Prove the set of invertible elements of an algebra is closed under multiplication by nonzero scalars.

To prove that the set of invertible elements of an algebra \( A \) is closed under multiplication by nonzero scalars, let's consider the following:

Let \( A \) be an algebra over a field \( F \), and let \( \lambda \in F \) be a nonzero scalar. Suppose \( a \in A \) is an invertible element. We need to show that \( \lambda a \) is also invertible in \( A \).

### Proof:

1. **Existence of an Inverse for \( a \)**:
    - Since \( a \) is invertible, there exists an element \( a^{-1} \in A \) such that:
      \[
      a a^{-1} = a^{-1} a = 1_A
      \]
      where \( 1_A \) is the multiplicative identity in \( A \).

2. **Consider the Element \( \lambda a \)**:
    - We need to find an element \( b \in A \) such that:
      \[
      (\lambda a) b = b (\lambda a) = 1_A.
      \]
    - Let's propose \( b \) in the form of:
      \[
      b = \frac{1}{\lambda} a^{-1},
      \]
      where \( \frac{1}{\lambda} \) is the multiplicative inverse of \( \lambda \) in the field \( F \) (since \( \lambda \) is nonzero, its inverse exists in \( F \)).

3. **Verification**:
    - We now check if this \( b \) indeed serves as the inverse of \( \lambda a \):
      \[
      (\lambda a) \left( \frac{1}{\lambda} a^{-1} \right) = \lambda \left( a \frac{1}{\lambda} \right) a^{-1} = \lambda \left( \frac{1}{\lambda} (a a^{-1}) \right) = \lambda \left( \frac{1}{\lambda} \cdot 1_A \right) = \lambda \cdot \frac{1}{\lambda} \cdot 1_A = 1_A.
      \]
    - Similarly:
      \[
      \left( \frac{1}{\lambda} a^{-1} \right) (\lambda a) = \frac{1}{\lambda} \left( a^{-1} \lambda \right) a = \frac{1}{\lambda} \left( \lambda (a^{-1} a) \right) = \frac{1}{\lambda} \left( \lambda \cdot 1_A \right) = \frac{1}{\lambda} \cdot \lambda \cdot 1_A = 1_A.
      \]

4. **Conclusion**:
    - Therefore, \( b = \frac{1}{\lambda} a^{-1} \) is indeed the inverse of \( \lambda a \).
    - Hence, \( \lambda a \) is invertible in \( A \).

Thus, we have shown that if \( a \) is invertible in the algebra \( A \), then \( \lambda a \) is also invertible for any nonzero scalar \( \lambda \) from the field \( F \).

### Summary:
The set of invertible elements of an algebra \( A \) is closed under multiplication by nonzero scalars.

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**Relative Questions:**

1. What is the definition of an algebra over a field?
2. How do you prove that the product of two invertible elements is invertible?
3. What is the importance of the field \( F \) being a field in this proof?
4. How does the invertibility of a scalar affect the invertibility of elements in an algebra?
5. Can you give an example of an algebra and its set of invertible elements?
6. How does this proof change if \( \lambda \) were zero?
7. What properties of the field \( F \) are used in this proof?
8. Can you prove the closure under addition for the set of invertible elements?

**Tip:** When working with algebras, always check the properties of the underlying field, as these properties often facilitate proofs and operations within the algebra.

Explore how to prove that the set of invertible elements in an algebra is closed under multiplication by nonzero scalars. This proof involves algebraic structures, fields, and the properties of invertible elements. Suitable for advanced undergraduate students and anyone interested in algebraic structures.

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