In "Linear algebra done right", the author proves by induction that given a vector space $V$, and a linear operator $T$ on that vector space, $V$ is equal to the direct sum of the generalized eigenspaces of $T$.
However, the proof starts saying that if there is only 1 eigenvalue, the thesis is obvious. I cannot understand why it is obvious that, if there is only 1 eigenvalue, than there is an eigenspace equal to $V$
Let $\lambda$ the unique eigenvalue then the characteristic polynomial of $T$ is
$$\chi_T(X)=(X-\lambda)^n$$ where $n=\dim V$ and so by the Cayley-Hamilton theorem $\chi_T(T)=0$ hence $$V=\ker (T-\lambda I)^n$$