Bound on determinant sum if some of rows and columns are zeros Let $A$ be positive definite and $B$ be positive semidefinite. From the Minkowski inequality,
$$ \det [A + B] \geq \det [A] + \det [ B ] $$
Now, consider the case the $i$th row and column of $B$ are all zeros. Then, $\det [B] = 0$. Let $B’$ denote $B$ with the $i$th row and column deleted. Then, I believe we now have the following inequality:
$$ \det [ A + B ] \geq \det [ A ] + \det [ B’ ] $$
I am not sure how to prove this, but this seems true. My logic is that the following.
The determinant of a matrix is the volume spanned by the columns of the matrix. So, if we have a row and column of zeros, then adding the matrices together increases the volume of A by no more than the volume of $\det [B’]$. At least, this is my intuition. I’m struggling to prove this hypothesis.
So, any suggestion on proving this?
 A: This is not true. Consider
$$A=\begin{bmatrix}1&\epsilon \\\epsilon&\epsilon\end{bmatrix},\ B=\begin{bmatrix}x&0\\0&0\end{bmatrix}$$
for some $0<\epsilon<1$. The matrix $A$ is positive definite since its diagonal entries and determinant are positive, and $B$ is positive semidefinite if $x\geq 0$. Then
$$\det(A+B)=(1+x)\epsilon-\epsilon^2=\det(A)+x\epsilon = \det(A)+\epsilon\det(B')<\det(A)+\det(B').$$
A: As the other answer has pointed out, the inequality is not true, but in general, we do have
$$
\det(A+B)\ge\det(A)+a_{ii}\det(B').
$$
Without loss of generality, assume that $i=1$. Let
$$
A=\pmatrix{a_{11}&v^\ast\\ v&A'},\ B=\pmatrix{0&0\\ 0&B'}.
$$
Since $A$ is positive definite, the Schur complement $A/a_{11}=A'-\frac{1}{a_{11}}vv^\ast$ is positive definite. Therefore
\begin{aligned}
\det(A+B)
&=\det\pmatrix{a_{11}&v^\ast\\ v&A'+B'}\\
&=a_{11}\det\left(A'+B'-\frac{1}{a_{11}}vv^\ast\right)\\
&\ge a_{11}\left[\det\left(A'-\frac{1}{a_{11}}vv^\ast\right)+\det(B')\right]\\
&=\det(A)+a_{11}\det(B').
\end{aligned}
