TITLE: Dimension of a subspace defined by multiple restrictions.
QUESTION [9 upvotes]: Let $$V=\{(x_1,x_2,x_3,\dots,x_{100})\in\mathbb{R}^{100}\,|\, x_1=x_2=x_3 \text{ and } x_{51}=x_{52}=x_{53}= \dots=x_{100}\}$$ What is $\dim V$?

If W is a subspace of vector space $V$ then $$\dim W = \dim V - \text{number of linearly independent restrictions}$$
In our case $\dim V= 100$ and the number of linearly independent restriction is $2$. Therefore $\dim W = 100-2=98$. Am I correct?? I think not. Please help.

REPLY [2 votes]: $V=<v_1, v_2, e_4, e_5, \ldots, e_{50}>$, where
$v_1=(1,1,1,0, \ldots,0)$,  $v_2=(\underbrace{0, \ldots, 0}_{\text{$50$ zeros}}, 1,1,\ldots 1)$  and $e_i=(0, 0, \ldots \underbrace{1}_{i}, \ldots ,0)).$.
So, $ \dim V=49.$