so does $\pmatrix{A\\ B}=\pmatrix{\ast&0\\ \ast&B_2}$. Since the change of bases amounts to a transformation in the form of $\pmatrix{A\\ B}\mapsto\pmatrix{U&0\\ 0&U}\pmatrix{A\\ B}V$ for some invertible $U$ and $V$, the $\pmatrix{A\\ B}$ before change also has deficient column rank.

Answer 2

Denote $$M_Q = \begin{pmatrix} I_n & A \\ Q & B\end{pmatrix}, \quad \operatorname{col}(A,B) = \pmatrix{A\\B}. $$

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The statement does hold with the additional assumption that $\ker(A) \subseteq \ker B$, i.e. the row-space of $A$ contains that of $B$.

Suppose for contradiction that $\operatorname{col}(A,B)$ has full rank. Let $U$ denote the column space of $A,B$. Let $P$ denote a matrix whose columns form a basis of $U^\perp$. By using column operations on $P$, we can bring $P$ to its column-echelon form, which is $$ P = \pmatrix{I_n\\ Q} $$ for some matrix $Q_*$. Because the columns of $P$ form a basis of $U^\perp$ and the columns of $\operatorname{col}(A,B)$ form a basis of $U$, we conclude that the columns of $M_{Q_*}$ form a basis of $\Bbb R^n$, which means that $M_{Q_*}$ is invertible and $\det(M_{Q_*}) \neq 0$.

Thus, $\operatorname{col}(A,B)$ indeed fails to have full rank if $\det(M_Q) = 0$ for all $Q$.