Consider a direct sum vector space V = V 1 ⊕ V 2 , and let X be a matrix acting on V 1 . I recently derived the equation in the title of this question, where 1 denotes identity matrix on V 2 , in the following way. Since 1 = 1 n for all n ≥ 0, and, A B ⊕ C D = ( A ⊕ C ) ( B ⊕ D )for all A,B,C,D, we have: e X ⊕ 1 = ( ∑ n = 0 ∞ X n n ! ) ⊕ 1 = ∑ n = 0 ∞ ( X n n ! ⊕ 1 n ) = ∑ n = 0 ∞ ( X ⊕ 1 ) n n ! = e X ⊕ 1 .

Question

Consider a direct sum vector space   V  =      V    1    ⊕      V    2  , and let X be a matrix acting on       V    1  . I recently derived the equation in the title of this question, where 1 denotes identity matrix on       V    2  , in the following way. Since  1  =      1    n  for all   n  ≥  0, and,  A  B  ⊕  C  D  =  (  A  ⊕  C  )  (  B  ⊕  D  )for all A,B,C,D, we have:                              e                      X                          ⊕        1                            =                  (                      ∑                          n              =              0                                      ∞                                                          X              n                                      n              !                                )                ⊕        1                                          =                  ∑                      n            =            0                                ∞                                    (                                    X              n                                      n              !                                ⊕                      1            n                    )                                                  =                  ∑                      n            =            0                                ∞                                                (            X            ⊕            1                          )              n                                            n            !                                                            =                  e                      X            ⊕            1                          .

Taxinov · Accepted Answer

Let   X  =      [                            2                      ]   and   1  =      I    1    =      [                            1                      ]  . Then       e    X    =      [                                        e            2                                ]   and       e    X    ⊕  1  =      [                                        e            2                                    0                                      0                          1                      ]  , while   X  ⊕  1  =      [                            2                          0                                      0                          1                      ]   which has the exponential       e          X      ⊕      1        =      [                                        e            2                                    0                                      0                                      e            1                                ]  Essentially you use an identity of the form  (  A  +      A    ′    )  ⊕  B  =  (  A  ⊕  B  )  +  (      A    ′    ⊕  B  ) but this isn&#039;t true in general as the right hand side would have a copy of 2B in the bottom right. It is true if B=0!

Why is it not true that e^(X) o+ 1 = e^(X o+ 1)?

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