A linear transformation is injective iff it preserves linear independence

up:: Linear Transformation

Let $T:U\to V$ be a linear transformation.

Injective $⟹$ Preserves linear independence

Let $T$ be Injective. Then, since A linear transformation is injective iff its kernel is trivial, $\ker T=\{0\}$.

Let $\{v_i\}$ be a Hamel Basis in $V$. Then

$$T(0)=T\left(\sum _i{\alpha }_i{v}_i\right)=\sum _i{\alpha }_{i}T(v_i)$$

can only happen if all ${\alpha }_i=0$. Thus, $\{T(v_i)\}$ is Linearly Independent in $W$.

Preserves linear independence $⟹$ Injective

Per hypothesis, $T$ preserves linear independence from any linearly independent subset $\{v_i{\}}_{i=1}^n\subset V$. In particular, since Every vector space which has a linearly independent set has a Hamel Basis containing this set, it will preserve linear independence from a Hamel Basis $\mathcal{B}$ (which Spans $V$).

For $\mathcal{B}=\{b_i{\}}_i$, we have that ${\sum }_i{\alpha }_i{b}_i=0⟺{\alpha }_i=0$, which is maintained by $T$. Thus,

$$\sum _{i}T({\alpha }_i{b}_i)=0⟺{\alpha }_i=0$$

Thus, $T(v)=0⟺v=0$, and thus $T$ is injective.

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References

• Um Curso de Álgebra Linear, Flávio Ulhoa Coelho & Mary Lilian Lourenço. Editora EDUSP.