Discussion
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?
Answer: D
The FALSE statement is \(\lambda = d + b\) is an eigenvalue.
Multiply \(A\) by the vector of ones:
\[ A\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} a + b \\ c + d \end{bmatrix} = (a + b)\begin{bmatrix} 1 \\ 1 \end{bmatrix} \]
because \(a + b = c + d\). So \([1\ \ 1]^{T}\) is an eigenvector with eigenvalue \(a + b\), and the first two statements are true.
The eigenvalues of a 2 x 2 matrix add up to its trace \(a + d\), so the other eigenvalue is \((a + d) - (a + b) = d - b\), and the third statement is true too. \(d + b\) could equal \(d - b\) only if \(b = 0\), which is excluded, and could equal \(a + b\) only in the special case \(d = a\). In general it is not an eigenvalue, so that statement is the false one.
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