Vector spaces are isomorphic iff their dimensions are same

Dependencies:

  1. Linear transformation
  2. Coordinatization over a basis
  3. Basis change is an isomorphic linear transformation

Let $U$ and $V$ be vector spaces. Then they are isomorphic iff there is a bijection from a basis of $U$ to a basis of $V$. The isomorphism is the basis changer function.

This means that if $U$ and $V$ are finite-dimensional vector spaces, they are isomorphic iff $\dim(U) = \dim(V)$.

This also means that a finite-dimensional vector space cannot be isomorphic to an infinite-dimensional vector space, since there cannot be a bijection from a finite basis to an infinite basis.

Proof of 'only-if' part

Let $P$ be a basis of $U$ and $Q$ be a basis of $V$. Let $\phi$ be a bijection from $P$ to $Q$. This means that a basis-changer $T: U \mapsto V$ exists from $P$ to $Q$. The basis changer is an isomorphic linear transformation. Therefore, $U$ is isomorphic to $V$.

Proof of 'if' part

Let $B$ be a basis of $U$. Let $T: U \mapsto V$ be an isomorphism. This means $T$ is one-to-one, therefore, $T$ is a bijection from $B$ to $T(B)$.

Part 1: $\operatorname{span}(T(B)) = T(U)$

Let $T(u) \in T(U)$, where $u \in U$. Since every element in $U$ is representable as a linear combination of $B$, $u = \sum_{b \in B} a_b b$.

\begin{align} T(u) &= T\left(\sum_{b \in B} a_b b \right) \\ &= \sum_{b \in B} a_b T(b) \tag{$\because T$ is a linear transformation} \\ &\in \operatorname{span}(T(B)) \end{align}

Therefore, $T(U) \subseteq \operatorname{span}(T(B))$.

Let $\sum_{b \in B} a_b T(b)$ be an element of $\operatorname{span}(T(B))$. Then \[ \sum_{b \in B} a_b T(b) = T\left( \sum_{b \in B} a_b b \right) \in T(U) \]

Therefore, $\operatorname{span}(T(B)) \subseteq T(U) \Rightarrow \operatorname{span}(T(B)) = T(U)$.

Part 2: $T(B)$ is linearly independent

\[ \sum_{b \in B} a_b T(b) = 0 \Rightarrow T\left( \sum_{b \in B} a_b b \right) = 0 \]

\[ T(0) = T(0 + 0) = T(0) + T(0) \implies T(0) = 0 \]

Since $T$ is a bijection, $\sum_{b \in B} a_b b = 0$. Since $B$ is linearly independent, $\forall b \in B, a_b = 0$. Therefore, $T(B)$ is linearly independent.

Part 3: Conclusion

Since $\operatorname{span}(T(B)) = T(U)$ and $T(B)$ is linearly independent, $T(B)$ is a basis of $T(U)$. Since $T$ is onto, $T(U) = V$.

Therefore, $T$ is a bijection from a basis of $U$ ($B$) to a basis of $V$ ($T(B)$).

Dependency for:

  1. Canonical decomposition of a linear transformation

Info:

Transitive dependencies:

  1. /linear-algebra/vector-spaces/condition-for-subspace
  2. /linear-algebra/matrices/gauss-jordan-algo
  3. /sets-and-relations/equivalence-relation
  4. Group
  5. Ring
  6. Polynomial
  7. Integral Domain
  8. Comparing coefficients of a polynomial with disjoint variables
  9. Field
  10. Vector Space
  11. Linear independence
  12. Span
  13. Linear transformation
  14. Semiring
  15. Matrix
  16. Stacking
  17. System of linear equations
  18. Product of stacked matrices
  19. Matrix multiplication is associative
  20. Reduced Row Echelon Form (RREF)
  21. Matrices over a field form a vector space
  22. Row space
  23. Elementary row operation
  24. Every elementary row operation has a unique inverse
  25. Row equivalence of matrices
  26. Row equivalent matrices have the same row space
  27. RREF is unique
  28. Identity matrix
  29. Inverse of a matrix
  30. Inverse of product
  31. Elementary row operation is matrix pre-multiplication
  32. Row equivalence matrix
  33. Equations with row equivalent matrices have the same solution set
  34. Basis of a vector space
  35. Linearly independent set is not bigger than a span
  36. Homogeneous linear equations with more variables than equations
  37. Rank of a homogenous system of linear equations
  38. Rank of a matrix
  39. Coordinatization over a basis
  40. Basis changer
  41. Basis change is an isomorphic linear transformation