# 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

• What is $\ \Sigma\$ in the second bullet point? Do you mean $\ \Delta= A^{-1}\$? And does $\ \left(A_{d:,d:}\right)\$ signify the last $\ D-d+1\$ rows and columns of $\ A\$, or the first $\ d\$ rows and columns (or something else entirely)? Commented Apr 8, 2019 at 19:09
• Yes sorry, originally it is a cov. matrix that's why I used $\Sigma$ but I changed the notation for more general clarity. Fixed. The abused subindex notation: $A_{:,d}$ is the $d^{th}$ column, $A_{d, :}$ the $d^{th}$ row and $A_{d:, d:}$ all the rows and columns starting from (and including) the $d^{th}$ ones. Commented Apr 8, 2019 at 19:12
• @lonzaleggiera edited the post with the clarification. Commented Apr 8, 2019 at 19:16
• $(A^{-1})_{pp}=(A_{pp}-A_{pq}A_{qq}^{-1}A_{qp})^{-1}$, where $p,q$ are index sets that combine to make 1:n. Look into Schur complements. Commented Apr 9, 2019 at 0:37
• @NickAlger you made my day with this amazing answer! It covers all my questions, including the extra ones, so if you want to post it as an answer I will be glad to accept it. Thanks a lot Commented Apr 9, 2019 at 17:18

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

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)$$