The characteristic equation is \[\det\matrix{3-\lambda & 0 \\ -15 & -2-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 3-\lambda & 0 \\ -15 & -2-\lambda \end{array} \right\vert &= (3-\lambda)(-2-\lambda)-0\cdot-15 \\
&= (3-\lambda)(-2-\lambda)
\end{aligned}\] So, the eigenvalues are \(\lambda_1 = 3\) and \(\lambda_2 = -2\).

Let \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{3 & 0 \\ -15 & -2}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=3\vec{v}\), this 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 \[A-3 I = \matrix{3 & 0 \\ -15 & -2} - \matrix{3 & 0 \\0& 3 }=\matrix{ 0 & 0 \\ -15 & -5}\] This can be done as follows:
\[\begin{aligned}
\matrix{0&0\\-15&-5\\}&\sim\matrix{-15&-5\\0&0\\}&{\blue{\begin{array}{c}R_2\\R_1\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}-{{1}\over{15}}R_1\\\phantom{x}\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{1\\-3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\-3}\right\rangle\).

Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{3 & 0 \\ -15 & -2}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-2\vec{v}\), that 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 \[A+2 I = \matrix{3 & 0 \\ -15 & -2} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 5 & 0 \\ -15 & 0}\] This can be done as follows:
\[\begin{aligned}
\matrix{5&0\\-15&0\\}&\sim\matrix{1&0\\-15&0\\}&{\blue{\begin{array}{c}{{1}\over{5}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&0\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+15R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{0\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{0\\1}\right\rangle\).
If possible, we avoided fractions in the solution.