Prove that finite dimensional $V$ is the direct sum of its generalized eigenspaces $V_\lambda$ Let $T$ be a linear operator on a finite dimensional complex vector space $V$. Prove that $V$ is the direct sum of its generalized eigenspaces.
I already proved that every eigenspace $V_\lambda$ is a $T$ invariant subspace of $V$.
I can find a proof that the generalized eigenspaces are linearly independent.
Can anyone help direct me towards a reasonably straightforward method of proving the rest of this?
Can I prove that the dimensions of the eigenspaces sum to the dimension of $V$? That would complete this, but I can't think of how to do that.
 A: A proof uses the very general following result (known in France as the lemme des noyaux (= kernels lemma):
Lemma: Let $E$ a $K$ vector space (not necessarily finite dimensional), $f$ an endomorphism of $E$, $P_1, \dots, P_n$ pairwise coprime polynomials in $K[x]$. Then
$$\ker\bigl(P_1(f)\bigr)\oplus\dots\oplus \ker\bigl(P_n(f)\bigr)=\ker\bigl((P_1\dots P_n)(f)\bigr)$$
Apply this lemma to the characteristic polynomial $\,\chi_f(x)=(x-\lambda_1)^{m_1}\dots(x-\lambda_r)^{m_r}$. The generalised eigenspaces are precisely the $\,\ker(f-\lambda_i)^{m_i}$s $\,(i=1,\dots,r)$ and $\,\ker\chi_f(x)=\ker0=V$ by Hamilton-Cayley.
Proof of the lemma (sketch):
By induction of the number of factors: we have to prove that if $P$  and $Q$ are coprime polynomials, $\ker P(f)\oplus\ker Q(f)=\ker (P\circ Q)(f) $.
As $(P\circ Q)(f)=(Q\circ P)(f)$, each of $\ker P(f),\ker Q(f)$ is contained in $(P\circ Q)(f)$, hence  $\ker P(f)+\ker Q(f)\subset (P\circ Q)(f)$.
Conversely, since $P$ and $Q$ are coprime, we have a Bézout identity: $U(x)P(x)+ V(x)Q(x)=1$ for some polynomials $U(x), V(x)$. Substituting $f$ for $x$, this translates as:
$$U(f)\circ P(f)+ V(f)\circ Q(f)= \operatorname{Id}_V.$$
Now let $x$ be in $\ker(P\circ Q)(f)$. We have 
$$x=U(f)\circ P(f)(x)+ V(f)\circ Q(f)(x)\tag{1}$$
Let's denote $x_2=U(f)\circ P(f)(x)$, $\,x_1=V(f)\circ Q(f)(x)$. It is easy to check $\,P(x_1)=0$ and $\,Q(x_2)=0$. Thus $x=x_1+x_2$ lies in $\ker P(f)+\ker Q(f)$.
Finally, the sum is direct, if $x\in\ker P(f)\cap\ker Q(f)$, we have $x=0$  from $\,(1)$.
The inductive step is straightforward (left as  an exercise! :o) —.
