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### Problem Recap:
The question is:  
**Let \( A \) be an \( n \times n \) square matrix. Prove that \( A^{-1} \) exists if and only if 0 is not an eigenvalue of \( A \).**

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### Proof:
To show:  
\( A^{-1} \) exists \( \iff \) 0 is **not** an eigenvalue of \( A \).

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#### 1. **Forward direction**:  
Assume \( A^{-1} \) exists. We must prove that 0 is not an eigenvalue of \( A \).

- Recall: \( A^{-1} \) exists if and only if \( A \) is invertible.
- A matrix \( A \) is invertible if and only if **determinant \( \det(A) \neq 0 \)**.
- A number \( \lambda \) is an **eigenvalue** of \( A \) if there exists a non-zero vector \( \mathbf{v} \) such that:
  \[ A \mathbf{v} = \lambda \mathbf{v}. \]

- If \( 0 \) were an eigenvalue of \( A \), then for some non-zero \( \mathbf{v} \), we would have:
  \[ A \mathbf{v} = 0 \cdot \mathbf{v} = \mathbf{0}. \]

- This would mean that \( A \) has a **non-trivial solution** to the homogeneous equation \( A \mathbf{v} = \mathbf{0} \), implying that \( A \) is **singular** (i.e., not invertible).  
  But since we assumed \( A \) is invertible, this contradiction shows that **0 cannot be an eigenvalue**.

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#### 2. **Reverse direction**:  
Now, assume that 0 is not an eigenvalue of \( A \). We must prove that \( A \) is invertible.

- Since 0 is not an eigenvalue, the equation \( A \mathbf{v} = 0 \) has only the **trivial solution** \( \mathbf{v} = \mathbf{0} \).
- This implies that the matrix \( A \) is **injective** (one-to-one) and thus has **full rank**.
- A square matrix \( A \) with full rank has \( \det(A) \neq 0 \), which implies that \( A \) is invertible, meaning \( A^{-1} \) exists.

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### Conclusion:
We have shown both directions:
- If \( A^{-1} \) exists, then 0 is not an eigenvalue of \( A \).
- If 0 is not an eigenvalue of \( A \), then \( A^{-1} \) exists.

Thus, \( A^{-1} \) exists if and only if 0 is not an eigenvalue of \( A \).

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Would you like any part of the proof clarified or expanded upon?

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### 5 Related Questions:
1. What does it mean for a matrix to be singular or non-singular?
2. How are eigenvalues and determinants of a matrix related?
3. Can a matrix with all non-zero eigenvalues be non-invertible?
4. How do you compute the inverse of a matrix using Gaussian elimination?
5. What are some real-world applications of matrix inverses?

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**Tip:**  
A matrix is invertible if and only if its determinant is non-zero, which makes checking invertibility easy for smaller matrices!

Let A be an n×n square matrix. Prove A⁻¹ exists if and only if 0 is not an eigenvalue of A.

Learn how to prove that the inverse of an n×n matrix A exists if and only if 0 is not an eigenvalue of A. This proof explores the relationship between matrix inversion, eigenvalues, and determinants, based on key concepts in linear algebra. Ideal for undergraduate mathematics students.

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