Linear Transformation has Finite Rank iff Domain Quotiented by Kernel is Finite Dimensional
Theorem
Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a linear transformation.
Let $\ker T$ be the kernel of $T$.
Let $X/\ker T$ be the quotient vector space of $X$ modulo $\ker T$.
Then $T$ has finite rank if and only if $X/\ker T$ is finite dimensional.
Proof
Let $q : X \to X/\ker T$ be the quotient mapping.
Sufficient Condition
Suppose that $X/\ker T$ is finite dimensional.
Let $\set {\map q {x_1}, \map q {x_2}, \ldots, \map q {x_n} }$ be a basis for $X/\ker T$.
Let $x \in X$.
Then there exists $\alpha_1, \alpha_2, \ldots, \alpha_n \in \GF$ such that:
- $\ds \map q x = \sum_{i \mathop = 1}^n \alpha_i \map q {x_i}$
From Quotient Mapping is Linear Transformation, we have:
- $\ds \map q {x - \sum_{i \mathop = 1}^n \alpha_i x_i} = 0$
So there exists $f \in \ker T$ such that:
- $\ds x = f + \sum_{i \mathop = 1}^n \alpha_i x_i$
Then:
- $\ds T x = \sum_{i \mathop = 1}^n \alpha_i T x_i$
since $T$ is linear.
So the range of $T$ is generated by $\set {T x_1, T x_2, \ldots, T x_n}$.
$\Box$
Necessary Condition
Suppose that the range of $T$ is finite dimensional.
Let:
- $\set {T x_1, \ldots, T x_n}$
be a basis for the range of $T$.
Let $x \in X$.
Then there exists $\alpha_1, \ldots, \alpha_n \in \GF$ such that:
- $\ds T x = \sum_{i \mathop = 1}^n \alpha_i T x_i$
Then:
- $\ds \map T {x - \sum_{i \mathop = 1}^n \alpha_i x_i} = 0$
That is:
- $\ds x - \sum_{i \mathop = 1}^n \alpha_i x_i \in \ker T$
so that:
- $\ds \map q {x - \sum_{i \mathop = 1}^n \alpha_i x_i} = 0$
Then:
- $\ds \map q x = \sum_{i \mathop = 1}^n \alpha_i \map q {x_i}$
from Quotient Mapping is Linear Transformation.
That is:
- $\map q x \in \span \set {\map q {x_1}, \ldots, \map q {x_n} }$
Since $x \in X$ was arbitrary, we deduce that $X/\ker T$ is generated by $\set {\map q {x_1}, \map q {x_2}, \ldots, \map q {x_n} }$.
So $X/\ker T$ is finite dimensional.
$\blacksquare$