The characteristic equation is \[\det\matrix{-3-\lambda & 0 \\ 0 & 5-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -3-\lambda & 0 \\ 0 & 5-\lambda \end{array} \right\vert &= (-3-\lambda)(5-\lambda)-0\cdot0 \\
&= (-3-\lambda)(5-\lambda)
\end{aligned}\] So, the eigenvalues are \(\lambda_1 = 5\) and \(\lambda_2 = -3\).

Let \(\lambda = 5\) be an eigenvalue of the matrix \[A=\matrix{-3 & 0 \\ 0 & 5}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=5\vec{v}\), this is, \[(A-5 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-5 I\). We can do this through row reduction of the matrix \[A-5 I = \matrix{-3 & 0 \\ 0 & 5} - \matrix{5 & 0 \\0& 5 }=\matrix{ -8 & 0 \\ 0 & 0}\] This can be done as follows:
\[\begin{aligned}
\matrix{-8&0\\0&0\\}&\sim\matrix{1&0\\0&0\\}&{\blue{\begin{array}{c}-{{1}\over{8}}R_1\\\phantom{x}\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 5\) equals \(\left\{ r \cv{0\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{0\\1}\right\rangle\).

Let \(\lambda = -3\) be an eigenvalue of the matrix \[A=\matrix{-3 & 0 \\ 0 & 5}\] 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 \[A+3 I = \matrix{-3 & 0 \\ 0 & 5} - \matrix{-3 & 0 \\0& -3 }=\matrix{ 0 & 0 \\ 0 & 8}\] This can be done as follows:
\[\begin{aligned}
\matrix{0&0\\0&8\\}&\sim\matrix{0&8\\0&0\\}&{\blue{\begin{array}{c}R_2\\R_1\end{array}}}\\\\ &\sim\matrix{0&1\\0&0\\}&{\blue{\begin{array}{c}{{1}\over{8}}R_1\\\phantom{x}\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -3\) equals \(\left\{ r \cv{-1\\0} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-1\\0}\right\rangle\).
If possible, we avoided fractions in the solution.