Given Information
We are given with a matrix A: \[ A = \left[ \begin{array} { c c } { 1 } & { 5 } \\ { - 2 } & { 3 } \end{array} \right] \] We have to find the eigenvalues and a basis for each eigenspace

Step-1: The characteristic equation
Write the characteristic Equation: \[\begin{array}{l} det (A - \lambda I) = 0\\ \left| {\begin{array}{*{20}{c}} {1 - \lambda }&5\\ { - 2}&{3 - \lambda } \end{array}} \right| = 0\\ (1 - \lambda )(3 - \lambda ) + 10 = 0\\ {\lambda ^2} - 4\lambda + 13 = 0 \end{array}\]

Step-2: The eigenvalues
Solve the characteristic polynomial: \[\begin{array}{l} {\lambda ^2} - 4\lambda + 13 = 0\\ \lambda = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(1)(13)} }}{{2(1)}}\\ = \dfrac{{4 \pm \sqrt {16 - 52} }}{2}\\ = \dfrac{{4 \pm \sqrt { - 36} }}{2}\\ = \dfrac{{4 \pm 6i}}{2}\\ = 2 \pm 3i\\ \lambda = 2 - 3i,\,\,2 + 3i \end{array}\] So, eigenvalues are $2 - 3i,\,\,\,{\rm{and}}\,\,2 + 3i$

Step-3: Eigenvectors for $2 - 3i$
Write the equation form \[\begin{array}{l} (A - \lambda I){\bf{x}} = 0\\ (A - \left( {2 - 3i} \right)I){\bf{x}} = 0\\ \left| {\begin{array}{*{20}{c}} {1 - \lambda }&5\\ { - 2}&{3 - \lambda } \end{array}} \right|\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} {3i - 1}&5\\ { - 2}&{1 + 3i} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0 \end{array}} \right] \end{array}\] The augmented matrix: \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} {3i - 1}&5&0\\ { - 2}&{1 + 3i}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{5}{{3i - 1}}}&0\\ { - 2}&{1 + 3i}&0 \end{array}} \right]::\,\,\left\{ {{R_1} = \dfrac{{{R_1}}}{{3i - 1}}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{{3i + 1}}{{ - 2}}}&0\\ { - 2}&{1 + 3i}&0 \end{array}} \right]::\,\,\left\{ {{\rm{Rationalize}}\,\,{\rm{the}}\,\,{\rm{term}}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 2&{ - \left( {3i + 1} \right)}&0\\ { - 2}&{1 + 3i}&0 \end{array}} \right]::\,\,\left\{ {{{\rm{R}}_1} = {R_1} \times 2} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 2&{ - \left( {3i + 1} \right)}&0\\ 0&0&0 \end{array}} \right]::\,\,\left\{ {{{\rm{R}}_2} = {R_1} + {R_2}} \right\} \end{array}\] The system of Equations: \[\begin{array}{l} 2{x_1} + ( - 3i - 1){x_2} = 0\\ {x_2} = \left( {\dfrac{2}{{3i + 1}}} \right){x_1} \end{array}\] The Eigenvector: Choose $x_1 = 1+3i$

\[{v_1} = \left[ {\begin{array}{*{20}{c}} {1 + 3i}\\ 2 \end{array}} \right]\]

Step-4: Eigenvectors for $2 + 3i$
Write the equation form \[\begin{array}{l} (A - \lambda I){\bf{x}} = 0\\ (A - \left( {2 + 3i} \right)I){\bf{x}} = 0\\ \left| {\begin{array}{*{20}{c}} {1 - \lambda }&5\\ { - 2}&{3 - \lambda } \end{array}} \right|\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} { - 1 + 3i}&5\\ { - 2}&{1 + 3i} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0 \end{array}} \right] \end{array}\] The augmented matrix: \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} { - 1 + 3i}&5&0\\ { - 2}&{1 + 3i}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 1 + 3i}&5&0\\ 0&0&0 \end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} + 2{R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} {1 - 3i}&{ - 5}&0\\ 0&0&0 \end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} + 2{R_1}} \right\} \end{array}\] The system of Equations: \[\begin{array}{l} (1 - 3i){x_1} - 5{x_2} = 0\\ {x_1} = \dfrac{{5{x_2}}}{{1 - 3i}} \end{array}\] The Eigenvector: Choose $x_2 = 1-3i$

\[{v_2} = \left[ {\begin{array}{*{20}{c}} 5\\ {1 - 3i} \end{array}} \right]\]