Is the inverse of a truncated matrix the same as the truncation of the inverse? Given:

*

*$A \in \mathbb{R}^{D\times D}$ as a symmetric, square, positive definite, non-singular matrix

*$\Delta := A^{-1}$
Is the following statement true?
$(A_{d:,d:})^{-1} = \Delta_{d:,d:}$
Or in words, is truncating and finding the inverse identical to finding the inverse and then truncating?
Note about abused subindex notation: $A_{:,i}$ means the $i^{th}$ column,  $A_{i, :}$ the $i^{th}$ row and $A_{i:, j:}$ the submatrix resulting from starting at the $i^{th}$ row and $j^{th}$ column (included).

Bonus questions:

*

*If the equality holds, would it hold for any $A_{d1:d2, d3:d4}$ submatrix?


*What would be the case if any of the $A$ features change? (asymmetric, semidefinite, rectangular, singular or quasi-zero eigenvalues...)
Cheers!
Andres
 A: As Nick Alger pointed out (big thanks to him), the relation between a truncated matrix and its inverse is regulated by the Schur complement, as follows:
If we partition a matrix $A$ into four submatrices as follows,
$
A =\begin{pmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{pmatrix}
$
The inverse can be expressed as follows:
$
\Delta := A^{-1} =\begin{pmatrix}
\Delta_{11} & \Delta_{12} \\
\Delta_{21} & \Delta_{22}
\end{pmatrix}
=
\begin{pmatrix}
(\frac{A}{A_{22}})^{-1} & (\frac{A}{A_{22}})^{-1} A_{12} (A_{22})^{-1}  \\
(A_{22})^{-1} A_{21} (\frac{A}{A_{22}})^{-1} & (\frac{A}{A_{11}})^{-1}
\end{pmatrix}
$
Where the fractions express the so-called Schur complements of their respective blocks:
$
\frac{A}{A_{22}} = A_{11} - A_{12} (A_{22})^{-1}A_{21} = (\Delta_{11})^{-1}\\
\frac{A}{A_{11}} = A_{22} - A_{21} (A_{11})^{-1}A_{12} = (\Delta_{22})^{-1}
$

*

*Especially note that in general the following is not true: $(A_{ij})^{-1} = \Delta_{ij}$.

*Also note that this will hold for any submatrix, since the rows and columns can be "shuffled" at convenience.


Further properties:

*

*If $A$ is non-singular, this works both ways (i.e. $A$ and $\Delta$ can be swapped above). If $A$ or $A_{22}$ are singular, the generalized Schur complement can be applied.

*$det(A) = det(A_{22}) det(\frac{A}{A_{22}})$

*If $A$ is symmetric, so is its inverse. The $M_{ii}$ partitions will be symmetric and $M_{ij} = M_{ji}^T$.

*If $A$ is positive definite, so will be $\Delta$ and any $M_{ii}$ partition.


I think that summarizes pretty much all the questions I had on the matter!
Cheers
Andres
A: In general the answer is no. Let  $\ A=\pmatrix{2&1\\1&2}\ $.  Then $\ \Delta=A^{-1}=\frac{1}{3}\pmatrix{2&-1\\-1&2}\ $, $\left(A_{2:,2:}\right)^{-1}=\left(\frac{1}{2}\right)\ne\ \Delta_{2:,2:}=\left(\frac{2}{3}\right)\ $.
Reply to questions: I have to confess that since the statement didn't seem likey to me to be true, I didn't read through your calculations very carefully. That said, I didn't pick up any errors either.
The example I gave was a special case of $\ A=I_D + aa^\top\ $, which has inverse $\ \Delta=I_D-\frac{aa^\top}{1+a^\top a}\ $.  For this more general case $\ A_{d:,d:}=I_{D+1-d}+a_{d:} a^\top_{d:}\ $, which has inverse $\ \left(A_{d:,d:}\right)^{-1}=I_{D+1-d}-\frac{a_{d:} a^\top_{d:}}{1+a^\top_{d:}a_{d:}}\ $, while $\ \Delta_{d:,d:}=I_{D+1-d}-\frac{a_{d:} a^\top_{d:}}{1+a^\top a}\ $. So, in this case, there is indeed a fairly simple relation between $\ \left(A_{d:,d:}\right)^{-1}\ $ and $\ \Delta_{d:,d:}\ $, namely,
$$ \left(1+a^\top_{d:}a_{d:}\right)\left(I_{D+1-d}-\left(A_{d:,d:}\right)^{-1}\right)= \left(1+a^\top a\right)\left(I_{D+1-d}-\Delta_{d:,d:}\right)$$
