Question

The two eigenvalues of the matrix \(\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\) are:

• A. 2, 0
• B. 1, 1
• C. 1, 2
• D. 1, 0

Hint:

Use \(\det(A-\lambda I)=0\); the matrix is singular, so one eigenvalue is 0 and the other equals the trace.

Answer 1

The Correct Option is A

Step 1: The eigenvalues \(\lambda\) satisfy the characteristic equation \(\det(A - \lambda I) = 0\), where \(A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\).

Step 2: Form the determinant:
\[\det\begin{pmatrix} 1-\lambda & 1 \\ 1 & 1-\lambda \end{pmatrix} = (1-\lambda)^2 - 1 = 0.\]
Step 3: Expand:
\[\lambda^2 - 2\lambda + 1 - 1 = \lambda^2 - 2\lambda = 0 \;\Rightarrow\; \lambda(\lambda - 2) = 0.\]
Step 4: Hence \(\lambda = 0\) or \(\lambda = 2\).
\[\boxed{\lambda = 2,\; 0}\]