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

Notice that

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

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).