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    Possible rank triples for three matrices summing to zero

    P-BKF07-6B

    Area
    Prelim
    Seen at
    BERKELEY
    Years
    2007
    Status
    draft

    Given a positive integer \(n\), determine all possible triples \[ (\operatorname{rank}A,\operatorname{rank}B,\operatorname{rank}C) \] as \(A,B,C\) range over real \(n\times n\) matrices satisfying \[ A+B+C=0. \]

    Solution

    Let \[ a\coloneqq\operatorname{rank}A, \qquad b\coloneqq\operatorname{rank}B, \qquad c\coloneqq\operatorname{rank}C. \]

    1. Every realizable triple \((a,b,c)\) satisfies\[ 0\le a,b,c\le n \]and\[ a\le b+c, \qquad b\le c+a, \qquad c\le a+b. \]

    The bounds \(0\le a,b,c\le n\) hold for ranks of \(n\times n\) matrices. Since \(C=-(A+B)\), \[ \operatorname{im}C \subseteq \operatorname{im}A+\operatorname{im}B. \] Therefore \[ c\le a+b. \] The other two inequalities follow in the same way from \(A=-(B+C)\) and \(B=-(C+A)\).

    2. Conversely, let \((a,b,c)\in\{0,\ldots,n\}^3\) satisfy thethree inequalities in step

    1. After permuting the three matrices ifnecessary, assume\[ c\ge a, \qquad c\ge b. \]

    The conditions and the equation \(A+B+C=0\) are symmetric under permuting \(A,B,C\). Thus one may label a largest member of \(\{a,b,c\}\) by \(c\).

    3. Define diagonal matrices \(A\) and \(B\) by\[ A_{ii}= \begin{cases} 1,&1\le i\le a,\\ 0,&a<i\le n, \end{cases} \]and\[ B_{ii}= \begin{cases} 1,&c-b<i\le c,\\ 0,&\text{otherwise}. \end{cases} \]Then\[ \operatorname{rank}A=a, \qquad \operatorname{rank}B=b, \qquad \operatorname{rank}(A+B)=c. \]

    The first two rank equalities are immediate from the numbers of nonzero diagonal entries. The inequality \(c\le a+b\) is equivalent to \[ c-b\le a. \] Hence the support intervals \[ \{1,\ldots,a\}, \qquad \{c-b+1,\ldots,c\} \] overlap or meet so that their union is exactly \(\{1,\ldots,c\}\). On the overlap the corresponding diagonal entry of \(A+B\) is \(2\), which is nonzero over \(\RR\); elsewhere in the union it is \(1\). Thus \(A+B\) has exactly \(c\) nonzero diagonal entries, and \(\operatorname{rank}(A+B)=c\).

    4. Setting \(C\coloneqq-(A+B)\) realizes the triple \((a,b,c)\).

    By construction, \[ A+B+C=0. \] Step 3 gives \[ \operatorname{rank}C =\operatorname{rank}(A+B) =c, \] while the ranks of \(A\) and \(B\) are \(a\) and \(b\).

    5. The possible rank triples are exactly\[ \boxed{ \left\{ (a,b,c)\in\{0,\ldots,n\}^3: a\le b+c,\ b\le c+a,\ c\le a+b \right\}. } \]

    Step 1 gives the necessary conditions, and steps <1>2–<1>4 realize every triple satisfying them.

    6. Q.E.D.

    Step 5 is the requested classification.