why the quotient space is finite $X/\ker T$ Let $T:X\rightarrow Y$ be a linear operator from Banach space to Banach space, if $Y$ is finite dimensional, show $X/\ker T$ is finite dimensional, moreover has same dimension with $Y$.
Any help is appreciated.

What you seem to be looking for is a proof of the first isomorphism theorem.
Let $\{k_i\}_{i \in I}$ be a basis of $\ker T$. This extends to a basis $\{k_i\}_{i \in I} \cup \{x_j\}_{j \in J}$ of $X$ (where $\operatorname{span} \{k_i\}_{i \in I} \cap \operatorname{span} \{x_j\}_{j \in J} = \{0\}$). Hence $\{ x_j + \ker T\}_{j \in J}$ is a basis of $X / (\ker T)$. We write $[x_j]$ for the element $x_j + \ker T$ of the quotient space, and we get an induced operator $T' \colon X / (\ker T) \to Y$ via $T'[x_j] = T x_j$. The heart of the proof is the following statement:

Since $\{[x_j]\}_{j \in J}$ is linearly independent in $X / (\ker T)$, so is $\{T'[x_j]\}_{j \in J}$ in $Y$.
Proof: By definition of linear independence, we must show that if any finite linear combination of vectors in $\{T'[x_j]\}_{j \in J}$ is the $0$ vector, then the coefficients are all $0$. We have $a_1 T'[x_{j_1}] + \cdots + a_n T'[x_{j_n}] = 0 \implies T\left( a_1 x_{j_1} + \cdots + a_n x_{j_n} \right) = 0$, which implies that $a_1 x_{j_1} + \cdots + a_n x_{j_n} \in \ker T = \operatorname{span} \{k_i\}_{i \in I}$. Since $\operatorname{span} \{k_i\}_{i \in I} \cap \operatorname{span} \{x_j\}_{j \in J} = \{0\}$, we must have $a_1 = \cdots = a_n = 0$, completing the proof.

Since $Y$ is finite dimensional, the above statement implies that $J$ is finite, and hence $X / (\ker T)$ is finite dimensional.
As @Mathematician42 says, it's not necessarily the case that $\dim\left( X / (\ker T) \right) = \dim Y$; we could have some basis vector of $Y$ which is never hit by $T$ (or $T'$).
