Find the eigenvalues of the following matrix where $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$ as well as the associated eigenvector 
Find the eigenvalues of the following matrix $\begin{bmatrix}-12&-6&0&0\\8&2&0&0\\0&0&-14&-9\\0&0&42&25\end{bmatrix}$
The eigenvalues are $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$ . Find the associated eigenvector for $\lambda_1, \lambda_2, \lambda_3, \lambda_4$

So should I start out by doing $A - I_m \lambda$ where $I_m$ is the identity matrix? So I would have $\begin{bmatrix}-12&-6&0&0\\8&2&0&0\\0&0&-14&-9\\0&0&42&25\end{bmatrix} - \begin{bmatrix}\lambda&0&0&0\\0&\lambda&0&0\\0&0&\lambda&0\\0&0&0&\lambda\end{bmatrix} = \begin{bmatrix}-12-\lambda&-6&0&0\\8&2-\lambda&0&0\\0&0&-14-\lambda&-9\\0&0&42&25-\lambda\end{bmatrix}$ then would it be correct to then put the matrix in either Upper Triangular Form or Lower Triangular Form so that I can then take the product of the diagonal which is the determinant and then solve for the eigenvalues $\lambda$?

Note that your matrix is the block diagonal matrix
$$
M = 
\left[\begin{array}{rr}
A & 0\\ 0 & B
\end{array}\right]
$$
where
\begin{align*}
A &=
\left[\begin{array}{rr}
-12 & -6 \\
8 & 2
\end{array}\right]
&
B &=
\left[\begin{array}{rr}
-14 & -9 \\
42 & 25
\end{array}\right]
\end{align*}
Next, note that the eigenvalues and corresponding eigenvectors of $A$ are
\begin{align*}
\lambda_1 &= -4 &
v_1 &= (3,-4) &
\lambda_2 &= -6 &
v_2 &= (1,-1)
\end{align*}
and the eigenvalues and corresponding eigenvectors of $B$ are
\begin{align*}
\lambda_3 &= 7 &
v_3 &= (3,-7) &
\lambda_4 &= 4 &
v_4 &= (1,-2)
\end{align*}
The eigenvalues and corresponding eigenvectors of $M$ are thus
\begin{align*}
\lambda_1 &= -4 &
v_1 &= (3,-4,0,0) &
\lambda_2 &= -6 &
v_2 &= (1,-1,0,0) \\
\lambda_3 &= 7 &
v_3 &= (0,0,3,-7) &
\lambda_4 &= 4 &
v_4 &= (0,0,1,-2)
\end{align*}
