I'm reading a proof of the Nilpotent Splitting Theorem (which states that given a nilpotent operator $T$ of order $m$ and a vector $x\in V$ such that $T^{m-1}x\not=0$, then there exists a direct complement of the span of the set $\lbrace Tx,\ldots T^{m-1}x\rbrace$, which is T-invariant).
At one point it asserts that it is clear that the set $\lbrace Tx,\ldots T^{m-1}x\rbrace$ spans $Im(T)$, but I can't see why this must be true. I know that set is linearly independent and obviously it is included in $Im(T)$, so it would be enough to show that $dim(Im(T))<m$.
Why is this true? I couldn't arrive at a contradiction by assuming the
opposite.
Added:
The Nilpotent-Splitting Theorem:
Suppose $V$ is a real or complex vector space (not necessarily finite dimensional), and $T ∈ L(V)$ is nilpotent of index $m$. Let $x$ be a vector in $V$ with $T^{m−1}x \not= 0$. Let $V_1$ be the span of the list $(x, T x, . . . , T^{m−1}x)$ (so $V_1$ is T-invariant). Then there is a subspace $W_1$ of V that is T-invariant, such that $V = V_1 ⊕ W_1$.
Proof:
The proof proceeds by induction on m, the index of nilpotence.
The case $m = 1$. In this case $T$ is the zero-operator on $V$ , so $x$ is any non-zero vector in $V$ , $V_1$ = span (x), and we can take $W_1$ to be any subspace of $V$ that is complementary to $V_1$ (every subspace of V is invariant for the zero-operator).
The Induction Hypothesis. Suppose $m > 1$ and suppose the result is true for all operators that are nilpotent of index $m − 1$. The proof now proceeds in four steps.
Step I: Pushing Down. We focus on $ran(T)$, which you can easily check is an invariant subspace for $T$ on which $T$ is nilpotent of index $m−1$, and which is clearly the span of the linearly independent list $(T x,... ,T^{m−1}x)=(y, ... ,T^{m−2}y)$, where $y = T x$. So we may apply the induction hypothesis to the restriction of T to ran T, with $V_1$ replaced by the subspace $Y_1 = span (y, ... ,T^{m−2} y)$.
The result is a T-invariant subspace $Y_2$ of $ran (T)$ such that $ran(T) = Y_1 ⊕ Y_2$
And the proof keeps on.
As stated, the claim is false. For example, take $$ T = \pmatrix{0&1\\&0 \\ &&0&1\\&&&0} $$ is nilpotent with $T^2 = 0$. $x = (0,1,0,0)$ satisfies $Tx \neq 0$, but $\{Tx\}$ fails to span the image of $T$, which is $2$-dimensional.
Of course, we still have a satisfactory direct complement, namely the span of $\{(1,0,0,0),(0,0,1,0),(0,0,0,1)\}$.
It seems that, in this step, the author is actually considering the restriction of $T$ to $V_1$, which is the span of $\{x,Tx,\dots,T^{m-1}x\}$. With this adjustment, the claim makes sense.