By constructing a counterexample, show that the union of two subspaces is not in general, a subspace?

I don't really have a clue on how to start this. I really only want a hint on how to start his

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- Oct 9th 2010, 03:39 PMacevipaUnion of two subspaces
By constructing a counterexample, show that the union of two subspaces is not in general, a subspace?

I don't really have a clue on how to start this. I really only want a hint on how to start his - Oct 9th 2010, 03:42 PMJhevon
your question as posed is strange. you usually use counter examples to disprove something, while here it seems like you want to prove something. check the wording and ask again.

as for the hint: begin at the beginning, haha. What does it mean to be a subspace? and in what context are you describing subspaces? - Oct 9th 2010, 03:46 PMacevipa
Yeah sorry, I just edited my question

So could we take two subspaces just as an example

So $\displaystyle A=\{y=x\}$ and $\displaystyle B=\{y=-x\}$ are subspaces of $\displaystyle \mathbb{R}^2$

$\displaystyle \begin{pmatrix}2\\2\end{pmatrix}\in A\cup B$

$\displaystyle \begin{pmatrix}1\\-1\end{pmatrix}\in A\cup B$

$\displaystyle \begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}1\ \-1\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}\no t\in A\cup B$

Therefore, not closed under addition.

Is this proof good enough - Oct 9th 2010, 08:31 PMJhevon
Not a "proof", but yes, as a counter-example, it works. provided you have the definition for addition in $\displaystyle \displaystyle \mathbb R ^2$ given.

I was thinking of an example along the lines of:

$\displaystyle \displaystyle A$ = the set of all 2x2 matrices over the reals of the form $\displaystyle \displaystyle \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$

and $\displaystyle \displaystyle B$ = the set of all 2x2 matrices over the reals of the form $\displaystyle \displaystyle \begin{pmatrix} a & a + b \\ a + b & b \end{pmatrix}$

Both with standard matrix addition and scalar multiplication operations. - Oct 9th 2010, 08:33 PMacevipa
- Oct 9th 2010, 08:36 PMJhevon
yes

Quote:

What do you mean when you say "provided you have the definition for addition in $\displaystyle \mathbb{R}^2$ given".

- Oct 9th 2010, 08:39 PMacevipa
- Oct 9th 2010, 08:43 PMJhevon