The characteristic equation is \[\det\matrix{1-\lambda & 0 \\ 3 & 2-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 1-\lambda & 0 \\ 3 & 2-\lambda \end{array} \right\vert &= (1-\lambda)(2-\lambda)-0\cdot3 \\
&= (1-\lambda)(2-\lambda)
\end{aligned}\] So, the eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 1\).

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

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