1. **Problem Statement:** We have three signals $s_1(t) = 1$, $s_2(t) = t$, and $s_3(t) = t^2$ defined on $[0,1]$. We want to verify their linear independence, compute their norms under the inner product $\langle f,g \rangle = \int_0^1 f(t)g(t) dt$, apply Gram–Schmidt to get an orthonormal set, and interpret the results. 2. **Linear Independence:** To check if $s_1, s_2, s_3$ are linearly independent, suppose $a s_1 + b s_2 + c s_3 = 0$ for all $t \in [0,1]$. This means: $$a + b t + c t^2 = 0 \quad \forall t \in [0,1].$$ Since this is a polynomial equal to zero everywhere, all coefficients must be zero: $$a=0, \quad b=0, \quad c=0.$$ Thus, $s_1, s_2, s_3$ are linearly independent. 3. **Norms under the inner product:** The norm is $\|s_i\| = \sqrt{\langle s_i, s_i \rangle}$. - $\|s_1\|^2 = \int_0^1 1 \cdot 1 dt = 1$ - $\|s_2\|^2 = \int_0^1 t^2 dt = \frac{1}{3}$ - $\|s_3\|^2 = \int_0^1 t^4 dt = \frac{1}{5}$ 4. **Gram–Schmidt Process:** We construct orthonormal vectors $e_1, e_2, e_3$ from $s_1, s_2, s_3$. - Step 1: $e_1 = \frac{s_1}{\|s_1\|} = 1$ - Step 2: Orthogonalize $s_2$ against $e_1$: $$u_2 = s_2 - \langle s_2, e_1 \rangle e_1 = t - \int_0^1 t \cdot 1 dt \cdot 1 = t - \frac{1}{2}.$$ Norm: $$\|u_2\|^2 = \int_0^1 (t - \frac{1}{2})^2 dt = \int_0^1 (t^2 - t + \frac{1}{4}) dt = \frac{1}{3} - \frac{1}{2} + \frac{1}{4} = \frac{1}{12}.$$ Normalize: $$e_2 = \frac{u_2}{\|u_2\|} = \frac{t - \frac{1}{2}}{\sqrt{\frac{1}{12}}} = \sqrt{12} \left(t - \frac{1}{2}\right) = 2 \sqrt{3} \left(t - \frac{1}{2}\right).$$ - Step 3: Orthogonalize $s_3$ against $e_1$ and $e_2$: $$u_3 = s_3 - \langle s_3, e_1 \rangle e_1 - \langle s_3, e_2 \rangle e_2.$$ Calculate inner products: $$\langle s_3, e_1 \rangle = \int_0^1 t^2 \cdot 1 dt = \frac{1}{3},$$ $$\langle s_3, e_2 \rangle = \int_0^1 t^2 \cdot 2 \sqrt{3} (t - \frac{1}{2}) dt = 2 \sqrt{3} \left( \int_0^1 t^3 dt - \frac{1}{2} \int_0^1 t^2 dt \right) = 2 \sqrt{3} \left( \frac{1}{4} - \frac{1}{6} \right) = 2 \sqrt{3} \cdot \frac{1}{12} = \frac{\sqrt{3}}{6}.$$ So: $$u_3 = t^2 - \frac{1}{3} - \frac{\sqrt{3}}{6} \cdot 2 \sqrt{3} (t - \frac{1}{2}) = t^2 - \frac{1}{3} - \frac{1}{3} (t - \frac{1}{2}) = t^2 - \frac{1}{3} - \frac{1}{3} t + \frac{1}{6} = t^2 - \frac{1}{3} t - \frac{1}{6}.$$ Norm: $$\|u_3\|^2 = \int_0^1 \left(t^2 - \frac{1}{3} t - \frac{1}{6}\right)^2 dt = \frac{1}{180}.$$ Normalize: $$e_3 = \frac{u_3}{\|u_3\|} = \frac{t^2 - \frac{1}{3} t - \frac{1}{6}}{\sqrt{\frac{1}{180}}} = \sqrt{180} \left(t^2 - \frac{1}{3} t - \frac{1}{6}\right) = 6 \sqrt{5} \left(t^2 - \frac{1}{3} t - \frac{1}{6}\right).$$ 5. **Interpretation:** The Gram–Schmidt process transformed the original non-orthogonal signals $s_1, s_2, s_3$ into an orthonormal set $e_1, e_2, e_3$ with respect to the inner product defined by integration over $[0,1]$. This orthonormal set can be used as a basis for signal representation, simplifying analysis and processing by ensuring signals are mutually orthogonal and normalized. **Final orthonormal set:** $$e_1 = 1,$$ $$e_2 = 2 \sqrt{3} \left(t - \frac{1}{2}\right),$$ $$e_3 = 6 \sqrt{5} \left(t^2 - \frac{1}{3} t - \frac{1}{6}\right).$$