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Let \(\lambda \) is a scaler, \(A\) is an \(n \times n\) matrix and \({\bf{x}}\) is an eigenvector corresponding to \(\lambda \), \(\lambda \) is said to an eigenvalue of the matrix \(A\) if there exists a nontrivial solution \({\bf{x}}\) of \(A{\bf{x}} = \lambda {\bf{x}}\).

Denote the given matrix by \(A\).

\(A = \left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\)

According to the definition of eigenvalue, \(\lambda = 2\) is the eigenvalue of the matrix \(A\), if satisfies the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

Substitute \(\lambda = 2\) into \(A{\bf{x}} = \lambda {\bf{x}}\).

\(A{\bf{x}} = 2{\bf{x}}\)

The obtained equation is equivalent to \(\left( {A - 2I} \right){\bf{x}} = 0\), which is a homogeneous equation.

So, first, solve the matrix \(\left( {A - 2I} \right)\) by using \(A = \left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\).

\(\begin{array}{c}\left( {A - 2I} \right) = \left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right) - 2\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right) - \left( {\begin{array}{*{20}{c}}2&0\\0&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&2\\3&6\end{array}} \right)\end{array}\)

It can be observed that the columns of the obtained matrix are linearly dependent, where elements of the second column twice the multiple of elements of the first column, which can be written as:

\(\left( {A - 2I} \right) = \left( {\begin{array}{*{20}{c}}1&{2\left( 1 \right)}\\3&{2\left( 3 \right)}\end{array}} \right)\)

Hence, \(\left( {A - 2I} \right){\bf{x}} = 0\) has a nontrivial solution, so \(\lambda = 2\) is an eigenvalue of the given matrix \(\left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\).