1. **State the problem:** We are given a set of vectors \( M = \{(1,0,1,0), (1,1,1,1), (-1,2,0,1)\} \) and asked to orthogonalize this basis using the Gram-Schmidt process. 2. **Recall the Gram-Schmidt process:** Given vectors \( v_1, v_2, v_3 \), we construct an orthogonal set \( u_1, u_2, u_3 \) by: $$ \begin{aligned} &u_1 = v_1 \\ &u_2 = v_2 - \mathrm{proj}_{u_1}(v_2) = v_2 - \frac{\langle v_2, u_1 \rangle}{\langle u_1, u_1 \rangle} u_1 \\ &u_3 = v_3 - \mathrm{proj}_{u_1}(v_3) - \mathrm{proj}_{u_2}(v_3) = v_3 - \frac{\langle v_3, u_1 \rangle}{\langle u_1, u_1 \rangle} u_1 - \frac{\langle v_3, u_2 \rangle}{\langle u_2, u_2 \rangle} u_2 \end{aligned} $$ where \( \langle a, b \rangle \) is the dot product of vectors \(a\) and \(b\). 3. **Calculate \( u_1 \):** $$ u_1 = v_1 = (1,0,1,0) $$ 4. **Calculate \( u_2 \):** First, compute \( \langle v_2, u_1 \rangle = (1)(1) + (1)(0) + (1)(1) + (1)(0) = 1 + 0 + 1 + 0 = 2 \) Compute \( \langle u_1, u_1 \rangle = 1^2 + 0^2 + 1^2 + 0^2 = 1 + 0 + 1 + 0 = 2 \) Projection: $$ \mathrm{proj}_{u_1}(v_2) = \frac{2}{2} u_1 = 1 \times (1,0,1,0) = (1,0,1,0) $$ Then: $$ u_2 = v_2 - \mathrm{proj}_{u_1}(v_2) = (1,1,1,1) - (1,0,1,0) = (0,1,0,1) $$ 5. **Calculate \( u_3 \):** Compute \( \langle v_3, u_1 \rangle = (-1)(1) + (2)(0) + (0)(1) + (1)(0) = -1 + 0 + 0 + 0 = -1 \) Compute \( \langle v_3, u_2 \rangle = (-1)(0) + (2)(1) + (0)(0) + (1)(1) = 0 + 2 + 0 + 1 = 3 \) Compute \( \langle u_2, u_2 \rangle = 0^2 + 1^2 + 0^2 + 1^2 = 0 + 1 + 0 + 1 = 2 \) Projection terms: $$ \mathrm{proj}_{u_1}(v_3) = \frac{-1}{2} u_1 = -\frac{1}{2} (1,0,1,0) = \left(-\frac{1}{2}, 0, -\frac{1}{2}, 0\right) $$ $$ \mathrm{proj}_{u_2}(v_3) = \frac{3}{2} u_2 = \frac{3}{2} (0,1,0,1) = \left(0, \frac{3}{2}, 0, \frac{3}{2}\right) $$ Then: $$ u_3 = v_3 - \mathrm{proj}_{u_1}(v_3) - \mathrm{proj}_{u_2}(v_3) = (-1,2,0,1) - \left(-\frac{1}{2}, 0, -\frac{1}{2}, 0\right) - \left(0, \frac{3}{2}, 0, \frac{3}{2}\right) $$ Calculate stepwise: $$ (-1,2,0,1) - \left(-\frac{1}{2}, 0, -\frac{1}{2}, 0\right) = \left(-1 + \frac{1}{2}, 2 - 0, 0 + \frac{1}{2}, 1 - 0\right) = \left(-\frac{1}{2}, 2, \frac{1}{2}, 1\right) $$ Then subtract \( \left(0, \frac{3}{2}, 0, \frac{3}{2}\right) \): $$ u_3 = \left(-\frac{1}{2}, 2, \frac{1}{2}, 1\right) - \left(0, \frac{3}{2}, 0, \frac{3}{2}\right) = \left(-\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}\right) $$ 6. **Final orthogonal basis:** $$ \boxed{ \left\{ (1,0,1,0), \quad (0,1,0,1), \quad \left(-\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}\right) \right\} $$ This set is orthogonal because each vector is orthogonal to the others by construction. **Summary:** We used the Gram-Schmidt process to convert the original basis into an orthogonal basis by projecting and subtracting components along previous vectors.