For the matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), if the relation \(a+b=c+d\) holds and \(a,b,c,d \neq 0\), then which one of the following statements about \(A\) is FALSE?
Show Hint
- \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\) is an eigenvector
- \(\lambda = a+b\) is an eigenvalue
- \(\lambda = d-b\) is an eigenvalue
- \(\lambda = d+b\) is an eigenvalue
The Correct Option is D
Solution and Explanation
Multiply \(A\) by the column vector \([1,1]^T\):
\[ A\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}a+b\\c+d\end{bmatrix} \]
We are told \(a+b=c+d\), so both entries of the result are the same number, \(a+b\). This means
\[ A\begin{bmatrix}1\\1\end{bmatrix} = (a+b)\begin{bmatrix}1\\1\end{bmatrix} \]
By definition this makes \([1,1]^T\) an eigenvector of \(A\), with eigenvalue \(\lambda_1 = a+b\). So statements (A) and (B) are both TRUE.
Step 2: Find the second eigenvalue using the trace.
For a 2x2 matrix, the sum of the two eigenvalues always equals the trace (the sum of the diagonal entries):
\[ \lambda_1 + \lambda_2 = a+d \]
We already know \(\lambda_1 = a+b\), so
\[ \lambda_2 = (a+d) - (a+b) = d - b \]
Step 3: Compare this with the remaining statements.
Statement (C) says \(\lambda = d-b\) is an eigenvalue. That is exactly the \(\lambda_2\) we just found, so (C) is TRUE.
Statement (D) says \(\lambda = d+b\) is an eigenvalue. But the real second eigenvalue is \(d-b\), not \(d+b\). Since \(b \neq 0\), \(d+b\) and \(d-b\) are two different numbers, so \(d+b\) cannot equal \(\lambda_2\) in general. It also cannot equal \(\lambda_1=a+b\) unless \(a=d\), which is not guaranteed by the given condition.
Final Answer:
Statement (D), \(\lambda = d+b\) is an eigenvalue, is the FALSE one. \[ \boxed{\lambda = d+b \text{ is NOT an eigenvalue}} \]
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