I'm trying to find the Schur factorisation of the matrix $A\in\mathbb{R}^3$. Where
$$A = \frac{1}{100}\begin{bmatrix}92 && 0 && -144\\0 && 100 && 0\\-144 && 0 && 8\end{bmatrix}$$
I've found the eigenvalues to be $\lambda = -1,1,2$.
I understand the next step is to select one (I'll pick $\lambda = 1$ for convenience) and calculate the normalised vector $\mathbf{A - I}$ where I get to the point
$$\mathbf{A - I} = \frac{1}{100}\begin{bmatrix}-8 && 0 && -144\\ 0 && 0 && 0\\-144 && 0 && -92\end{bmatrix}$$
But I'm unsure on how to normalise this matrix as usually we only have a 1x2 matrix in previous examples.