Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Vector Space
Theorem
Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $T : X \to Y$ be a linear transformation.
Then:
- $\dim T \sqbrk X \le \dim X$
Proof
From Vector Space has Basis, there exists a basis $\BB$ for $X$.
By Image of Generating Set of Vector Space under Linear Transformation is Generating Set of Image, $T \sqbrk \BB$ is a generator for $T \sqbrk X$.
From Generator of Vector Space Contains Basis, there exists a basis $\BB'$ for $T \sqbrk X$ such that $\BB' \subseteq T \sqbrk \BB$.
From Cardinality of Image of Mapping not greater than Cardinality of Domain, we have $\card {T \sqbrk \BB} \le \card \BB$.
Also, since $\BB' \subseteq T \sqbrk \BB$, we have:
- $\card {\BB'} \le \card {T \sqbrk \BB}$
Since $\BB'$ is a basis for $T \sqbrk X$, we have:
- $\card {\BB'} = \dim T \sqbrk X$
and since $\BB$ is a basis for $X$ we have:
- $\card \BB = \dim X$
So we obtain:
\(\ds \dim T \sqbrk X\) | \(=\) | \(\ds \card {\BB'}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \card {T \sqbrk \BB}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \card \BB\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dim X\) |
$\blacksquare$