# Theoretical question about the rank and existence of an inverse of a block matrix,

Let A and B be two $n \times n$ square matrices with complex coefficients, and consider the $2n \times 2n$ matrix $M$ given by

$$M = \begin{bmatrix} A & A \\ A & B \\ \end{bmatrix}$$

1. Determine the rank of $M$ in terms of $A$ and $B$.

2. What is the condition for $M$ to have an inverse $M^{-1}$? Compute $M^{-1}$ when it exists.

Any ideas on how to get started on this problem are welcome.

Thanks,

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If $A$ is rank deficient, then $M$ is also rank deficient. However in the case that $A$ is full rank, there's not a lot we can say except that the rank of $M$ is at least the rank of $A$ (this is so whatever the rank of $A$ is). – hardmath Jan 4 at 23:18
You can do Gaussian elimination to compute the inverse, except instead of working with individual entries, instead you work with entire blocks. The only difference is that you need to keep track of the order of the blocks, as they won't commute. – Nick Alger Jan 4 at 23:18
Hi @hardmath, thanks so much, yes, I agree with you. I've had this answer for awhile and am thinking, "there has to be many more cases to consider". It looks like you also think that there is not much more to say...hmmm... – User001 Jan 4 at 23:21
Hi @NickAlger, would it also be correct, if I augmented this $2nx2n$ matrix with a $2nx2n$ identity matrix I, and then row-reduce until I get a $2nx2n$ identity matrix on the left side? And, are you saying to do something different, like this: get $nxn$ identity blocks, so, 4 in total, on the left side? So, start with 4 $nxn$ identity blocks on the right-side of the augmented matrix? Thanks so much, – User001 Jan 4 at 23:25
@User001 If you want to go the augmented route, then you can augment with the 2n-by-2n identity. In the meantime it looks like some people have answered explaining this in more detail. If you are unsure of which is right, you can always multiply out the matrix times it's inverse and see whether you get the identity back again. – Nick Alger Jan 4 at 23:49

## 4 Answers

\begin{align} [ M \mid I ] &= \left[ \begin{array}{cc|cc} A & A & I & 0 \\ A & B & 0 & I \end{array} \right] \\ &\to \left[ \begin{array}{cc|cc} A & A & I & 0 \\ 0 & B-A & -I & I \end{array} \right] \\ & \to \left[ \begin{array}{cc|cc} I & I & A^{-1} & 0 \\ 0 & I & -(B-A)^{-1} & (B-A)^{-1} \end{array} \right] \\ & \to \left[ \begin{array}{cc|cc} I & 0 & A^{-1} +(B-A)^{-1} & -(B-A)^{-1} \\ 0 & I & -(B-A)^{-1} & (B-A)^{-1} \end{array} \right] = [ I \mid M^{-1} ] \end{align}

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@mvw....wow...this is too cool...! :-) Thanks so much! – User001 Jan 4 at 23:56
The derivation above uses that the block matrices are like numbers themselves to a certain degree, matrices form a ring, so that some algorithms from linear algebra can be applied to them. – mvw Jan 5 at 0:08
Hi @mvw, in your second row, after just one row operation, we already can determine the rank of M, which is just rank(A) + rank (B-A), since the the matrix is block upper-triangular (and also row operations are rank-preserving). Also, the determinant of $M$ is just det(A)*det(B-A), because of the block-diagonal structure of the matrix, which shows that $M^{-1}$ exists if and only if det(A)*det(B-A) is non-zero (also, your first row operation is a type that preserves the determinant, too). – User001 Jan 5 at 0:33
And finally, computation of the inverse is what you have shown, with block row-operations on an augmented matrix [M | I]. Do I have it all correct? Thanks so much @mvw, – User001 Jan 5 at 0:33
I just wanted to get the inverse. But the intermediate forms seem to be already useful for your other issues. – mvw Jan 5 at 0:48

Note that $$M= \underbrace{\begin{bmatrix}I&0\\I&I\end{bmatrix}}_{=P} \underbrace{\begin{bmatrix}A&0\\0&B-A\end{bmatrix}}_{=M^\prime} \underbrace{\begin{bmatrix}I&I\\0&I\end{bmatrix}}_{=Q}$$ But $P$ and $Q$ are invertible so $\DeclareMathOperator{rank}{rank}\rank M=\rank M^\prime$. Since $M^\prime$ is block-diagonal we have $$\rank M=\rank A+\rank(B-A)$$

To see why $\rank M=\rank M^\prime$, note the two facts:

Fact 1. Let $A$ be an $m\times n$ matrix. If $B$ is an $n\times k$ matrix of rank $n$, then $\rank(AB)=\rank(A)$.

In our case take $A=M$ and $B=Q^{-1}$. This implies that $\rank(MQ^{-1})=\rank(M)$.

Fact 2. Let $A$ be an $m\times n$ matrix. If $C$ is an $l\times m$ matrix of rank $m$, then $\rank(CA)=\rank(A)$.

In our case take $A=M^\prime$ and $C=P$. This implies that $\rank(PM^\prime)=\rank(M^\prime)$.

Since $MQ^{-1}=PM^\prime$ we see that $$\rank(M)=\rank(MQ^{-1})=\rank(PM^\prime)=\rank(M^\prime)$$ For more about rank, see wikipedia. Also, try to prove these facts yourself!

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Hi @BrianFitzpatrick, really cool factorization :-). Can I ask you a naive follow-up question? Since $P$ and $Q$ are both invertible, why does determining the rank of $M$ just reduce to checking the rank of $M'$? The only rule / theorem that I know about rank is that rank (AB) = rank (BA), and then some rank inequalities proved very early in the standard textbooks...thanks, – User001 Jan 4 at 23:39
...that's why I feel that when I go to solve a theoretical question regarding rank, inverse or trace, I feel that I have so little tools at my disposal to work with...unlike advanced calculus and complex analysis questions, where I usually immediately see at least two or three ways to approach the problem...@BrianFitzpatrick, thanks, – User001 Jan 4 at 23:41
@User001 See the properties of rank here: en.wikipedia.org/wiki/Rank_(linear_algebra)#Properties – Brian Fitzpatrick Jan 4 at 23:42
Hi @BrianFitzpatrick, with the matrix $M$, and following all of the comments and answers I've gotten so far, I block-row-reduced, but stopped after just one operation, as I think it may be enough: I knocked out the lower left block matrix A. The entries left are A,A on the first row and B in the lower-right corner. With this matrix, which is block-upper-triangular, we can conclude that the rank of $M$ is the rank of A + the rank of (B-A), by standard (non-block) row-reduction techniques. Is my thinking correct? – User001 Jan 5 at 0:13
Also, from this reduced matrix, the determinant of $M$ is just det(A)*det(B-A) (I believe it works because of the zero block matrix below or above the diagonal), as the row operation of adding a scalar multiple of a column / row to another column / row does not change the determinant. Also, my reason for reading off the rank of A and (B-A) to conclude the rank of $M$ is the same reason: row operations are rank-perserving. What do you think? Thanks, @BrianFitzpatrick, – User001 Jan 5 at 0:13

In this case we have rank($M$) = rank($A$) + rank($B-A$). So we can say a lot about the rank of $M$, just not in terms of rank($A$) and rank($B$).

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Hi @hardmath, why rank($B-A$)? Thanks, – User001 Jan 4 at 23:33
Block row elimination: subtract top row blocks from bottom row blocks (or the equivalent block column elimination). – hardmath Jan 4 at 23:36

First note that $$\begin{bmatrix} A & A \\ A & B \\ \end{bmatrix} = \begin{bmatrix} A & 0 \\ 0 & I \\ \end{bmatrix} \begin{bmatrix} I & 0 \\ A & I \\ \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & (B-A) \\ \end{bmatrix} \begin{bmatrix} I & I \\ 0 & I \\ \end{bmatrix}$$ (You can derive this by doing the steps in Gaussian elimination and then writing them as multiplications by "primitive" operation matrices.)

So you can immediately see that $M$ is invertible if and only if both $A$ and $B-A$ are invertible.

Now in that product, the second and fourth matrices are of full rank. Thus the product of the three rightmost matrices will have the same rank deficiency as that of $B-A$. And then the rank deficiency of $M$ is always somewhere between the sum of the deficiencies of $A$ and $B-A$ as an upper bound, and the minimum of those two deficiencies as a lower bound.

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