Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 25/Solution 1

Notice that

$\det (A-I\lambda )=\det (\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]-\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right])=-{\lambda }^3$

and the roots of this polynomial are $\lambda =0$. This is a triple root and so this is the only eigenvalue. Next, for any vector v, we have

$Av=0=0v$

and hence every nonzero vector v is an eigenvector (the zero vector is excluded by definition).