- #1

- 29

- 0

1

**u**=

**u**

1(x,y,z)=(x,y,z)

1(1,2,3) = (1,2,3)

1 (1,0,...,1) = (1,0,...,1)

1(x,y,z)=(x,y,z)

1(1,2,3) = (1,2,3)

1 (1,0,...,1) = (1,0,...,1)

I don't see how you could ever end up in a situation where you could end up with

1

**V**≠

**V**

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- #1

- 29

- 0

1

1(x,y,z)=(x,y,z)

1(1,2,3) = (1,2,3)

1 (1,0,...,1) = (1,0,...,1)

I don't see how you could ever end up in a situation where you could end up with

1

- #2

Mark44

Mentor

- 34,665

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It can happen if you have a vector space with an unusual kind of scalar multiplication. Keep in mind that a vector space consists of a set of vectors over some field (often, the field of real numbers ##\mathbb{R}## or the field of complex numbers ##\mathbb{C}##), together with operations for vector addition and for multiplication by a scalar.

1u=u

1(x,y,z)=(x,y,z)

1(1,2,3) = (1,2,3)

1 (1,0,...,1) = (1,0,...,1)

I don't see how you could ever end up in a situation where you could end up with

1V≠V

If scalar multiplication is defined like this

##k \cdot <x, y> = <kx, 0>##

then there is no scalar k for which ##k \cdot <x, y> = <x, y>##.

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