This is an exercise of Advanced linear algebra of Roman:
Let $V$ a finite-dimensional vector space over an infinite field. Prove that if $S_1,\ldots,S_k$ are subspaces of $V$ of equal dimension ever exists a subspace $T$ of $V$ such that $V=S_j\oplus T$ for all $j=1,\ldots,k$.
Can someone take a look to the proof below?
Let $\dim S_j=m$ for all $j=1,\ldots,k$ and $\dim V=n$. By induction on $n-m$ it can be seen that if $m=n$ then $T=\{0\}$, so the statement holds.
Now suppose that it holds for all $\ell$ such that $n-\ell\le m+1$, thus we need to prove that it also holds for $m$.
Because every $S_j$ of dimension $m$ is contained in some $S^*_j$ of dimension $m+1$ then $S_j\oplus T^*$ is well-defined, for $T^*$ the common complement of the $S^*_j$, and $\dim(S_j\oplus T^*)=n-1$, thus we need to find some vector $v\in V$ such that $T:=\langle v\rangle\oplus T^*$ and $v\notin S_j\oplus T^*$ for all $j=1,\ldots,k$.
Setting $R_j:=S_j\oplus T^*$ for all $j=1,\ldots,k$ it is enough to find some common complement to these family of $R_j$ provided that $\dim R_j=n-1$ for all $j=1,\ldots,k$, what exists by hypothesis, so we are done.
Observe that the induction holds for all $m\le n$ because when $m<0$ the statement is vacuously true.
Im not sure that this exercise is correctly done because I dont used the fact that it is a vector space over an infinite field. This means that the result holds also for vector spaces over finite fields or, by the contrary, there is something wrong in this proof?
(I found in internet a solution for this exercise slightly different that uses the fact that $V$ it is a vector space over an infinite field.)
UPDATE: here it is the assertion that if the dimension of the field is finite then this doesnt hold in general (it just hold whenever $m<|\Bbb F|)$, but why? What is wrong in the above proof?