Question

Two \(n\times n\) matrices \(A\) and \(B\) have a common eigenvalue \(2\), and the same corresponding nonzero eigenvector.
Which of the following options is/are correct?
(Note: \(I\) is the \(n\times n\) identity matrix.)

• A. Determinant \((A-2I)=0\)
• B. Determinant \((B-2I)=0\)
• C. Determinant \((A+B-2I)=0\)
• D. Determinant \((A+B-4I)=0\)

Hint:

Use Av=2v and Bv=2v directly: check which of the four combinations sends the shared eigenvector v to the zero vector.

Answer 1

The Correct Option is A, B, D

Step 1: Write down what is given.
There is a nonzero vector \(v\) such that
\[ Av=2v \quad\text{and}\quad Bv=2v \]
because both matrices share the eigenvalue \(2\) with the same eigenvector.

Step 2: Analyze option (A).
\[ (A-2I)v=Av-2v=2v-2v=0 \]
Since \(v\) is nonzero and \((A-2I)v=0\), the matrix \(A-2I\) is singular. So
\[ \det(A-2I)=0 \]
Option (A) is correct.

Step 3: Analyze option (B).
By the exact same reasoning,
\[ (B-2I)v=Bv-2v=2v-2v=0 \]
so \(B-2I\) is also singular and
\[ \det(B-2I)=0 \]
Option (B) is correct.

Step 4: Analyze option (C).
\[ (A+B-2I)v=Av+Bv-2v=2v+2v-2v=2v \]
Since \(v\) is nonzero, \((A+B-2I)v\) equals \(2v\), which is not the zero vector. This shows \(v\) is not in the null space of \(A+B-2I\), so nothing in the given data forces this matrix to be singular. Option (C) is incorrect.

Step 5: Analyze option (D).
\[ (A+B-4I)v=Av+Bv-4v=2v+2v-4v=0 \]
Since \(v\) is nonzero and \((A+B-4I)v=0\), the matrix \(A+B-4I\) is singular, so
\[ \det(A+B-4I)=0 \]
Option (D) is correct.

Step 6: Final conclusion.
The common eigenvector \(v\) shows directly that \(A-2I\), \(B-2I\), and \(A+B-4I\) are all singular, while \(A+B-2I\) has no such guarantee.
\[ \boxed{\text{(A), (B), and (D) are correct}} \]