Finding Eigenvectors when we have lots of zeroes

Question

\begin{array}{cc} 0 & 0 \\ 0 & 8 \\ \end{array}

I have $\lambda_1=8$ and $\lambda_2=0$ but cannot find $V_1$ or $V_2$

I try

\begin{array}{cc} \lambda & 0 \\ 0 & 8-\lambda \\ \end{array} Sub in $\lambda_1$ \begin{array}{cc} \lambda & 0 \\ 0 & \lambda-8 \\ \end{array} Now \begin{array}{cc} -8 & 0 \\ 0 & 0 \\ \end{array} So what is V_1? i dont understand. thank you much

Answer 1

You just solve $Av=0$ and $Av=8v$ respectively. But tha is trivial: in a basis of eigenvectors, the matrix will become a diagonal matrix, with the eigenvalues as diagonal elements. And you matrix is already diagonal. Hence the eigenvalues are $0$ and $8$, and the eigenvectors are $e_1=\begin{pmatrix}1\\0\end{pmatrix}$, $\enspace e_2=\begin{pmatrix}0\\1\end{pmatrix}$.