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Matrices and rank inequality exercise

  1. Jun 9, 2014 #1
    The problem statement

    Let ##A ∈ K^{m×n}## and ##B ∈ K^{n×r}##
    Prove that min##\{rg(A),rg(B)\}≥rg(AB)≥rg(A)+rg(B)−n##

    My attempt at a solution

    (1) ##AB=(AB_1|...|AB_j|...|AB_r)## (##B_j## is the ##j-th## column of ##B##), I don't know if the following statement is correct: the columns of ##AB## are a linear combination of the columns of ##B##, then ##rg(AB)≤rg(B)##.

    (2)In a similar way, ##AB= \begin{bmatrix} —A_1B— \\ \vdots \\ —A_jB— \\ \vdots \\—A_mB— \end{bmatrix}## (##A_j## denotes the ##j-th## row of ##A##), so the rows of ##AB## are a linear combination of the rows of ##A##, from here one deduces ##rg(AB)≤rg(A)##.

    From (1) and (2) it follows ##rg(AB)≤min\{rg(A),rg(B)\}##.

    This is what I've done so far. I am having doubts with, for example (1), this statement I've conjectured: the columns of ##AB## are a linear combination of the columns of ##B##, then ##rg(AB)≤rg(B)##, but wouldn't this be the case iff ##AB=(α_1B_1|...|α_jB_j|...|α_rB_r)## with ##α_1,...,α_n ∈ K## instead of ##(AB_1|...|AB_j|...|AB_r)## ? This is a major doubt I have, the same goes for (2).

    I need help to show the inequality ##rg(AB)≥rg(A)+rg(B)−n##
     
  2. jcsd
  3. Jun 13, 2014 #2

    LCKurtz

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    Not sure I will answer your question, but I would suggest you might get more responses if you defined your terms. What is ##K^{mxn}##? What does ##rg(A)## mean?
     
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