Science:Math Exam Resources/Courses/MATH307/December 2008/Question 06 (b)/Solution 1

Recall the definition of the eigenvalue, we know that $-i$ is one of Q’s eigenvalues when $det(Q+iI)=0$, and $Q+iI$ is invertible if and only if $det(Q+iI)\neq 0$. From part a, we know that Q is a symmetric matrix. A symmetric matrix will only have real eigenvalues and since $-i$ is not real, it is not an eigenvalue of Q and $det(Q+iI)$ is never equal to zero.

Therefore, the matrix $Q+iI$ is invertible.