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Assume $A,B$ are $n \times n$ matrices.

Prove that the rank of the block matrix that is defined as the following $$ \pmatrix{A& AB\\B&B+B^2} $$ is equal to $ \mbox{rank} A +\mbox{rank} B$.

I have absolutely no clue...

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3 Answers 3

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Denoting the identity matrix by $I_n$, we have $$M=\pmatrix{A&AB\\B&B+B^2}=\underbrace{\pmatrix{A&0\\0&B}}_{D}\underbrace{\pmatrix{I_n&0\\I_n&I_n}}_{S}\underbrace{\pmatrix{I_n&B\\0&I_n}}_{T}.$$ Now it is easy to see that $S, T$ are both invertible, while $D$ has rank equal to rank $A$ + rank $B$, hence $M$ as well. This shows also that the range of $M\in M_{2n}(K)$, seen as a linear operator on $K^n\oplus K^n$, is $\mbox{im }M=\mbox{im } D=\mbox{im }A\oplus \mbox{im }B$.

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  • $\begingroup$ (+1) This is absolutely the way to see what's going on here. $\endgroup$ Aug 15, 2013 at 14:27
  • $\begingroup$ I really like your solution. two things are not clear to me: 1. I don't see how T is clearly invertible. 2. how did you deduce that, because S and T are invertible, and the rank of D is A+B, the rank of the multiplication is A+B? $\endgroup$
    – CnR
    Aug 15, 2013 at 15:16
  • $\begingroup$ @CnR 1. The determinants of $S$ and $T$ is $1$ (they are both triangular with $1$ on the diagonal). It is also easy to find the inverses if you prefer: just replace $B$ by $-B$, and you get the inverse of $T$. Do the same trick with $S$. See transvection matrices for more about this type of matrices. $\endgroup$
    – user85486
    Aug 15, 2013 at 15:21
  • $\begingroup$ I see, and 2: how did you deduce that, because S and T are invertible, and the rank of D is A+B, the rank of the multiplication is A+B? $\endgroup$
    – CnR
    Aug 15, 2013 at 15:23
  • $\begingroup$ @CnR 2. The rank of a (square or rectangular) matrix is not affected by multiplication by an invertible matrix, left or right. For a linear algebraic approach of this, in this case, this is right multiplication. The rank of $M$ is the dimension of the range, that is the vector space $M(K^{2n})$. Now if $S$ is invertible, you have $S(K^{2n})=K^{2n}$ (ie $S$ is surjective). So $MS(K^{2n})=M(K^{2n})$. That is the range of $MS$, whence its rank, is the same as that of $M$. $\endgroup$
    – user85486
    Aug 15, 2013 at 15:26
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Row/column reduction of block matrices works here. First substract $B$ times the first column from the second column. Then subtract the second column from the first, giving you the matrix

$$C = \pmatrix{A& 0\\ 0 & B}$$

This is where the matrices in the answer by Humanity come from. Note that the row operations will not change the rank of the matrix. Hence the matrix has same rank as $C$, and it is not too difficult to see that $C$ has rank equal to $\operatorname{rank} A + \operatorname{rank} B$.

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Let $C,D$ be the reduced row echelon form of $A$ respectively $B$. Let $E,F$ be invertible matrices so that $EA=C$ and $FB=D$.

Then

$$\pmatrix{E& 0\\0&F}\pmatrix{A& AB\\B&B+B^2}=\pmatrix{EA& EAB\\FB&FB+FB^2}$$

Now note that all the leading ones in this matrix appear either in $EA$ or in $FB$. [If a row is $0$ in $EA$, what can you say about the same row in $EAB$?, same for $FB$].

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