Show that (D-A)ⁿ annihilates x^n-1e^Ax.

Solution: (proof by induction)

(n=1)

D e^Ax=A e^Ax, so (D-A) e^Ax=A e^Ax-A e^Ax=0 so (D-A) annihilates e^Ax

(n>1)

We assume the theorem is true for the n-1 case, that is, (D-A)^n-1 annihilates x^n-2e^Ax.

Knowing that (D-A)^n-1x^n-2e^Ax=0, we want to show that (D-A)ⁿ x^n-1e^Ax=0.

Note that (D-A)ⁿ x^n-1e^Ax=(D-A)ⁿ x^n-2xe^Ax and with algebra, =(D-A)^n-1x^n-2(D-A)xe^Ax

Examining (D-A)xe^Ax, we get (D-A)xe^Ax=Axe^Ax+e^Ax-Axe^Ax=e^Ax.

Substituting this back into (D-A)^n-1x^n-2(D-A)xe^Ax, we get (D-A)^n-1x^n-2e^Ax which is zero by our assumption.

QED.

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