Answer on Question #61910 – Math – Linear Algebra
Question
Apply the Gram-Schmidt diagonalisation process to find an orthonormal basis for the subspace of C 4 C^4 C 4
{ ( 1 , − i , 0 , 1 ) , ( − 1 , 0 , i , 0 ) , ( − i , 0 , 1 , − 1 ) } . \{(1, -i, 0, 1), (-1, 0, i, 0), (-i, 0, 1, -1)\}. {( 1 , − i , 0 , 1 ) , ( − 1 , 0 , i , 0 ) , ( − i , 0 , 1 , − 1 )} . Solution
For vectors in C 4 C^4 C 4 < x , y = " " > = x 1 y ‾ 1 + x 2 y ‾ 2 + x 3 y ‾ 3 + x 4 y ‾ 4 , <x, y=""> = x_1 \overline{y}_1 + x_2 \overline{y}_2 + x_3 \overline{y}_3 + x_4 \overline{y}_4, < x , y = "" >= x 1 y 1 + x 2 y 2 + x 3 y 3 + x 4 y 4 ,
Basis:
u 1 = v 1 = ( 1 , − i , 0 , 1 ) ; < u 1 , u 1 > = 1 2 + 1 2 + 0 2 + 1 2 = 3 ; ∥ u 1 ∥ = ⟨ u 1 , u 1 ⟩ = 3 ; v 2 = ( − 1 , 0 , i , 0 ) ; < v 2 , u 1 > = − 1 ⋅ 1 + 0 ⋅ i + i ⋅ 0 + 0 ⋅ 1 = − 1 ; u 2 = v 2 − p r o j u 1 v 2 = v 2 − ⟨ v 2 , u 1 ⟩ ⟨ u 1 , u 1 ⟩ u 1 = ( − 1 , 0 , i , 0 ) − ( − 1 3 ) ( 1 , − i , 0 , 1 ) = = ( − 1 + 1 3 , 0 − i 3 , i , 1 3 ) = ( − 2 3 , − i 3 , i , 1 3 ) ; < u 2 , u 2 > = ( − 2 3 ) 2 + ( − 1 3 ) 2 + 1 2 + ( 1 3 ) 2 = 15 9 = 5 3 ; ∥ u 2 ∥ = ⟨ u 2 , u 2 ⟩ = 5 3 ; < v 3 , u 1 > = − i ⋅ 1 + 0 ⋅ i + 1 ⋅ 0 + ( − 1 ) ⋅ 1 = − i − 1 = − ( i + 1 ) ; < v 3 , u 2 > = − i ⋅ ( − 2 3 ) + 0 ⋅ i 3 + 1 ⋅ ( − i ) + ( − 1 ) ⋅ 1 3 = − i 3 − 1 3 = − i + 1 3 ; u 3 = v 3 − p r o j u 1 v 3 − p r o j u 2 v 3 = v 3 − ⟨ v 3 , u 1 ⟩ ⟨ u 1 , u 1 ⟩ u 1 − ⟨ v 3 , u 2 ⟩ ⟨ u 2 , u 2 ⟩ u 2 = ( − i , 0 , 1 , − 1 ) + + i + 1 3 ( 1 , − i , 0 , 1 ) + i + 1 3 × 5 3 ( − 2 3 , − i 3 , i , 1 3 ) = ( − i , 0 , 1 , − 1 ) + i + 1 3 ( 1 , − i , 0 , 1 ) + + i + 1 15 ( − 2 , − i , 3 i , 1 ) = ( − i + i + 1 3 − 2 i + 2 15 , 0 − i × i + 1 3 − i × i + 1 15 , 1 + 0 + 3 i × i + 1 15 , − 1 + i + 1 15 , − 1 + i + 1 15 ) = ( − 15 i + 5 i + 5 − 2 i − 2 15 , − 5 i 2 − 5 i − i 2 − i 15 , 1 + i 2 + i 5 , − 15 + 5 i + 5 + i + 1 15 ) = \begin{aligned}
u_1 &= v_1 = (1, -i, 0, 1); \\
&< u_1, u_1 >= 1^2 + 1^2 + 0^2 + 1^2 = 3; \\
&\|u_1\| = \sqrt{\langle u_1, u_1 \rangle} = \sqrt{3}; \\
&v_2 = (-1, 0, i, 0); \\
&< v_2, u_1 >= -1 \cdot 1 + 0 \cdot i + i \cdot 0 + 0 \cdot 1 = -1; \\
&u_2 = v_2 - proj_{u_1} v_2 = v_2 - \frac{\langle v_2, u_1 \rangle}{\langle u_1, u_1 \rangle} u_1 = (-1, 0, i, 0) - \left(-\frac{1}{3}\right)(1, -i, 0, 1) = \\
&= \left(-1 + \frac{1}{3}, \quad 0 - \frac{i}{3}, i, \frac{1}{3}\right) = \left(-\frac{2}{3}, -\frac{i}{3}, i, \frac{1}{3}\right); \\
&< u_2, u_2 >= \left(-\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 + 1^2 + \left(\frac{1}{3}\right)^2 = \frac{15}{9} = \frac{5}{3}; \\
&\|u_2\| = \sqrt{\langle u_2, u_2 \rangle} = \sqrt{\frac{5}{3}}; \\
&< v_3, u_1 >= -i \cdot 1 + 0 \cdot i + 1 \cdot 0 + (-1) \cdot 1 = -i - 1 = -(i + 1); \\
&< v_3, u_2 >= -i \cdot \left(-\frac{2}{3}\right) + 0 \cdot \frac{i}{3} + 1 \cdot (-i) + (-1) \cdot \frac{1}{3} = -\frac{i}{3} - \frac{1}{3} = -\frac{i+1}{3}; \\
&u_3 = v_3 - proj_{u_1} v_3 - 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 = (-i, 0, 1, -1) + \\
&\quad + \frac{i+1}{3}(1, -i, 0, 1) + \frac{i+1}{3 \times \frac{5}{3}}\left(-\frac{2}{3}, -\frac{i}{3}, i, \frac{1}{3}\right) = (-i, 0, 1, -1) + \frac{i+1}{3}(1, -i, 0, 1) + \\
&\quad + \frac{i+1}{15}(-2, -i, 3i, 1) = \left(-i + \frac{i+1}{3} - \frac{2i+2}{15}, \quad 0 - i \times \frac{i+1}{3} - i \times \frac{i+1}{15}, 1 + 0 + 3i \times \frac{i+1}{15}, -1 + \frac{i+1}{15}, -1 + \frac{i+1}{15}\right) = \left(\frac{-15i + 5i + 5 - 2i - 2}{15}, \quad \frac{-5i^2 - 5i - i^2 - i}{15}, 1 + \frac{i^2 + i}{5}, \frac{-15 + 5i + 5 + i + 1}{15}\right) = \\
\end{aligned} u 1 = v 1 = ( 1 , − i , 0 , 1 ) ; < u 1 , u 1 >= 1 2 + 1 2 + 0 2 + 1 2 = 3 ; ∥ u 1 ∥ = ⟨ u 1 , u 1 ⟩ = 3 ; v 2 = ( − 1 , 0 , i , 0 ) ; < v 2 , u 1 >= − 1 ⋅ 1 + 0 ⋅ i + i ⋅ 0 + 0 ⋅ 1 = − 1 ; u 2 = v 2 − p ro j u 1 v 2 = v 2 − ⟨ u 1 , u 1 ⟩ ⟨ v 2 , u 1 ⟩ u 1 = ( − 1 , 0 , i , 0 ) − ( − 3 1 ) ( 1 , − i , 0 , 1 ) = = ( − 1 + 3 1 , 0 − 3 i , i , 3 1 ) = ( − 3 2 , − 3 i , i , 3 1 ) ; < u 2 , u 2 >= ( − 3 2 ) 2 + ( − 3 1 ) 2 + 1 2 + ( 3 1 ) 2 = 9 15 = 3 5 ; ∥ u 2 ∥ = ⟨ u 2 , u 2 ⟩ = 3 5 ; < v 3 , u 1 >= − i ⋅ 1 + 0 ⋅ i + 1 ⋅ 0 + ( − 1 ) ⋅ 1 = − i − 1 = − ( i + 1 ) ; < v 3 , u 2 >= − i ⋅ ( − 3 2 ) + 0 ⋅ 3 i + 1 ⋅ ( − i ) + ( − 1 ) ⋅ 3 1 = − 3 i − 3 1 = − 3 i + 1 ; u 3 = v 3 − p ro j u 1 v 3 − p ro j u 2 v 3 = v 3 − ⟨ u 1 , u 1 ⟩ ⟨ v 3 , u 1 ⟩ u 1 − ⟨ u 2 , u 2 ⟩ ⟨ v 3 , u 2 ⟩ u 2 = ( − i , 0 , 1 , − 1 ) + + 3 i + 1 ( 1 , − i , 0 , 1 ) + 3 × 3 5 i + 1 ( − 3 2 , − 3 i , i , 3 1 ) = ( − i , 0 , 1 , − 1 ) + 3 i + 1 ( 1 , − i , 0 , 1 ) + + 15 i + 1 ( − 2 , − i , 3 i , 1 ) = ( − i + 3 i + 1 − 15 2 i + 2 , 0 − i × 3 i + 1 − i × 15 i + 1 , 1 + 0 + 3 i × 15 i + 1 , − 1 + 15 i + 1 , − 1 + 15 i + 1 ) = ( 15 − 15 i + 5 i + 5 − 2 i − 2 , 15 − 5 i 2 − 5 i − i 2 − i , 1 + 5 i 2 + i , 15 − 15 + 5 i + 5 + i + 1 ) = </x,>
= ( − 12 i + 3 15 , 6 − 6 i 15 , 4 + i 5 , 6 i − 9 15 ) = ( 1 − 4 i 5 , 2 − 2 i 5 , 4 + i 5 , 2 i − 3 5 ) ; < u 3 , u 3 > = = 1 5 2 ( ( 1 − 4 i ) ( 1 + 4 i ) + ( 2 − 2 i ) ( 2 + 2 i ) + ( 4 − i ) ( 4 + i ) + ( 2 i − 3 ) ( 2 i + 3 ) ) = = 1 25 ( 1 + 16 + 4 + 4 + 16 + 1 − 4 − 9 ) = 29 25 ; ∥ u 3 ∥ = ⟨ u 3 , u 3 ⟩ = 29 25 = 29 5 . \begin{array}{l}
= \left(\frac{-12i+3}{15}, \frac{6-6i}{15}, \frac{4+i}{5}, \frac{6i-9}{15}\right) = \left(\frac{1-4i}{5}, \frac{2-2i}{5}, \frac{4+i}{5}, \frac{2i-3}{5}\right); \\
< u_3, u_3 > = \\
= \frac{1}{5^2} \left((1 - 4i)(1 + 4i) + (2 - 2i)(2 + 2i) + (4 - i)(4 + i) + (2i - 3)(2i + 3)\right) = \\
= \frac{1}{25} (1 + 16 + 4 + 4 + 16 + 1 - 4 - 9) = \frac{29}{25}; \\
\| u_3 \| = \sqrt{\langle u_3, u_3 \rangle} = \sqrt{\frac{29}{25}} = \frac{\sqrt{29}}{5}.
\end{array} = ( 15 − 12 i + 3 , 15 6 − 6 i , 5 4 + i , 15 6 i − 9 ) = ( 5 1 − 4 i , 5 2 − 2 i , 5 4 + i , 5 2 i − 3 ) ; < u 3 , u 3 >= = 5 2 1 ( ( 1 − 4 i ) ( 1 + 4 i ) + ( 2 − 2 i ) ( 2 + 2 i ) + ( 4 − i ) ( 4 + i ) + ( 2 i − 3 ) ( 2 i + 3 ) ) = = 25 1 ( 1 + 16 + 4 + 4 + 16 + 1 − 4 − 9 ) = 25 29 ; ∥ u 3 ∥ = ⟨ u 3 , u 3 ⟩ = 25 29 = 5 29 .
Checking:
⟨ u 2 , u 1 ⟩ = ( − 2 3 , − i 3 , i , 1 3 ) ⋅ ( 1 , − i , 0 , 1 ) = − 2 3 + 1 3 + 1 3 = 0 ; ⟨ u 3 , u 1 ⟩ = ( 1 − 4 i 5 , 2 − 2 i 5 , 4 + i 5 , 2 i − 3 5 ) ⋅ ( 1 , − i , 0 , 1 ) = 1 5 ( 1 − 4 i + i ( 2 − 2 i ) + ( 2 i − 3 ) ) = = 0 ; ⟨ u 3 , u 2 ⟩ = ( 1 − 4 i 5 , 2 − 2 i 5 , 4 + i 5 , 2 i − 3 5 ) ⋅ ( − 2 3 , − i 3 , i , 1 3 ) = 1 15 ( − 2 ( 1 − 4 i ) + i ( 2 − 2 i ) − 3 i ( 4 + i ) + ( 2 i − 3 ) ) = 0. \begin{array}{l}
\langle u_2, u_1 \rangle = \left(-\frac{2}{3}, -\frac{i}{3}, i, \frac{1}{3}\right) \cdot (1, -i, 0, 1) = -\frac{2}{3} + \frac{1}{3} + \frac{1}{3} = 0; \\
\langle u_3, u_1 \rangle = \left(\frac{1-4i}{5}, \frac{2-2i}{5}, \frac{4+i}{5}, \frac{2i-3}{5}\right) \cdot (1, -i, 0, 1) = \frac{1}{5} \left(1 - 4i + i(2 - 2i) + (2i - 3)\right) = \\
= 0; \\
\langle u_3, u_2 \rangle = \left(\frac{1-4i}{5}, \frac{2-2i}{5}, \frac{4+i}{5}, \frac{2i-3}{5}\right) \cdot \left(-\frac{2}{3}, -\frac{i}{3}, i, \frac{1}{3}\right) = \frac{1}{15} (-2(1 - 4i) + i(2 - 2i) - 3i(4 + i) + (2i - 3)) = 0.
\end{array} ⟨ u 2 , u 1 ⟩ = ( − 3 2 , − 3 i , i , 3 1 ) ⋅ ( 1 , − i , 0 , 1 ) = − 3 2 + 3 1 + 3 1 = 0 ; ⟨ u 3 , u 1 ⟩ = ( 5 1 − 4 i , 5 2 − 2 i , 5 4 + i , 5 2 i − 3 ) ⋅ ( 1 , − i , 0 , 1 ) = 5 1 ( 1 − 4 i + i ( 2 − 2 i ) + ( 2 i − 3 ) ) = = 0 ; ⟨ u 3 , u 2 ⟩ = ( 5 1 − 4 i , 5 2 − 2 i , 5 4 + i , 5 2 i − 3 ) ⋅ ( − 3 2 , − 3 i , i , 3 1 ) = 15 1 ( − 2 ( 1 − 4 i ) + i ( 2 − 2 i ) − 3 i ( 4 + i ) + ( 2 i − 3 )) = 0.
So vectors u 1 , u 2 , u 3 u_1, u_2, u_3 u 1 , u 2 , u 3
Orthonormal basis:
e 1 = u 1 ∥ u 1 ∥ = 1 3 ( 1 , − i , 0 , 1 ) = ( 1 3 , − i 3 , 0 , 1 3 ) e_1 = \frac{u_1}{\|u_1\|} = \frac{1}{\sqrt{3}} (1, -i, 0, 1) = \left(\frac{1}{\sqrt{3}}, -\frac{i}{\sqrt{3}}, 0, \frac{1}{\sqrt{3}}\right) e 1 = ∥ u 1 ∥ u 1 = 3 1 ( 1 , − i , 0 , 1 ) = ( 3 1 , − 3 i , 0 , 3 1 ) e 2 = u 2 ∥ u 2 ∥ = 3 5 ( − 2 3 , − i 3 , i , 1 3 ) = ( − 2 15 , − i 15 , 3 i 15 , 1 15 ) e_2 = \frac{u_2}{\|u_2\|} = \sqrt{\frac{3}{5}} \left(-\frac{2}{3}, -\frac{i}{3}, i, \frac{1}{3}\right) = \left(-\frac{2}{\sqrt{15}}, -\frac{i}{\sqrt{15}}, \frac{3i}{\sqrt{15}}, \frac{1}{\sqrt{15}}\right) e 2 = ∥ u 2 ∥ u 2 = 5 3 ( − 3 2 , − 3 i , i , 3 1 ) = ( − 15 2 , − 15 i , 15 3 i , 15 1 ) e 3 = u 3 ∥ u 3 ∥ = 5 29 ( 1 − 4 i 5 , 2 − 2 i 5 , 4 + i 5 , − 3 + 2 i 5 ) = ( 1 − 4 i 29 , 2 − 2 i 29 , 4 + i 29 , − 3 + 2 i 29 ) . e_3 = \frac{u_3}{\|u_3\|} = \frac{5}{\sqrt{29}} \left(\frac{1-4i}{5}, \frac{2-2i}{5}, \frac{4+i}{5}, \frac{-3+2i}{5}\right) = \left(\frac{1-4i}{\sqrt{29}}, \frac{2-2i}{\sqrt{29}}, \frac{4+i}{\sqrt{29}}, \frac{-3+2i}{\sqrt{29}}\right). e 3 = ∥ u 3 ∥ u 3 = 29 5 ( 5 1 − 4 i , 5 2 − 2 i , 5 4 + i , 5 − 3 + 2 i ) = ( 29 1 − 4 i , 29 2 − 2 i , 29 4 + i , 29 − 3 + 2 i ) .
Answer: { ( 1 3 , − i 3 , 0 , 1 3 ) , ( − 2 15 , − i 15 , 3 i 15 , 1 15 ) , ( 1 − 4 i 29 , 2 − 2 i 29 , 4 + i 29 , − 3 + 2 i 29 ) } \left\{ \left(\frac{1}{\sqrt{3}}, -\frac{i}{\sqrt{3}}, 0, \frac{1}{\sqrt{3}}\right), \left(-\frac{2}{\sqrt{15}}, -\frac{i}{\sqrt{15}}, \frac{3i}{\sqrt{15}}, \frac{1}{\sqrt{15}}\right), \left(\frac{1-4i}{\sqrt{29}}, \frac{2-2i}{\sqrt{29}}, \frac{4+i}{\sqrt{29}}, \frac{-3+2i}{\sqrt{29}}\right) \right\} { ( 3 1 , − 3 i , 0 , 3 1 ) , ( − 15 2 , − 15 i , 15 3 i , 15 1 ) , ( 29 1 − 4 i , 29 2 − 2 i , 29 4 + i , 29 − 3 + 2 i ) }
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#339914 on Dec 2023