Vi utför Gram-Schmidt-metoden på kolonnerna $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ av $A$ :

\mathbf{u}_1 = \frac{1}{\sqrt{1 + 4 + 0 + 4}} \mathbf{v}_1 = \frac{1}{3} \begin{bmatrix} 1 \\ -2 \\ 0 \\ 2 \end{bmatrix} \overline{\mathbf{u}_2} = \mathbf{v}_2 - (\mathbf{v}_2 \cdot \mathbf{u}_1) \mathbf{u}_1 = \begin{bmatrix} 1 \\ 3 \\ 2 \\ -2 \end{bmatrix} - \frac{1}{9} (-9) \begin{bmatrix} 1 \\ -2 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 2 \\ 0 \end{bmatrix} \mathbf{u}_2 = \frac{1}{\sqrt{4 + 1 + 4 + 0}} \begin{bmatrix} 2 \\ 1 \\ 2 \\ 0 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 2 \\ 1 \\ 2 \\ 0 \end{bmatrix} \begin{aligned} \overline{\mathbf{u}_3} &= \mathbf{v}_3 - (\mathbf{v}_3 \cdot \mathbf{u}_1) \mathbf{u}_1 - (\mathbf{v}_3 \cdot \mathbf{u}_2) \mathbf{u}_2 \\ &= \begin{bmatrix} 0 \\ 1 \\ -2 \\ 1 \end{bmatrix} - 0 - \left(-\frac{1}{3}\right) \begin{bmatrix} 2 \\ 1 \\ 2 \\ 0 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 2 \\ 4 \\ -4 \\ 3 \end{bmatrix} \end{aligned} \mathbf{u}_3 = \frac{1}{4 + 16 + 16 + 9} \begin{bmatrix} 2 \\ 4 \\ -4 \\ 3 \end{bmatrix} = \frac{1}{\sqrt{45}} \begin{bmatrix} 2 \\ 4 \\ -4 \\ 3 \end{bmatrix}.