The matrix $ A := \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 4 \end{bmatrix}$
has the eigenvalues $1$ and $4$. I am looking for the eigenspace corresponding to the eigenvalue $4$. So I need to find the solutions of $(A - \lambda I)v = 0$ with $\lambda = 4$, i.e.
$(A - 4I)v = \begin{bmatrix} -3 & 1 & 0 \\ 0 & -3 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$ or equivalently $\begin{bmatrix} 3 & -1 & 0 \\ 0 & 3 & -1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$
So $3y = z$ and $3x = y$, therefore $x = 9z$
Let $z = t$. Then
$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = t \begin{bmatrix} 9 \\ 3 \\ 1 \end{bmatrix}$
So the eigenspace is $\operatorname{span} \{ \begin{bmatrix} 9 & 3 & 1 \end{bmatrix}^T \}$.
But I have checked two online calculators and they give this eigenspace as $\operatorname{span} \{\begin{bmatrix} 1 & 3 & 9 \end{bmatrix}^T \},$ so where am I going wrong?