Does $\det(A-BC) = \det (A)$?

Question

Suppose that $A$ is $a \times a$, $B$ is $a \times b$ and $C$ is $b \times a$ where $a>b$. Does the following hold?

$$\det(A-BC) = \det (A)$$

I reckon that this is so since $\det(BC) = 0$ because $B$ has more rows than $C$.

Answer 1

Even if $\det BC=0$, it's a mistake to think $\det (A-BC)=\det A -\det BC$.

Answer 2

Here's an explicit counterexample with $a = 2$ and $b = 1$: $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix}1 \\ 0 \end{bmatrix}, \qquad C = \begin{bmatrix} 1 & 0 \end{bmatrix}. $$ Then $$ BC = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \qquad \Rightarrow A - BC = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $$ and $\det(A - BC) = 0 \neq \det(A)$.

The flaw in your reasoning is provided by J.G.'s answer: in general, $\det(A + B) \neq \det(A) + \det(B)$.

Answer 3

One way we can make this happen is by controlling the ranks of the matrices.

We know that $rank(BC) \leq \min(rank(B), rank(C))$ and that $rank(A-BC) \leq rank(A) + rank(BC) \leq rank(A) + \min(rank(B), rank(C))$

So if we keep the ranks of all the matrices small relative to the dimension of $A$, we'll definitely have that $A-BC$ and $A$ are both singular, and consequently $0 = det(A-BC) = det(A)$.

If we don't control the ranks or even allow $A$ to be invertible, it might hold under some very special circumstances. As pointed out in the comments, the matrix determinant lemma is where you should look for answers. For example, one nice case that jumps out at a glance is when $A$ is the identity matrix and $CB = 0$ (which implies $det(BC) =0$ independent of dimension considerations, since $det$ is multiplicative). In that case the matrix determinant lemma shows $det(A-BC) = 1 = det(A)$.