Dimension of a subspace defined by multiple restrictions. 
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.
 A: Each 'equal sign' is a different, linearly independent restriction. So you have $2 + 49 = 51$ restrictions.
Another way to think about this is in terms of degrees of freedom. Choosing a value for $x_1$ fills in $3$ of the coordinates, and choosing a value for $x_{51}$ fills in another $50$ coordinates. For the other $47$ coordinates, you are free to make independent choices. Thats a total of $49=100-51$ choices, which is the dimension.
A: You have more than two restrictions:
$$
x_2=x_1\\
x_3=x_1\\
x_{52}=x_{51}\\
x_{53}=x_{51}\\
...\\
x_{100}=x_{51}$$
A: $V=<v_1, v_2, e_4, e_5, \ldots, e_{50}>$, 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.$
A: A general vector in $V$ looks like
$$
\begin{bmatrix}
x\\x\\x\\y_4\\y_5\\\vdots\\ y_{50}\\z\\z\\\vdots\\z
\end{bmatrix}=x\begin{bmatrix}
1\\1\\1\\0\\0\\\vdots\\ 0\\0\\0\\\vdots\\0
\end{bmatrix}
+z\begin{bmatrix}
0\\0\\0\\0\\0\\\vdots\\ 0\\1\\1\\\vdots\\1
\end{bmatrix}
+\sum_{i=4}^{50}y_i\begin{bmatrix}
0\\0\\0\\\vdots\\0\\1\\ 0\\0\\0\\\vdots\\0
\end{bmatrix}
$$
Where $1$ in the last sum is at the $i^{\text{th}}$ place.
So there are $49$ linearly independent vectors needed to span this set. Hence the dimension is $49$.
