Subsection 27.3

Orthogonality (27.26-27.29)

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Problem 27.26

In a complex arithmetic space with the standard scalar product, find a basis in the orthogonal complement of the subspace given by the system of linear equations:

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(1)

x1+ix2=0x_{1} + ix_{2} = 0;

(2)

x1+ix2+(1−i)x3=0;−ix1+(2+i)x2−x3=0x_{1} + ix_{2} + (1 - i) x_{3} = 0; - ix_{1} + (2 + i) x_{2} - x_{3} = 0;

(3)

a homogeneous system with matrix A377\mathbf{A}_{377}. A377=[11−i01+i3i0−i1]\mathbf{A}_{377} = \left[\begin{smallmatrix} 1 & 1-i & 0 \\ 1+i & 3 & i \\ 0 & -i & 1 \end{smallmatrix}\right].

(4)

a homogeneous system with matrix A368\mathbf{A}_{368}. A368=[1+i2i−211+i3i−311+i4i−41]\mathbf{A}_{368} = \left[\begin{smallmatrix} 1+i \sqrt{2} & i-\sqrt{2} & 1 \\ 1+i \sqrt{3} & i-\sqrt{3} & 1 \\ 1+i \sqrt{4} & i-\sqrt{4} & 1 \end{smallmatrix}\right].

(5)

a homogeneous system with matrix A372\mathbf{A}_{372}. A372=[12i1−i−i2−1−i36i3−3i]\mathbf{A}_{372} = \left[\begin{smallmatrix} 1 & 2 i & 1-i \\ -i & 2 & -1-i \\ 3 & 6 i & 3-3 i \end{smallmatrix}\right].

Problem 27.27

In a complex arithmetic space with the standard scalar product, find a basis in the orthogonal complement of the subspace spanned by the following vectors:

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(1)

[1i]\left[\begin{smallmatrix} 1 \\ i \end{smallmatrix}\right]

(2)

[−i11+i]\left[\begin{smallmatrix} -i \\ 1 \\ 1+i \end{smallmatrix}\right]

(3)

[1−i1],[i10]\left[\begin{smallmatrix} 1 \\ - i \\ 1\end{smallmatrix}\right], \left[\begin{smallmatrix} i \\ 1 \\ 0 \end{smallmatrix}\right]

(4)

the columns of the matrix A372\mathbf{A}_{372}. A372=[12i1−i−i2−1−i36i3−3i]\mathbf{A}_{372} = \left[\begin{smallmatrix} 1 & 2 i & 1-i \\ -i & 2 & -1-i \\ 3 & 6 i & 3-3 i \end{smallmatrix}\right].

(5)

the columns of the matrix A368\mathbf{A}_{368}. A368=[1+i2i−211+i3i−311+i4i−41]\mathbf{A}_{368} = \left[\begin{smallmatrix} 1+i \sqrt{2} & i-\sqrt{2} & 1 \\ 1+i \sqrt{3} & i-\sqrt{3} & 1 \\ 1+i \sqrt{4} & i-\sqrt{4} & 1 \end{smallmatrix}\right].

Problem 27.28

Using the orthogonalization process, construct an orthonormal basis in the linear span of the given vectors of a complex arithmetic space with the standard scalar product:

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(1)

[1i]\left[\begin{smallmatrix} 1 \\ i\end{smallmatrix}\right], [11]\left[\begin{smallmatrix} 1 \\ 1\end{smallmatrix}\right];

(2)

[2−ii]\left[\begin{smallmatrix} 2 -i \\ i\end{smallmatrix}\right], [4−i2−3i]\left[\begin{smallmatrix} 4-i \\ 2-3i\end{smallmatrix}\right];

(3)

[1i1]\left[\begin{smallmatrix} 1 \\ i \\ 1\end{smallmatrix}\right], [2−ii−12]\left[\begin{smallmatrix} 2-i \\ i-1 \\ 2\end{smallmatrix}\right];

(4)

[1i1]\left[\begin{smallmatrix} 1 \\ i \\ 1\end{smallmatrix}\right], [i10]\left[\begin{smallmatrix} i \\ 1 \\ 0\end{smallmatrix}\right], [−101]\left[\begin{smallmatrix} -1 \\ 0 \\ 1\end{smallmatrix}\right];

(5)

[1+i2+i1−i]\left[\begin{smallmatrix} 1+i \\ 2+i \\ 1-i\end{smallmatrix}\right], [−24+i1−i]\left[\begin{smallmatrix} -2 \\ 4+i \\ 1-i\end{smallmatrix}\right], [12+i2−i]\left[\begin{smallmatrix} 1 \\ 2+i \\ 2-i\end{smallmatrix}\right].

Problem 27.29

In a complex arithmetic space with the standard scalar product, find the orthogonal projection of the vector xx onto the linear span of the vectors a1,…,aka_{1}, \ldots , a_{k}:

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(1)

x=[1−i],a=[1−1]x = \left[\begin{smallmatrix} 1 \\ -i \end{smallmatrix}\right], \quad a = \left[\begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right]

(2)

x=[2+i02−i],a=[−1i1+i]x = \left[\begin{smallmatrix} 2+i \\ 0 \\ 2-i \end{smallmatrix}\right], \quad a = \left[\begin{smallmatrix} -1 \\ i \\ 1+i \end{smallmatrix}\right]

(3)

x=[2+ii2−i],a1=[1i1],a2=[i0−i]x = \left[\begin{smallmatrix} 2+i \\ i \\ 2-i \end{smallmatrix}\right], \quad a_{1} =\left[\begin{smallmatrix} 1 \\ i \\ 1\end{smallmatrix}\right], \quad a_{2} = \left[\begin{smallmatrix} i \\ 0 \\ -i \end{smallmatrix}\right]

(4)

x=[1+i1+i1],a1=[−1i1]x = \left[\begin{smallmatrix} 1+i \\ 1+i \\ 1 \end{smallmatrix}\right], \quad a_{1} = \left[\begin{smallmatrix} -1 \\ i \\ 1 \end{smallmatrix}\right], a2=[1+i1−i0]a_{2} = \left[\begin{smallmatrix} 1+i \\ 1-i \\ 0 \end{smallmatrix}\right]

(5)

x=c130,a1=c134,a2=c132x = \mathbf{c}_{130}, \quad a_{1} = \mathbf{c}_{134}, \quad a_{2} = \mathbf{c}_{132}. c130=[−2+i1+i1],c134=[1−2i],c132=[103i]\mathbf{c}_{130} = \left[\begin{smallmatrix} -2+i \\ 1+i \\ 1 \end{smallmatrix}\right], \mathbf{c}_{134} = \left[\begin{smallmatrix} 1 \\ -2 \\ i \end{smallmatrix}\right], \mathbf{c}_{132} = \left[\begin{smallmatrix} 1 \\ 0 \\ 3 i \end{smallmatrix}\right].