There is only one linear transformation which maps between bases of equal size

up:: Linear Transformation

Given Vector Spaces $U,V$ with Hamel Basis $B_U\equiv \{u_i{\}}_{i=1}^n\subset U,B_V\equiv \{v_i{\}}_{i=1}^n\subset V$. Then there is a unique linear transformation $T:U\to V$ such that

$$\forall u_i\in B_U:T(u_i)=v_i\in B_V$$

each element in $B_U$ gets mapped to $B_V$.

Proof

Since $B_U$ is a basis for $U$, then each vector $u\in U$ can be written as a Linear Combination

$$u=\sum _i{\alpha }_i{u}_i$$

If we seek a linear transformation, it must preserve these combinations, as well as the hypothesis that $T(u_i)=v_i$.

Therefore, define $T:U\to V$ such that

$$\forall u=\sum _i{\alpha }_i{u}_i\in U:T(u)\stackrel{def}{=}\sum _i{\alpha }_i{v}_i$$

$T$ is linear

Having $u,v\in U$ with respective linear combinations ${\sum }_i{\alpha }_i{u}_i,{\sum }_i{\beta }_i{u}_i$, we have that

$$u+\lambda v=\sum _i({\alpha }_i+\lambda {\beta }_i)u_i$$

whose image under $T$ is

$$T(u+\lambda v)=\sum _i({\alpha }_i+\lambda {\beta }_i)v_i=\sum _i{\alpha }_i{v}_i+\lambda \sum _i{\beta }_i{v}_i=T(u)+\lambda T(v)$$

$T$ is unique

Let $S:U\to V$ be a linear transformation which also preserves basis elements $S(u_i)=v_i$.

Let $u={\sum }_i{\alpha }_i{u}_i\in U$. Then

$$S(u)=\sum _i{\alpha }_{i}S(u_i)=\sum _i{\alpha }_i{v}_i\stackrel{def}{=}T(u)$$

Thus, $\forall u\in U:T(u)=S(u)$, and thus $T=S$.

Corollaries

• Every vector space with same dimension are isomorphic to each other

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References

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