I am given the following matrix of a linear operator A, and asked to diagonalize it. $$ A = \begin{pmatrix} -2 & 2 & 6 \\ 5 & 1 & -6 \\ -5 & 2 & 9\\ \end{pmatrix} $$ Solving for $det(A-\lambda I) = 0$, I have the following eigenvalues:
$\lambda = 3 , \lambda = 3, \lambda = 2$ (counting multiplicities).
For $\lambda = 3$ , $Ker(A-\lambda I)$ has the basis $\begin{pmatrix} \frac25 \\ 1 \\ 0 \end{pmatrix} $ and $\begin{pmatrix} \frac 65 \\ 0 \\ 1 \\ \end{pmatrix} $ , which are the eigenvectors corresponding to $\lambda = 3$.
On the other hand, for $\lambda = 2$ , this is where I have some confusion.
We want to find the nullspace of $(A-2I) = \begin{pmatrix} -4 & 2 & 6 \\ 5 & -1 & -6 \\ -5 & 2 & 7 \\ \end{pmatrix}$. Computing the reduced echelon form of the matrix of $(A-2I)$ , I get $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$ (without row swap operations) and conclude that the nullspace $Ker(A-2I)$ is trivial. As such, I conclude the geometric multiplicities and algebraic multiplicities of the eigenvalues do not coincide and hence the linear operator A cannot be diagonalized.
However, in the solution given, row exchange operations are performed when computing the reduced echelon form of $(A-2I)$. Rows 1 and 2 are swapped such that eventually, the matrix $\begin{pmatrix} -4 & 2 & 6 \\ 5 & -1 & -6 \\ -5 & 2 & 7 \\ \end{pmatrix}$ is reduced to $\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix}$ (which eventually leads to A being diagonalizable).
Why is that so? I tried verifying my computations for basis vectors in the nullspace $Ker(A-2I)$ with several online calculators, and all of them indicate that $Ker(A-2I)$ is trivial and no row exchanges take place in computing this. Aren't these results conflicting? One says that $Ker(A-2I)$ is trivial, while the other eventually leads to $Ker(A-2I)$ having the basis $\begin{pmatrix} 1 \\ -1 \\ 1 \\ \end{pmatrix} $.
Thanks in advance for any help!