# Subspace spanned by subspace

• Feb 4th 2010, 09:27 PM
Mollier
Subspace spanned by subspace
Hi,

problem:
Suppose that x and y are vectors and M is a subspace in a vector space V;
let K be the subspace spanned by M and x, and let K be the subspace spanned by M and y.
Prove that if y is in K but not in M, then x is in K.

attempt:
So, K is the set of all linear combinations of elements in M and the vector y.
K is also the set of all linear combination of elements in M and the vector x.
This definition of K makes me think that x is always in K.
What am I missing here?

Thanks!
• Feb 5th 2010, 12:36 AM
flyingsquirrel
Hi,
Quote:

Originally Posted by Mollier
problem:
Suppose that x and y are vectors and M is a subspace in a vector space V;
let K be the subspace spanned by M and x, and let K be the subspace spanned by M and y.
Prove that if y is in K but not in M, then x is in K.

It does not make sense : $\displaystyle K$ is at the same time the subspace spanned by $\displaystyle M$ and $\displaystyle x$ and the subspace spanned by $\displaystyle M$ and $\displaystyle y$... but these subspaces may be different !

Here is what I think they meant :
Quote:

problem:
Suppose that $\displaystyle x$ and $\displaystyle y$ are vectors and $\displaystyle M$ is a subspace in a vector space $\displaystyle V$;
let $\displaystyle K_x$ be the subspace spanned by $\displaystyle M$ and $\displaystyle x$, and let $\displaystyle K_y$ be the subspace spanned by $\displaystyle M$ and $\displaystyle y$.
Prove that if $\displaystyle y$ is in $\displaystyle K_x$ but not in $\displaystyle M$, then $\displaystyle x$ is in $\displaystyle K_y$.
• Feb 5th 2010, 11:58 PM
Mollier
Hi!

We then have:

\displaystyle \begin{aligned} \mathbb{K}_x=&\;span\{\mathbb{M},x\}\\ \mathbb{K}_y=&\;span\{\mathbb{M},y\} \end{aligned}

If $\displaystyle y\in\mathbb{K}_x$ but $\displaystyle y\notin\mathbb{M}$ then $\displaystyle x\in\mathbb{M}$ and therfore $\displaystyle x\in\mathbb{K}_y$.

Is this any better?

Thanks.
• Feb 9th 2010, 07:47 PM
Mollier
Gentle bump.