Prove there are infinitely many $k$-dimensional subspaces of a finite-dimensional vector space May $V$ be an $n$ dimensional Vektorspace such that $\dim (V) =: n \ge 2$.
We shall prove, that there are infinitely many $k$-dimensional subspaces of $V$, $\forall k \in \{1, 2, ..., n-1\}$.
So first, I thought about using induction, the base step is not that hard, for $n=2$ we take two vectors, say $a$ and $b$ and define infinitely many 1-dimensional subspaces as span$\{a+jb\}$ for $j \in \mathbb N$.
It is easy to see those vector spaces are not all equal, but I kinda realised that induction is not the way to go, as I think $n$ is fixed.
Anyhow, then I thought about using finiteness of basis for $V$ to try to construct those subspaces (using vectors from basis). I failed to do so, so I'm just asking for a hint or any useful advice where to start with this.
 A: HINT: How many lines are there in $\Bbb R^2$ passing through $\{(0,0)\}$.
How many planes are there in $\Bbb R^3$ containing $\{(0,0,0)\}$.
A: Subspaces:$(a_1\times a_2\times...\times a_{k-1}\times(a_k+ma_{k+1}))$, $m\in N$ 
A: In fact, induction does work, even though there are other, more direct approaches. I am going to assume that $V$ is a vector space over some infinite field, otherwise your result is false.
If $n=2$, then $k=1$: consider then all the straight lines passing through the origin; it is obvious and it doesn't require a proof that they are infinitely many.
Assume the statement true for $n$ and consider $V$ with $\dim V = n+1$.


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*The case $k < n$: if there were only finitely many subspaces of dimension $k$, then let $W \subset V$ be a subspace of $\dim W = n$; by the induction hypothesis, $W$ has infinitely many subspaces of dimension $k$, but these are also subspaces of $V$, which were assumed to be finitely many - so we have obtained a contradiction. This is the part where we use induction.

*The case $k=n$: if you choose a basis, then each $v = (v_1, \dots, v_n)$ will give rise to the linear form $f_v (u) = u_1 v_1 + \dots + u_n v_n$. Notice that $\dim (\ker f_v) = n$ and that $\ker f_v = \ker f_w$ if and only if there exist $\lambda$ such that $v = \lambda w$. But there are infinitely many non-proportional vectors in $V$ (they are in bijective correspondence with the projective space of $V$), so there are infinitely many different linear forms, therefore infninitely many subspaces of dimension $n$.
