Dimension of Subspace in can $\Bbb{R}^5$

Question

Dimension of Subspace

For this question, I understand the minimum dimension of $W$ must be $2$ because of the rank nullity theorem. So $5-3 = 2$. However, Would the dimension of $W$ be at most $4$? I came this conclusion because of this list of vectors : $(1,0,0,0,0)$, $(0,0,1,0,0)$, $(0,1,0,1,0)$ and $(0,0,0,1,1)$ include $e_1$ and $e_3$ but do they avoid the spaces $e_2$, $e_4$, and $e_5$?

Answer 1

Your subspace $W$ has $e_1$ and $e_3$ already, which are obviously independent from the rest of the basis vectors. We can just consider how many independent vectors don't contain $e_2$, $e_4$, or $e_5$ in their span. So what we're left with is basically a question of "what's the highest dimensional subspace of $\Bbb R^n$ which doesn't include any $e_i$?". It's clear that $n$ doesn't work; any $n$ dimensional subspace of $\Bbb R^n$ is just $\Bbb R^n$ itself.

However, using the same strategy you used here, we can show in general that we can have $n-1$ dimension. To do this, choose one of the basis vectors $e_k$, then the new set of vectors is $v_i = e_k + e_i$ for $i\neq k$. Clearly there are $n-1$ of these, and a linear combination of these gives us

$$ \sum c_iv_i = e_k\sum c_i + \sum c_i e_i $$

Since the $e_i$ are all independent, this only vanishes if each $c_i$ is zero, so the $v_i$ are independent. Further, if we set this equal to any $e_j$ for $j\neq k$, it's clear that $c_j = 1$ while the rest of the $c_i = 0$, but this gives us a nonzero coefficient on $e_k$. Similarly, if we set it equal to $e_k$, we need the sum of $c_i$ to be $1$, which means that some $c_j \neq 0$ and so $e_j$ will have nonzero coefficient. Thus, the span of the $v_i$'s has dimension $n-1$ but contains none of the $e_i$'s.

So, we have that $W$ has dimension at most $4$.

Answer 2

Note $W$ cannot have dimension 5. Otherwise, $W=\mathbb{R}^5$ and $e_2,e_5,e_5$ would belong to W.

But here's a hint.

Consider the collection of points $(x_1,x_2,x_3,x_4,x_5)\in \mathbb{R}^5$ defined by $x_2-2x_4+x_5=0$.