Let $V,W$ be finite dimensional vector spaces over an infinite field $k$. Fix a positive integer $n \leq \dim(W), \dim(V)$.

Given $n$ injective linear maps $f_i:V\rightarrow W$, such that the $f_i$ are linearly independent as elements of $\operatorname{Hom}(V,W)$, is there necessarily a vector $v\in V$ with $f_i(v)$ linearly independent in $W$?

This holds for $n=1,2$ and if the $f_i$ commute and are diagonalisable, since then one may simultaneously diagonalise.

In terms of the conditions, we clearly require linear independence of the maps for the conclusion, and injectivity is required to exclude the case of $\dim(V) > \dim(W)$, where the images will never be linearly independent.

  • 1
    $\begingroup$ Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*\in V^*$ and an arbitrary nonzero $w\in W$. $\endgroup$
    – user856
    Dec 18, 2018 at 5:57

2 Answers 2


This fails for $n=3.$ Consider $f_i$ with matrix representations

$$ \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}, \begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}, \begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}. $$

Or in other words: $\mathrm{id}, \mathrm{id}+e_1e_2^*, \mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by $\{v,e_1\}.$


Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U \subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + \lambda g_{i}$ for sufficiently small $\lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $\lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v \in V$ under $f_{i}$ is in $U + \mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.

When can we form this construction? It is sufficient to have $m^{2} \geq n-2$, because for any basis $\beta = \{ u_{j} : j \}$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{\ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} \geq n-2$ for any $m \geq 0$. Since we also need $m \geq 1$, the construction thus works in general for any $n \geq 3$.


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