Seja a função < , > : R3 x R3 > R definida por = x1x2 + 5y1y2 + 2z1z2.

Dados dois vetores \(u = \left( {{x_1}, {y_1}, {z_1}} \right)\) e \(v = \left( {{x_2}, {y_2}, {z_2}} \right)\), verificam-se as propriedades:

• Simetria: \(\left\langle {v, u} \right\rangle = {x_2}{x_1} + 5{y_2}{y_1} + 2{z_2}{z_1} = {x_1}{x_2} + 5{y_1}{y_2} + 2{z_1}{z_2} = \left\langle {u, v} \right\rangle\)
• Positividade: \(\left\langle {u, u} \right\rangle = x_1^2 + 5y_1^2 + 2z_1^2 \Rightarrow \left\langle {u, u} \right\rangle \geqslant 0 \forall u \in {{\mathbb R}^3}\)
• Distributividade: se \(w = \left( {{x_3}, {y_3}, {z_3}} \right)\), \(\left\langle {u + w, v} \right\rangle = \left( {{x_1} + {x_3}} \right){x_2} + 5\left( {{y_1} + {y_3}} \right){y_2} + 2\left( {{z_1} + {z_3}} \right){z_2} = \left( {{x_1}{x_2} + 5{y_1}{y_2} + 2{z_1}{z_2}} \right) + \left( {{x_3}{x_2} + 5{y_3}{y_2} + 2{z_3}{z_2}} \right) = \left\langle {u, v} \right\rangle + \left\langle {w, v} \right\rangle\)
• Homogeneidade: \(\left\langle {\lambda u, v} \right\rangle = \left( {\lambda {x_1}} \right){x_2} + 5\left( {\lambda {y_1}} \right){y_2} + 2\left( {\lambda {z_1}} \right){z_2} = \lambda \left( {{x_1}{x_2} + 5{y_1}{y_2} + 2{z_1}{z_2}} \right) = \lambda \left\langle {u, v} \right\rangle\), \(\forall \lambda \in {\mathbb R}\)

Portanto, \(f\) é um produto interno.

b)

Seja \(u = \left( {1, 0, 0} \right)\), \(v = \left( {0, 1, 0} \right)\), \(w = \left( {0, 0, 1} \right)\). Esta base já é ortogonal, pois \(\left\langle {u, v} \right\rangle = \left\langle {u, w} \right\rangle = \left\langle {v, w} \right\rangle = 0\). Para normalizá-la, basta dividir cada vetor por um escalar igual à sua norma:

\[\dfrac{u}{{{{\left\| u \right\|}_2}}} = \dfrac{u}{{\sqrt {\left\langle {u, u} \right\rangle } }} = \dfrac{u}{{\sqrt {{1^2} + 5 \times {0^2} + 2 \times {0^2}} }} = u = \left( {1,0,0} \right)\]

\[\dfrac{v}{{{{\left\| v \right\|}_2}}} = \dfrac{v}{{\sqrt {\left\langle {v, v} \right\rangle } }} = \dfrac{v}{{\sqrt {{0^2} + 5 \times {1^2} + 2 \times {0^2}} }} = \dfrac{v}{{\sqrt 5 }} = \left( {0,\dfrac{1}{{\sqrt 5 }},0} \right)\]

\[\dfrac{w}{{{{\left\| w \right\|}_2}}} = \dfrac{w}{{\sqrt {\left\langle {w, w} \right\rangle } }} = \dfrac{v}{{\sqrt {{0^2} + 5 \times {0^2} + 2 \times {1^2}} }} = \dfrac{v}{{\sqrt 2 }} = \left( {0,0,\dfrac{1}{{\sqrt 2 }}} \right)\]

Portanto, \(\left\{ {\left( {1,0,0} \right), \left( {0,\dfrac{1}{{\sqrt 5 }},0} \right), \left( {0,0,\dfrac{1}{{\sqrt 2 }}} \right)} \right\}\) é uma base ortonormal.