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Let $V$ be a vector space and $dim(V)=n$.

Let $GL(V)$ denote the set of all invertible linear transformations from $V$ to itself.

Next we define $Gr(k,n)$ to be set of all $k-$dimensional subspaces of $V$ with $k<n$.

We now define the following action:

$GL(V) \times Gr(k,n) \to Gr(k,n)$ defined as $T.W\mapsto T(W) $

I'm asked to show that the action is transitive.

My thoughts: To show that the action is transitive we must show that, given $U$ $\in Gr(k,n)$ $\exists\;\; T'\in GL(V)$ such that $T'.W=U$

I know that if $W$ is a subspace of $V$ and if $T\in GL(V)$ then $T(W)$ is a subspace of $V$ and $dim(T(W))=k$

So basically as $T$ varies in $GL(V)$ so does the $T(W)$ vary in $V$. But how can I show that every $k-$dimensional subspace of $V$ is the image of some transformation?

Hints please!

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  • $\begingroup$ A basis of $W$ and a basis of $U$ have the same number of elements $\endgroup$ Aug 9, 2017 at 7:29
  • $\begingroup$ @Max If i define a map which carries the basis elements of $W$ to the basis elements of $U$, what guarantees that the map is Linear? $\endgroup$
    – Naive
    Aug 9, 2017 at 7:39
  • $\begingroup$ You define it to be linear. Don't you know the theorem that states that given a basis of a space $V$, and a family of vectors of a space $E$ of the same size, and a mapping from the basis to the family, there is one and only one linear map from $V\to E$ that coincides with said mapping on the basis ? $\endgroup$ Aug 9, 2017 at 9:13

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You could pick a basis for $W$ and $U$, and then construct a transformation that mapped one basis onto the other.

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  • $\begingroup$ I too had thought about this, but what confuses me is : If I do construct such a map from $W\to U$ where a basis element of $W$ is mapped to a basis element of $U$, what guarantees that the map is linear? $\endgroup$
    – Naive
    Aug 9, 2017 at 7:32
  • $\begingroup$ Given a map from a basis $\{b_1,\dots,b_k\}$ to any set of vectors $\{v_1,\dots,v_k\}$, you can define $T(\alpha_1b_1+\dots+\alpha_kb_k)=\alpha_1v_1+\dots+\alpha_kv_k$. It's easy to prove $T$ is linear. In your case, you'll need some additional basis vectors for $V$ (but any basis for a subspace can be extended to a basis for the whole space). You end up with two bases for $V$, and the $T$ I gave has all the properties you want (you'll need to prove this). $\endgroup$ Aug 9, 2017 at 7:58

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