# When calculating eigenvectors, is it incorrect to first convert the matrix $A$ into reduced row echelon form?

When calculating eigenvectors, is it incorrect to first convert the matrix $A$ into reduced row echelon form?

The following is an example of the phenomenon I am describing.

Let $A =$ $$\begin{matrix} 1 & 2 & 6 \\ 0 & 3 & 5 \\ 0 & 0 & 4 \\ \end{matrix}$$

This can be converted into reduced row echelon form $A =$ $$\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}$$

Therefore, since this is an upper-triangular matrix,

$\det(A-\lambda I) = (1 - \lambda)(1 - \lambda)(1 - \lambda)$

$\therefore \lambda = 1$

However, if we did not convert $A$ into reduced row echelon form first, we would get

$\det(A-\lambda I) = (1 - \lambda)(3 - \lambda)(4 - \lambda)$

$\therefore \lambda = 1, 3, 4$

These are different answers. However, if there are errors in my reasoning, they are not evident to me. I would appreciate it if someone could please clarify any errors in my reasoning, why the reasoning is erroneous, and what the correct reasoning should be.

Thank you.

Well, clearly it is wrong and your case demonstrates that it must be wrong, since the eigenvectors of the reduced echelon form are all vectors in $\mathbb R^3$, while this is obviously not true for the original matrix $A$ ($[1,1,1]^T$, for example is an eigenvector of the second, but not the first, matrix).
Converting a matrix into row echelon form is equivalent to finding an invertible matrix $B$ such that $BA=\overline A$ (where $\overline A$ is the row echelon form of $A$). But when you do that, cannot conclude, from just that equation, what the eigenvectors and eigenvalues of $A$ are. For example, if $A$ is equal to $\lambda \cdot I$, then the eigenvalues of $A$ are all $\lambda$, while the eigenvaleus of $\overline A$ are all $1$.
• You have my thanks. I never knew row operations do not preserve eigenvectors or eigenvalues - that is interesting. What do you mean the vectors of the original matrix are not vectors in $\mathbb{R}^3$? They look to be in $\mathbb{R}^3$ to me. Commented Nov 18, 2016 at 13:21
• @ThePointer Every vector in $\mathbb R^3$ is an eigenvector of the second matrix, but not every vector in $\mathbb R^3$ is the eigenvector of the first.