Prove that the basis of $M$ such that it does not contain any element from $L$ exists, if $L \subset M$

Question

Let $M$ be finite-dimensional vector space and let $L$ be a proper subspace of $M$ (so $L \ne M$). Prove that there exists a basis of $M$ such that it contains no element from $L$.

I'm not even sure how to start with this. How is this even possible if $L$ is a subspace of $M$?

Answer 1

Hint: Let $\{x_1,\dots,x_k\}$ be a basis of $L$; extend it to a basis $\{x_1,\dots,x_k,x_{k+1},\dots,x_n\}$ of $M$.

Now define, for $i=1,\dots,k$, $y_i=x_i+x_{k+1}$ and consider $$ \{y_1,\dots,y_k,x_{k+1},\dots,x_n\} $$

Answer 2

I present you with yet another example. Consider the line $L$ to be the subspace generated by the element $(1,1,0)$ Now notice that if $M$ is the plane generated by $(1,0,0)$ and $(0,1,0)$, $M$ contains this line, but $L$ and $M$ are not equal.

We can extend our original line by simply adding $(0,1,0)$ and obtain $\{(1,1,0),(0,1,0)\}$, this is a basis for $M$, but now it contains an element of $L$, oops! We can change this basis by the method described by egreg and obtain $\{(1,2,0), (0,1,0)\}$, Notice that this basis does not contain any elements of $L$, but it generates $M$.