An <eigenvector> of a <matrix> $A$ is a nonzero vector $v$ for which $Av=\lambda v$. The scalar $\lambda$ is its <eigenvalue>. For a fixed <eigenvalue>, the zero vector together with all corresponding <eigenvectors> forms the <eigenspace> $\ker(A-\lambda I)$. The requirement $v\ne0$ prevents every scalar from being an <eigenvalue> merely because $A0=\lambda0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 1 / 2A / a / Solution

 Solution (source code)