If $$AX=\lambda X\tag{1}$$ where A is a $n\times n$ square matrix, X is a $n\times 1$ column vector and $\lambda$ is a number, then X is called the eigenvector of $A$ with eigenvalue $\lambda$.
Now suppose, $$AU=U \begin{pmatrix}\lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0\\ 0 & 0& \lambda_3\end{pmatrix}\tag{2}$$ where A and U are both $3\times 3$ matrices. From this, how can we say that the columns of U are eigenvectors of $A$ with eigenvalues $\lambda_{1,2,3}?$ Does it mean the equation (2) can be broken up into three equations like (1)?
Best Answer
Let $u_i$ be the $i^{th}$ column of $U$. By direct computation, you can verify that the $i^{th}$ column of $$U\begin{bmatrix}\lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0\\ 0 & 0& \lambda_3\end{bmatrix}\tag{2}$$ is simply $\lambda_iu_i$. Therefore, for each column, we have $$Au_i = \lambda u_i$$
Yes, basically