I'm having some problems understanding this:

$$V = \mathbb R^3\text{ and }W = \{(x,y,z) \mid x+y+z=0\}$$

So I want $V/W$ and a basis to it.

$$\dim V = 3$$ $$\dim W = 2$$ $$\dim V/W = 1$$

But a basis to $W$ could be $BW = \{(1,0,-1), (0,1,-1)\}$.

To extend it to $V$, we add $(0,0,1)$, which is linearly independent to span $BW$. So a basis to $V/W$ could be $(0,0,1) + W$.

But that's also a plane, $(x,y,-x-y+1)$, which has dimension $2$.

What's wrong?


$(0,0,1) + W$ is an affine plane of $V$ that cannot be used to evaluate the dimension of $V/W$ which is not a subspace of $V$. $V/W$ is an equivalence class.

  • $\begingroup$ So the fact that it gives me only one plane, I get the dimension is 1. Because if it had more equivalence classes, then they would be each a parallel plane to W? $\endgroup$ – Ana Carolina Nicolau Jul 6 '15 at 1:07

Take a vector not in $W$, for example the vector $(1,1,1)$ orthogonal to $W$. The equivalence class of this vector, in the notation of your course $(1,1,1)+W$, is a basis for $V/W$. We do not need to worry about a basis for the whole space, or for $W$.

Your calculation of the dimension of $V/W$ is correct.


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