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: