## What is an endomorphism of a vector space?

In mathematics, an endomorphism is a morphism from a mathematical object to itself. An endomorphism that is also an isomorphism is an automorphism. For example, an endomorphism of a vector space V is a linear map f: V → V, and an endomorphism of a group G is a group homomorphism f: G → G.

### Is a vector space over Q?

We’ve just noted that R as a vector space over Q contains a set of linearly independent vectors of size n + 1, for any positive integer n. Hence R cannot have finite dimension as a vector space over Q. That is, R has infinite dimension as a vector space over Q.

#### How do you free a vector space?

{f:X→K|f−1(K\{0})is finite}. { f : X → 𝕂 | f – 1 ( 𝕂 \ { 0 } ) is finite } . by 0 . The vector space structure for C(X) is defined as follows….References.

Title | free vector space over a set |
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Synonym | vector space generated by a set |

Related topic | TensorProductBasis |

**Is C vector space over Q?**

Thus, C is a two-dimensional R-vector space (and, as any field, one-dimensional as a vector space over itself, C). If α is not algebraic, the dimension of Q(α) over Q is infinite.

**What is meant by endomorphism?**

Definition of endomorphism : a homomorphism that maps a mathematical set into itself — compare isomorphism.

## What is the difference between endomorphism and automorphism?

As nouns the difference between automorphism and endomorphism. is that automorphism is (mathematics) an isomorphism of a mathematical object or system of objects onto itself while endomorphism is (geology) the assimilation of surrounding rock by an intrusive igneous rock.

### Why Z is not a vector space?

The answer is, “Yes.” If V is a vector space over a field of positive characteristic, then as an abelian group, every element of V has finite order. If V is a vector space over a field of characteristic 0, then as an abelian group, V is divisible. The abelian group Z has neither of these properties.

#### Why Q R is not a vector space?

If Q is a vector space over R, there is a scalar multiplication * such that x*1 is a rational number for all real numbers x. The function f(x)=x*1 is an injective function, so Q has a subset of the same cardinality as R. This is not possible because Q is countable and R is uncountable.

**Is every free module a vector space?**

In mathematics, a free module is a module that has a basis – that is, a generating set consisting of linearly independent elements. Every vector space is a free module, but, if the ring of the coefficients is not a division ring (not a field in the commutative case), then there exist non-free modules.

**Is span a vector space?**

The linear span of a set of vectors is therefore a vector space.

## Is R over CA vector space?

a vector space over its over field. For example, R is not a vector space over C, because multiplication of a real number and a complex number is not necessarily a real number.

### How do you prove endomorphism?

Theorem

- Let (Z,+) be the additive group of integers.
- Then ϕ is a group endomorphism if and only if:
- We have that n=n⏞1+⋯+1 for any positive integer n.
- Thus ϕ:(Z,+)→(Z,+) is a group homomorphism from Z to Z.
- Hence by definition ϕ is a group endomorphism.