The characteristic equation is \[\det\matrix{19-\lambda & -42 \\ 7 & -16-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 19-\lambda & -42 \\ 7 & -16-\lambda \end{array} \right\vert &= (19-\lambda)(-16-\lambda)+42\cdot7 \\
&= (19-\lambda)(-16-\lambda)+294 \\ &= \lambda^2-3\,\lambda-10
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-3\,\lambda-10 = (\lambda+2)(\lambda-5) \] So, the eigenvalues are \(\lambda_1 = -2\) and \(\lambda_2 = 5\).

Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{19 & -42 \\ 7 & -16}\] 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 \[A+2 I = \matrix{19 & -42 \\ 7 & -16} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 21 & -42 \\ 7 & -14}\] This can be done as follows:
\[\begin{aligned}
\matrix{21&-42\\7&-14\\}&\sim\matrix{1&-2\\7&-14\\}&{\blue{\begin{array}{c}{{1}\over{21}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-2\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-7R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{-2\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-2\\-1}\right\rangle\).

Let \(\lambda = 5\) be an eigenvalue of the matrix \[A=\matrix{19 & -42 \\ 7 & -16}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=5\vec{v}\), that 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 \[A-5 I = \matrix{19 & -42 \\ 7 & -16} - \matrix{5 & 0 \\0& 5 }=\matrix{ 14 & -42 \\ 7 & -21}\] This can be done as follows:
\[\begin{aligned}
\matrix{14&-42\\7&-21\\}&\sim\matrix{1&-3\\7&-21\\}&{\blue{\begin{array}{c}{{1}\over{14}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-3\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-7R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 5\) equals \(\left\{ r \cv{3\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{3\\1}\right\rangle\).
If possible, we avoided fractions in the solution.