Suppose that $V$ is a $K$-vector space and $X, Y$ vector subspaces of $V$ with $\dim(X) \geq \dim(Y) \implies \exists$ a linear map $f : V \to V$ such that $f(X)=Y$.
I do not know what strategy to adapt to resolve this exercise.
I thought of restricting $f$ to $X \to V$ and showing that $\dim(\mathrm{Im} \; f)=\dim Y$, but it does not lead me anywhere.
I thought of treating first the case $\dim(X) = \dim(Y)=n$, if $Y \subset X$ we can say that $X=Y \implies f = \rm Id$, if not let $B_X = \{x_i \lvert i \in [0,n]\}$ a basis of $X$ and $B_Y = \{y_i \lvert i \in [0,n]\}$ a basis of $Y$. Let $f : X \to Y$ a map defined by the following $f(\sum\alpha_ix_i)=\alpha_iy_i$. We can verify that it is linear : let $a,b \in X$ with $a=\sum a_i x_i$ and $b=\sum b_i x_i$, $\lambda \in K$. $f(\lambda a + b)=\sum (\lambda a_i + b_i)y_i = \lambda \sum a_i f_i + \sum b_i f_i = \lambda f(a) + f(b)$. This map is obviously surjective.
For the case $\dim(X) > \dim(Y)$ I guess I can do the same with $B_X = \{x_i \lvert i \in [0,n]\}$ a basis of $X$ and $B_Y = \{y_i \lvert i \in [0,n-k]\}$ a basis of $Y$ and just send the $n-k$ first $x_i$ on $y_i$ and the additional $x_i$ on $0_V$ ? Honestly not sure about this (but I think of X as a plane in $\mathbb R^3$ and $Y$ a straight line then $\dim(X)=2<1=\dim(Y)$ and by sending a vector of $X$'s basis to any nonzero vector of $Y$ and the other vector of the $X$ basis to $0_V$ it seems to work but I need to check the linearity of $f$ in this case)
I think there is some more clever ways to do this, maybe by finding a single map $f : X \to Y$ that is surjective.