The characteristic equation is \[\det\matrix{6-\lambda & -2 \\ 3 & 1-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 6-\lambda & -2 \\ 3 & 1-\lambda \end{array} \right\vert &= (6-\lambda)(1-\lambda)+2\cdot3 \\
&= (6-\lambda)(1-\lambda)+6 \\ &= \lambda^2-7\,\lambda+12
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-7\,\lambda+12 = (\lambda-4)(\lambda-3) \] So, the eigenvalues are \(\lambda_1 = 4\) and \(\lambda_2 = 3\).

Let \(\lambda = 4\) be an eigenvalue of the matrix \[A=\matrix{6 & -2 \\ 3 & 1}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=4\vec{v}\), this is, \[(A-4 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-4 I\). We can do this through row reduction of the matrix \[A-4 I = \matrix{6 & -2 \\ 3 & 1} - \matrix{4 & 0 \\0& 4 }=\matrix{ 2 & -2 \\ 3 & -3}\] This can be done as follows:
\[\begin{aligned}
\matrix{2&-2\\3&-3\\}&\sim\matrix{1&-1\\3&-3\\}&{\blue{\begin{array}{c}{{1}\over{2}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-3R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 4\) equals \(\left\{ r \cv{1\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\1}\right\rangle\).

Let \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{6 & -2 \\ 3 & 1}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=3\vec{v}\), that is, \[(A-3 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-3 I\). We can do this through row reduction of the matrix \[A-3 I = \matrix{6 & -2 \\ 3 & 1} - \matrix{3 & 0 \\0& 3 }=\matrix{ 3 & -2 \\ 3 & -2}\] This can be done as follows:
\[\begin{aligned}
\matrix{3&-2\\3&-2\\}&\sim\matrix{1&-{{2}\over{3}}\\3&-2\\}&{\blue{\begin{array}{c}{{1}\over{3}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{2}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-3R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{2\\3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{2\\3}\right\rangle\).
If possible, we avoided fractions in the solution.