Why is this the inverse of an nxn identity matrix plus an nxn upper triangular matrix? ( I + N ) is ( I − N + N 2 − ⋯ )

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Why is this the inverse of an nxn identity matrix plus an nxn upper triangular matrix?  (  I  +  N  ) is   (  I  −  N  +      N    2    −  ⋯  )

madirans2m · Accepted Answer

Step 1First note that N is nilpotent, i.e., when computing the powers       N    2   ,       N    3   , and so on, the upper triangle gets smaller and smaller until we find that       N    n   is zero (and so are all higher powers). With this in mind,  (  I  +  N  )  (  I  −  N  +      N    2    −      N    3    ±  ⋯  ±      N    n    )  =  (  I  −  N  +      N    2    ∓  ⋯  +  (  −  1      )    n        N    n    )  +  N  (  I  −  N  +      N    2    ∓  ⋯  +  (  −  1      )    n        N    n    )  =  (  I  −  N  +      N    2    ∓  ⋯  +  (  −  1      )    n        N    n    )  +  (  N  −      N    2    ±  ⋯  +  (  −  1      )    n        N          n      +      1        )  =  Ibecause all other summands cancel in pairs (apart from       N          n      +      1       which is zero anyway)

Cameron Benitez · Accepted Answer

Step 1As we know from fifth grade&quot; should not be taken literally. What is alluded to is a formula for the sum of an infinite geometric series:  1  −  r  +      r    2    −      r    3    +      r    4    −  ⋯  =      1          1      +      r      if       |    r      |    &amp;lt;  1. . Maybe most people encounter that in high school.The extent to which it can be applied to matrices is more complicated than       |    r      |    &amp;lt;  1. . However, note that                                  (        I        +        N        )        (        I        −        N        +                  N          2                −                  N          3                +                  N          4                −        ⋯        )                            =                                    I        (        I        −        N        +                  N          2                −                  N          3                +                  N          4                −        ⋯        )                                                          +        N        (        I        −        N        +                  N          2                −                  N          3                +                  N          4                −        ⋯        )                            =                                    I        −        N        +                  N          2                −                  N          3                +                  N          4                −        ⋯                                                        I                                          +        N        −                  N          2                +                  N          3                −                  N          4                +                  N          5                −        ⋯            In the last row, one could say that each term cancels the one above it, so the sum is I. That is problematic since questions of convergence are involved. In particular, what would happen with numbers with absolute values exceeding 1 were there instead of matrices? To think about convergence one must first consider a finite sequence:                          I        −        N        +                  N          2                −                  N          3                +                  N          4                −        ⋯        ±                  N          k                                                                I                                          +        N        −                  N          2                +                  N          3                −                  N          4                +        ⋯        ∓                  N          k                ±                  N                      k            +            1                                              =                                    I        ±                  N                      k            +            1                              and then ask whether       N          k      +      1        →  0 as   k  →  ∞  . .

Why is this the inverse of an nxn identity matrix plus an nxn upper triangular matrix? (I+N) is (I−N+N2−⋯)

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