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Let $V$ be a finite dimensional vector space over a field $F$ of dimension $n$ and let $\{\alpha_1,\alpha_2,\ldots,\alpha_n\}$ be a basis of $V$. We know that fixing the images of $\{\alpha_1,\alpha_2,\ldots,\alpha_n\}$ determine a unique vector space homomorphism $\varphi$ from $V$ to $V$. Moreover if these images are linearly independent, then it induces a unique isomorphism $\varphi$.

Let $\{W_1,W_2,\ldots,W_r\}$ be $r$ subspaces of $V$ and let $\{U_1,U_2,\ldots,U_r\}$ be their prescribed images (subspaces). Obviously, there may not exist any vector space homomorphism $\varphi$ from $V$ to $V$ such that $\varphi(W_i)=U_i$.

What is the minimum number of subspaces of $V$ such that fixing their images determine a unique vector space isomorphism $\varphi$ from $V$ to $V$?

I strongly feel that the answer is $n$, the dimension of the space, but I could not prove it. Please suggest some way.

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  • $\begingroup$ A very weak version would be to ask the question with the added restriction that the $W_i$ all have the same dimension, $r$ say. I think that it is then a question about projective geometry, and suspect it may be easily answered -- might it be $\binom{n}{r}$? -- but I have forgotten most of the geometry I once knew. $\endgroup$ – ancientmathematician Mar 21 '17 at 8:38
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If one of the $W_i$'s has two linearly independent vectors, say $w_{i,1}, w_{i,2}$, then with any linear isomorphism $T$ you can define another isomorphism $S$ such that $Sw_{i,1}=Tw_{i,2}$ and $Sw_{i,2}=Tw_{i,1}$. This will not change the image of $W_i$ but it will change the isomorphism, so it is not uniquely defined by those images, so you are correct that the minimum number is $n$.

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    $\begingroup$ But interchanging $w_{i,1}$ and $w_{i,2}$ may disturb the images of other $W_j$'s. $\endgroup$ – cryptomaniac Mar 21 '17 at 8:15
  • $\begingroup$ My question can be rephrased as follows: given a linear isomorphism $\varphi$, what is the minimum number of subspaces whose images we should know to determine $\varphi$ uniquely? $\endgroup$ – cryptomaniac Mar 21 '17 at 8:18
  • $\begingroup$ @cryptomaniac - I don't see how, really. Although the $W_j$'s may have non trivial intersection, their image as a subspace is NOT disturbed by the operation I suggested. $\endgroup$ – uniquesolution Mar 21 '17 at 9:11

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