Linear Transformation of Vector Space Monomorphism

Theorem

Let $G$ and $H$ be a $K$-vector space.

Let $\phi: G \to H$ be a linear transformation.

Then $\phi$ is a monomorphism iff for every linearly independent sequence $\left \langle {a_n} \right \rangle$ of vectors of $G$, $\left \langle {\phi \left({a_n}\right)} \right \rangle$ is a linearly independent sequence of vectors of $H$.

Proof

• Suppose $\phi$ is a monomorphism.

Let $\left \langle {a_n} \right \rangle$ be a linearly independent sequence.

Let:

$\displaystyle \sum_{k=1}^n \lambda_k \phi \left({a_k}\right) = 0$

Then:

$\displaystyle \phi \left({\sum_{k=1}^n \lambda_k a_k}\right) = 0$

So by hypothesis:

$\displaystyle \sum_{k=1}^n \lambda_k a_k = 0$

Hence:

$\forall k \in \left[{1 . . n}\right]: \lambda_k = 0$

• Suppose that for every linearly independent sequence $\left \langle {a_n} \right \rangle$ of vectors of $G$, $\left \langle {\phi \left({a_n}\right)} \right \rangle$ is a linearly independent sequence of vectors of $H$.

Let $\phi \left({a_1}\right) = 0$. Then $a_1 = 0$, otherwise the sequence $\left \langle {a_1} \right \rangle$ of one term would be linearly independent but $\left \langle {\phi \left({a_1}\right)} \right \rangle$ would not.

Thus $\ker \left({\phi}\right) = \left\{{0}\right\}$ and by the Quotient Theorem for Group Epimorphisms $\phi$ is an isomorphism and therefore a monomorphism.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 28$: Theorem $28.6$