Find all of the eigenvalues and eigenvectors of the following matrix. $\begin{bmatrix} 2019 & 2 & 3 & 4 & 5 \\ 2 & 2022 & 6 & 8 & 10 \\ 3 & 6 & 2027 & 12 & 15 \\ 4 & 8 & 12 & 2034 & 20 \\ 5 & 10 & 15 & 20 & 2043\end{bmatrix}$
Alright, there has to be an easier way to do this than finding the nasty determinant. Can anyone get me going here?
As the comments by Arthur and ulaff.net, $$ A-2018I = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 4 & 6 & 8 & 10 \\ 3 & 6 & 9 & 12 & 15 \\ 4 & 8 & 12 & 16 & 20 \\ 5 & 10 & 15 & 20 & 25 \end{pmatrix} \overset{\text{Gaussian}}{\underset{\text{elimination}}{\Longrightarrow}} \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} $$ So we have a four dimensional eigenspace w.r.t. the eigenvalue $2018$, a basis of which is $$ \{ (2,-1,0,0,0), (3,0,-1,0,0), (4,0,0,-1,0), (5,0,0,0,-1) \} $$
As the comment by amd, $$ \begin{align*} \operatorname{trace}A &= (2018+1)+(2018+4)+(2018+9)+(2018+16)+(2018+25) \\ &= (2018 \times 5) + (1+4+9+16+25) \\ &= \sum_{i=1}^5\lambda_i = \text{the sum of all eigenvalues} \end{align*} $$ So we can find one remaining eigenvalue $2018+55=2073$, and its eigenvector is $(1,2,3,4,5)$ since $$ \begin{align*} A-2073I &= (A-2018I)-55I \\ &= \begin{pmatrix} 1-55 & 2 & 3 & 4 & 5 \\ 2 & 4-55 & 6 & 8 & 10 \\ 3 & 6 & 9-55 & 12 & 15 \\ 4 & 8 & 12 & 16-55 & 20 \\ 5 & 10 & 15 & 20 & 25-55 \end{pmatrix} \\ & \overset{\text{Gaussian}}{\underset{\text{elimination}}{\Longrightarrow}} \begin{pmatrix} -1+\frac{1}{55} & \frac{2}{55} & \frac{3}{55} & \frac{4}{55} & \frac{5}{55} \\ 2 & -1 & 0 & 0 & 0 \\ 3 & 0 & -1 & 0 & 0 \\ 4 & 0 & 0 & -1 & 0 \\ 5 & 0 & 0 & 0 & -1 \end{pmatrix} \end{align*} $$