Why $\dim V = \dim (\ker T^n) + \dim (\text{range} T^n)$ does this mean $V = \ker T^n \oplus \operatorname{range} T^n$?

Question

Let $V$ be a vector space and $T \in \mathcal{L}(V)$.

If

$$\dim V = \dim(\ker T^n \oplus \operatorname{range} T^n) = \dim (\ker T^n) + \dim (\operatorname{range} T^n),$$

why does this mean $V = \ker T^n \oplus \operatorname{range} T^n$?

I am self-studying Axler's Linear Algebra Done Right. I am a bit confused, could someone please guide me towards a rigorous proof. Thanks a lot!

Answer 1

Let $W \leq V$ be a subspace with $\dim W = \dim V = n$. Then $W = V$.

Proof: Assume to the contrary that $W \subsetneq V$. Then we can pick $w \in V \setminus W$. Take a basis $\{w_1, \dots, w_n\}$ of $W$. Then $\{w,w_1, \dots, w_n\}$ is a linearly independent set of $n+1$ elements in $V$, which is impossible since $\dim V = n$. Thus, we must have $W = V$. $\quad \square$

Now apply this with $W = \ker T^n \oplus \operatorname{range} T^n \leq V$.