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.

$\blacksquare$