Prove that an n × n upper triangular matrix is invertible if and only if all of its
diagonal entries are non-zero.
2


To prove that an \( n \times n \) upper triangular matrix is invertible if and only if all of its diagonal entries are non-zero, we will break the proof into two parts: 

1. **If an upper triangular matrix is invertible, then all of its diagonal entries are non-zero.**
2. **If all diagonal entries of an upper triangular matrix are non-zero, then it is invertible.**

### Part 1: If an upper triangular matrix is invertible, then all of its diagonal entries are non-zero.

Let \( A \) be an \( n \times n \) upper triangular matrix. Suppose \( A \) is invertible. For an \( n \times n \) matrix, \( A \), to be invertible, its determinant must be non-zero, i.e., \( \det(A) \neq 0 \).

The determinant of an upper triangular matrix is the product of its diagonal elements. Thus,

\[
\det(A) = a_{11} \cdot a_{22} \cdot \dots \cdot a_{nn}
\]

where \( a_{ii} \) represents the diagonal entries of \( A \). Since \( \det(A) \neq 0 \), this product must be non-zero. Therefore, each \( a_{ii} \neq 0 \). Hence, if the matrix is invertible, all diagonal entries must be non-zero.

### Part 2: If all diagonal entries of an upper triangular matrix are non-zero, then it is invertible.

Now, suppose that all diagonal entries of \( A \), an \( n \times n \) upper triangular matrix, are non-zero. We want to prove that \( A \) is invertible.

Since \( A \) is upper triangular, its determinant is again the product of its diagonal entries:

\[
\det(A) = a_{11} \cdot a_{22} \cdot \dots \cdot a_{nn}
\]

Given that all \( a_{ii} \neq 0 \), we know that the determinant of \( A \) is non-zero:

\[
\det(A) \neq 0
\]

If the determinant of a matrix is non-zero, the matrix is invertible. Therefore, \( A \) is invertible.

### Conclusion:

We have proved both directions:

- If an upper triangular matrix is invertible, all of its diagonal entries must be non-zero.
- If all the diagonal entries of an upper triangular matrix are non-zero, then the matrix is invertible.

Thus, an \( n \times n \) upper triangular matrix is invertible if and only if all of its diagonal entries are non-zero.

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Would you like to go over any specific part of the proof, or explore further details?

Here are some related questions for further exploration:

1. What is the determinant of a lower triangular matrix, and how does it relate to invertibility?
2. How do the diagonal entries affect the rank of an upper triangular matrix?
3. What happens if only some diagonal entries of an upper triangular matrix are zero?
4. Can a matrix with zero diagonal entries ever be invertible if it’s not upper triangular?
5. How does the proof change for block triangular matrices?

**Tip**: The determinant of triangular matrices (both upper and lower) is always the product of the diagonal entries, which makes them easier to analyze for invertibility.

Prove that an n × n upper triangular matrix is invertible if and only if all of its diagonal entries are non-zero.

Understand how to prove that an n x n upper triangular matrix is invertible if and only if all its diagonal entries are non-zero. This step-by-step explanation covers key linear algebra concepts such as matrix invertibility and determinant properties. Ideal for undergraduate students learning matrix theory.

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