Can you use row operations to reduce a matrix to either upper or lower triangular to find the eigenvalues?

Question

For example, we have the matrix $$\begin{bmatrix}-1&0&1\\2&6&-14\\1&0&-1 \end{bmatrix}$$ Using row operations $R_3+R_1$ and $R_2 + 2R_1$ and finally $-1R_1$, we arrive at $$\begin{bmatrix}1&0&-1\\0&6&-12\\0&0&0\end{bmatrix}$$ which is an upper diagonal matrix and as such to find the eigenvalues of this we can simply do $\det(A-\lambda I)$ to achieve $$\det\begin{bmatrix}1-\lambda&0&-1\\0&6-\lambda&-12\\0&0&-\lambda\end{bmatrix}$$ which because it's upper diagonal, we can simply multiply the diagonal entries to get $$(1-\lambda)(6-\lambda)(-\lambda)\\\lambda = 1,6,0$$ I believe I'm mistaken somewhere as the eigenvalues should be 2, 6, 0, but I don't know where I'm going wrong or where I'm mistaken. Can someone let me know where my understanding is mistaken.

Answer 1

One possibility is to correctly produce a sequence of similar matrices. This means writing your row operations as (a product of) matrices, call that $L.$ But you must also apply inverse column operations, that is multiply on the right by $L^{-1}. \; \;$ Your first two row operations are accomplished by multiplying on the left by

$$ L = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \\ \end{array} \right) $$

Try it!

We find $$ R = L^{-1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \\ \end{array} \right) $$

Let us name your matrix $M$

$$ M = \left( \begin{array}{rrr} -1 & 0 & 1 \\ 2 & 6 & -14 \\ 1 & 0 & -1 \\ \end{array} \right) $$

Then $LMR$ comes out

$$ \left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} -1 & 0 & 1 \\ 2 & 6 & -14 \\ 1 & 0 & -1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} -2 & 0 & 1 \\ 0 & 6 & -12 \\ 0 & 0 & 0 \\ \end{array} \right) $$

Since the resulting matrix is triangular, the diagonal elements are the eigenvalues.

Let me repeat that this matrix really is similar to the original.