I saw a problem today, saying the following:
Find the eigenvalues for following matrices by inspection $$ \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & -1 \\ 0 & 0 & 4 \end{bmatrix} $$
And of course my approach, intuitively, was to make use of the zeros presented, and to find the characteristic polynomial with only the $(\lambda - a)$ fields. However, I found it really inefficient as when it comes to 4-dimensional square matrices, especially for those with insufficient zeros:
$$ \begin{bmatrix} 1&0&0&0 \\ 2&3&0&0 \\ 4&5&6&0 \\ 7&8&9&10 \\ \end{bmatrix} $$So I went seek posts on MSE regarding this, and found this post
The post suggested an approach to easily finding an eigenvalue, which is simply try to minus a thing on its trace, and see if it creates identical rows
And here I found all of entries on the diagonal are eigenvalues by subtracting accordingly and we found 0 on its diagonal