# Thread: Union of two subspaces

1. ## Union 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

2. Originally Posted by acevipa
By constructing a counterexample, show that the union of two subspaces is 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
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?

3. Yeah sorry, I just edited my question

So could we take two subspaces just as an example

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

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

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

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

Is this proof good enough

4. Originally Posted by acevipa
Yeah sorry, I just edited my question

So could we take two subspaces just as an example

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

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

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

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

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

I was thinking of an example along the lines of:

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

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

Both with standard matrix addition and scalar multiplication operations.

5. Originally Posted by Jhevon
Not a "proof", but yes, as a counter-example, it works. provided you have the definition for addition in $\displaystyle \mathbb R ^2$ given.

I was thinking of an example along the lines of:

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

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

Both with standard matrix addition and scalar multiplication operations.
But with the question asked, is what I did correct?

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

6. Originally Posted by acevipa
But with the question asked, is what I did correct?
yes

What do you mean when you say "provided you have the definition for addition in $\mathbb{R}^2$ given".
i mean, at some point in your class or your text it should be written that you can add two coordinates in $\mathbb R ^2$ by adding their corresponding components. refrain from taking an intuitive definition if you can, not unless you know it works (by proving it does, or if your text says so and proves it)--it doesn't always.

7. Originally Posted by Jhevon
yes

i mean, at some point in your class or your text it should be written that you can add two coordinates in $\mathbb R ^2$ by adding their corresponding components. refrain from taking an intuitive definition if you can, not unless you know it works (by proving it does, or if your text says so and proves it)--it doesn't always.
We've already learnt that we can add two coordinate vectors by adding their corresponding components, so its not something I've just assumed that you can do

By the way, thanks for all your help!

8. Originally Posted by acevipa
We've already learnt that we can add two coordinate vectors by adding their corresponding components, so its not something I've just assumed that you can do

By the way, thanks for all your help!
ok, well no worries then. you're good