linear algebra - Eigenvalues of a Matrix Using Diagonal Entries

I just started learning about complex eigenvalues and eigenvalues and one example in the book I am using says that the matrix $A = \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}$. The book then says that the eigenvalues are the roots to the characteristic equation $\lambda^2 +1=0$. But from an earlier section I learned that the eigenvalues of a triangular matrix is the entries on the main diagonal. $A$ is triangular when I use the row interchange operation on the matrix and becomes $A = \begin{bmatrix}
1 & 0\\

0 & -1
\end{bmatrix}$. The diagonal entries are $1$ and $-1$ but according to the book, the eigenvalues are $i$ and $-i$.

When given a matrix $A$, can I not use row operations to get it into a row equivalent matrix which is in triangular form and list the diagonal entries as eigenvalues?

Answer

Consider the matrix product
$$
Av=\begin{pmatrix}-&a_1&-\\-&a_2&-\\&...&\\-&a_n&-\end{pmatrix}v=\begin{pmatrix}a_1 v\\a_2 v\\\vdots\\a_n v\end{pmatrix}=\lambda v
$$
compared to

$$
\begin{pmatrix}0&1&&\cdots&\\1&0&&\cdots&\\&&1&\\\vdots&\ddots&&\ddots\\&&&&1\end{pmatrix}Av=\begin{pmatrix}-&a_2&-\\-&a_1&-\\&...&\\-&a_n&-\end{pmatrix}v=\begin{pmatrix}a_2 v\\a_1 v\\\vdots\\a_n v\end{pmatrix}\neq q v,\forall q
$$

so you cannot re-use any eigen vectors.

---

So what about the eigen values. We have
$$

\det(A-\lambda I)=0
$$
and if $B$ equals $A$ with two rows interchanged we have
$$
\det(B-\lambda I)=0
$$
but the rows of $\lambda I$ has NOT been interchanged so the $\lambda$'s has basically been attached to different positions in different row vectors. I know none of this is very deep or general in nature, but it explains what is going on in your specific example.