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 $A$.

Proof:

To show:
$A^{-1}$ exists $\iff$ 0 is not an eigenvalue of $A$.

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.

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.

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 $A$.

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

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?

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