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: Set up the shared condition.
Let $v\neq0$ be the common eigenvector, so $Av=2v$ and $Bv=2v$.

Step 2: Test each option by applying it to $v$.
For option (A): $(A-2I)v=2v-2v=0$, so $v$ is a nonzero vector in the null space of $A-2I$, which makes $A-2I$ singular, and the determinant is $0$.
For option (B): the same argument with $B$ gives $(B-2I)v=0$, so $B-2I$ is singular too.

Step 3: Test option (D) by adding the two eigen-equations.
Adding $Av=2v$ and $Bv=2v$ gives $(A+B)v=4v$, so $(A+B-4I)v=0$. Again $v\neq0$ lies in the null space, forcing $\det(A+B-4I)=0$.

Step 4: Test option (C) the same way.
$(A+B-2I)v=(A+B)v-2v=4v-2v=2v$. Since $v\neq0$, this is not the zero vector, so there is no reason from the given data to conclude $\det(A+B-2I)=0$.

Step 5: Collect the valid options.
Only (A), (B), and (D) are guaranteed true from the shared eigenvalue and eigenvector.
\[ \boxed{\text{(A), (B), (D)}} \]