Question

Let \(A\) be a \(3 \times 3\) real matrix such that \(A \ne I_3\) and the sum of the entries in each row of \(A\) is 1. Then, which of the following statements is/are necessarily TRUE?

• A. \(A - I_3\) is an invertible matrix
• B. The set \(\{x \in \mathbb{R}^3 : (A - I_3)x = 0\}\) has at least two elements (\(x\) is a column vector)
• C. The characteristic polynomial, \(p(\lambda)\), of \(A + 2A^2 + A^3\) has \((\lambda - 4)\) as a factor
• D. \(A\) cannot be an orthogonal matrix

Hint:

For matrices where each row sums to 1, \([1,1,1]^T\) is always an eigenvector corresponding to eigenvalue 1. Such matrices are not invertible if \(A \ne I\).

Answer 1

The Correct Option is B, C

Step 1: Analyze row sum property.
Given that the sum of entries in each row of \(A\) is 1, we can write \[ A \begin{bmatrix} 1
1
1 \end{bmatrix} = \begin{bmatrix} 1
1
1 \end{bmatrix}. \] Hence, \( \lambda = 1 \) is an eigenvalue of \(A\), with eigenvector \(v = [1, 1, 1]^T.\)
Step 2: Examine \(A - I_3\).
Since \(A v = v\), we have \((A - I_3)v = 0\), i.e., \(v\) lies in the null space of \(A - I_3\). Therefore, \(A - I_3\) is not invertible and its null space contains at least one non-zero vector. Hence, (A) is false and (B) is true since the null space has at least two elements (0 and \(v\)).
Step 3: Check orthogonality.
If \(A\) were orthogonal, all eigenvalues would have absolute value 1. However, the condition that all row sums are 1 and \(A \ne I_3\) violates orthogonality, since orthogonal matrices with eigenvalue 1 must have other eigenvalues ±1 or complex, which would alter row sums. Hence, (D) is true.
Step 4: Check (C).
No general guarantee exists that the polynomial \(A + 2A^2 + A^3\) has \((\lambda - 4)\) as a factor without specific eigenvalues of \(A\). So (C) is not necessarily true. Final Answer: \[ \boxed{\text{(B) and (D) are true.}} \]