- #1

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Why not make dot(u,v)=transpose(u)v rather than transpose(v)u?

- Thread starter tgt
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- #1

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Why not make dot(u,v)=transpose(u)v rather than transpose(v)u?

- #2

Hurkyl

Staff Emeritus

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What's the difference?

- #3

- 2,111

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[tex]

(u|v) = u^{\dagger} v

[/tex]

and IMO it is lot better than the mathematicians' convention

[tex]

(u|v) = v^{\dagger} u

[/tex]

When something is done in a dumb way, the reason is usually "for historical reasons". I guess that's the answer to the OP question this time too.

- #4

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One way generates a scalar and the other way generates an NxN matrix. Which is which depends on whether the vector is a 1xN row vector or a Nx1 column vector.What's the difference?

- #5

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This response is not logical!One way generates a scalar and the other way generates an NxN matrix. Which is which depends on whether the vector is a 1xN row vector or a Nx1 column vector.What's the difference?

If we assume [itex]u[/itex] and [itex]v[/itex] to be Nx1 vectors, then both [itex]u^Tv[/itex] and [itex]v^Tu[/itex] give a single component 1x1 matrix.

tgt did not ask about why to use Nx1 or 1xN matrices, so it is better not to start switching between them now.

The truth is that there are two different conventions for complex inner products, and they are

[tex]

(u|v) = \sum_{k=1}^N u^*_k v_k

[/tex]

and

[tex]

(u|v) = \sum_{k=1}^N u_k v^*_k

[/tex]

so I thought it would be natural to guess that the original question was related to this issue.

- #6

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I see now.

- #7

HallsofIvy

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No, you have misread. If u and v are column vectors (most common convention), then uOne way generates a scalar and the other way generates an NxN matrix. Which is which depends on whether the vector is a 1xN row vector or a Nx1 column vector.

But the question was about the difference between u

- #8

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Yes, I misread the OP as u^{T}v versus uv^{T}, as opposed to u^{T}v versus v^{T}u.

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