If A is nxn matrix of rank 1, q=dat(1+A), w=trace A then is w-q=1?

Let's break this down:

We are given:

1. \( A \) is an \( n \times n \) matrix with rank 1.
2. \( q = \det(1 + A) \), where \( 1 \) denotes the identity matrix \( I_n \).
3. \( w = \text{trace}(A) \).

We are asked to verify whether \( w - q = 1 \).

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### Step 1: Structure of a Rank 1 Matrix

Since \( A \) is a rank-1 matrix, it can be written as:

\[
A = \mathbf{u} \mathbf{v}^T
\]

where \( \mathbf{u} \) and \( \mathbf{v} \) are \( n \)-dimensional vectors.

### Step 2: Trace of \( A \)

The trace of \( A \), \( w = \text{trace}(A) \), is the sum of the diagonal elements of \( A \). Given that \( A = \mathbf{u} \mathbf{v}^T \), the trace of \( A \) is:

\[
\text{trace}(A) = \mathbf{u}^T \mathbf{v}
\]

Thus, we have:

\[
w = \mathbf{u}^T \mathbf{v}
\]

### Step 3: Determinant \( \det(1 + A) \)

To find \( \det(1 + A) \), we can use the **Matrix Determinant Lemma**. For a rank-1 matrix \( A = \mathbf{u} \mathbf{v}^T \), the determinant of \( 1 + A \) is given by:

\[
\det(1 + A) = (1 + \mathbf{v}^T \mathbf{u})
\]

Thus, \( q = \det(1 + A) \) becomes:

\[
q = 1 + \mathbf{v}^T \mathbf{u}
\]

### Step 4: Relation between \( w \) and \( q \)

We now have:

- \( w = \mathbf{u}^T \mathbf{v} \)
- \( q = 1 + \mathbf{u}^T \mathbf{v} \)

Hence:

\[
w - q = \mathbf{u}^T \mathbf{v} - (1 + \mathbf{u}^T \mathbf{v}) = -1
\]

So, \( w - q = -1 \), not 1.

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### Conclusion:
The expression \( w - q = 1 \) is **incorrect**. The correct result is \( w - q = -1 \).

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Do you want further details on any part of the explanation or examples?

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Here are 5 related questions to expand your understanding:

1. How does the rank of a matrix influence its properties like trace and determinant?
2. What is the significance of the Matrix Determinant Lemma in linear algebra?
3. Can you derive the Matrix Determinant Lemma for a general low-rank matrix?
4. How would the result change if \( A \) had a rank greater than 1?
5. How do the properties of eigenvalues relate to this problem for rank-1 matrices?

**Tip:** When dealing with rank-1 matrices, remember they are highly structured and have simple forms for both determinants and eigenvalues.

If A is nxn matrix of rank 1, q=det(1+A), w=trace A then is w-q=1?

Explore the relationship between the trace and determinant of a rank-1 matrix in the equation w - q = 1. Learn about the Matrix Determinant Lemma, rank-1 matrices, and how these concepts apply to solving this problem. Suitable for undergraduate students studying linear algebra.

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