Powers of a rank-one operator ($K^m=\phi (a{)}^{m-1}K(m\ge 1)$)

Let $Kx=a\phi (x)$ on a vector space, where $\phi$ is a linear functional. Its image of a linear map lies in the span of $a$. Applying $K$ twice gives $K^{2}x=a\phi (a)\phi (x)=\phi (a)Kx$. Mathematical induction then proves the displayed power formula. If $\phi (a)=0$, $K^2=0$, even when $K\ne 0$; this is a nilpotent operator. If $\phi (a)\ne 0$, then $K/\phi (a)$ is an idempotent linear map. In a real inner product space, the special case $\phi (x)=b\cdot x$ gives $\phi (a)=a\cdot b$.

For the bounded rank-one operator $Kx=a\phi (x)$, insert the powers of a rank-one operator into the convergent operator exponential power series. Writing $c=\phi (a)$ gives the displayed formula for $c\ne 0$. At $c=0$ the quotient has limit $t$, so $e^{tK}=I+tK$. This includes the zero operator and the nonzero nilpotent operator cases without dividing by zero.