Given a non-zero vector v, find the value that should be assigned to α to make Pij = αvivj and Qij = δij − αvivj into parallel and orthogonal projection tensors, respectively, i.e. tensors that satisfy, respectively, Pijvj = vi, Pijuj = 0 and Qijvj = 0, Qijuj = ui, for any vector u that is

Question

Given a non-zero vector v, find the value that should be assigned to ([latex]\alpha[/latex]) to make

[latex]P_{i j}=\alpha v_{i} v_{j} \quad ext { and } \quad Q_{i j}=\delta_{i j}-\alpha v_{i} v_{j}[/latex]

into parallel and orthogonal projection tensors, respectively, i.e. tensors that satisfy, respectively, [latex]P_{i j} v_{j}=v_{i}, P_{i j} u_{j}=0[/latex] and [latex]Q_{i j} v_{j}=0, Q_{i j} u_{j}=u_{i}[/latex], for any vector u that is orthogonal to v.

Show, in particular, that [latex]Q_{i j}[/latex] is unique, i.e. that if another tensor [latex]T_{i j}[/latex] has the same properties as [latex]Q_{i j}[/latex] then [latex]\left(Q_{i j}-T_{i j} ight) w_{j}=0[/latex] for any vector [latex]\mathbf{w}[/latex].

Accepted Answer

Consider

[latex]\begin{aligned}& P_{i j} v_{j}=\alpha v_{i} v_{j} v_{j}=\alpha|\mathbf{v}|^{2} v_{i}, ext { and } \& P_{i j} u_{j}=\alpha v_{i} v_{j} u_{j}=\alpha v_{i}\left(v_{j} u_{j} ight)=0, ext { as } \mathbf{u} ext { is orthogonal to } \mathbf{v} .\end{aligned}[/latex]

For [latex]P_{i j} v_{j}=v_{i}[/latex] it is clearly necessary that [latex]\alpha=|\mathbf{v}|^{-2}[/latex].

With this choice,

[latex]\begin{aligned}& Q_{i j} v_{j}=\left(\delta_{i j}-\alpha v_{i} v_{j} ight) v_{j}=v_{i}-\alpha\left(v_{j} v_{j} ight) v_{i}=v_{i}-|\mathbf{v}|^{-2}\left(v_{j} v_{j} ight) v_{i}=0, ext { and } \& Q_{i j} u_{j}=\left(\delta_{i j}-\alpha v_{i} v_{j} ight) u_{j}=u_{i}-\alpha\left(v_{j} u_{j} ight) v_{i}=u_{i}-0 v_{i}=u_{i} .\end{aligned}[/latex]

Thus the one assigned value for [latex]\alpha[/latex] gives both [latex]P_{i j}[/latex] and [latex]Q_{i j}[/latex] the required properties. Let [latex]\mathbf{u}^{(1)}[/latex] and [latex]\mathbf{u}^{(2)}[/latex] be any two linearly independent non-zero vectors orthogonal to v. Then any vector w can be expressed as [latex]\lambda \mathbf{v}+\mu \mathbf{u}^{(1)}+v \mathbf{u}^{(2)}[/latex].

Now suppose that [latex]T_{i j}[/latex] has the same properties as [latex]Q_{i j}[/latex] and consider

[latex]\begin{aligned}\left(Q_{i j}-T_{i j} ight) w_{j} & =\left(Q_{i j}-T_{i j} ight)\left(\lambda v_{j}+\mu u_{j}^{(1)}+ν u_{j}^{(2)} ight) \& =\lambda 0+\mu u_{j}^{(1)}+ν u_{j}^{(2)}-\lambda T_{i j} v_{j}-\mu T_{i j} u_{j}^{(1)}-ν T_{i j} u_{j}^{(2)} \& =0+\mu u_{j}^{(1)}+ν u_{j}^{(2)}-0-\mu u_{j}^{(1)}-ν u_{j}^{(2)}=0 .\end{aligned}[/latex]

In this sense, [latex]Q_{i j}[/latex] is unique.

Question 26.11: Given a non-zero vector v, find the value that should be ass...

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