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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)?

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    $\begingroup$ Yes it does. Indeed, if you see equation (2) "columnwise" you get the three equations of (1). $\endgroup$ – Crostul Jan 3 '17 at 8:10
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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 $$Au_i = \lambda u_i$$

Does it mean the equation (2) can be broken up into three equations like (1)?

Yes, basically

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This follows from the definition of matrix multiplication. Given two matrices $A,B$ that are compatible for multiplication, then for the product matrix $AB$ its $j$th column is the same as matrix-vector product $AC_j$ with $C_j$ being the $j$th column of $B$.

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