7

I'm self studying linear algebra and now I'm starting with proofs and so on. I found this exercise and this is the way I prove it. I think it's correct but I'm not sure I mean, what do you think?

Is the set of matrices $ \begin{pmatrix} x && x+y \\ x-y && y \end{pmatrix} $ where $x, y \in R $ a subspace of $M_{2\times 2}$

I know that for the set to be a subspace it needs to be closed under vector addition and under scalar multiplication. So:

Given $M_1, M_2 \in M_{2\times 2}, x,y, \in R$. Then $M_1 + M_2$ will be:

$$\begin{align} \begin{pmatrix} x_1 && x_1+y_1\\ x_1-y_1 && y_1 \end{pmatrix} &+ \begin{pmatrix} x_2 && x_2+y_2\\ x_2-y_2 && y_2 \end{pmatrix} = \\& \begin{pmatrix} (x_1 + x_2) && (x_1+x_2)+(y_1+y_2)\\ (x_1+x_2)-(y_1+y2) && (y_1 + y_2) \end{pmatrix} \end{align}$$

Which has the same structure so it's closed under vector addition.

Now, Given $M \in M_{2\times 2}, x,y,r \in R$. Then $rM_1$ will be:

$$ r\begin{pmatrix} x && x+y\\ x-y && y \end{pmatrix} = \begin{pmatrix} rx && r(x+y)\\ r(x-y) && ry \end{pmatrix} $$

Which, given that $x,y,r \in R$ is also closed under vector multiplication.

So yes. The set is a subspace of $M_{2\times 2}$

Asaf Karagila
  • 393,674
Susana
  • 299
  • 2
    It also needs to be non empty. But this is easy. – Julien Mar 08 '13 at 21:24
  • But your proof is fine... – copper.hat Mar 08 '13 at 21:24
  • @julien Wow, ok, but, how can I prove that it is not empty? – Susana Mar 08 '13 at 21:25
  • Take $x=y=0$. This gives an element of the set. – Julien Mar 08 '13 at 21:25
  • 1
    @julien I do not think anyone will confuse a set of matrices parameterized by two copies of a field with the emptyset. (But wait! You forgot to check if the field is empty...! muffled sobs) – rschwieb Mar 08 '13 at 21:28
  • 1
    @rschwieb Sure...but I think it still has to be mentioned since it is part of the definition. Even if one only says: the set is obviously non empty. Now if we adopt Andreas approach, there is no more need to say that since you present it as the span of two vectors. – Julien Mar 08 '13 at 21:34
  • @julien I know you are completely justified in requiring that a vector space is nonempty. However, (nothing personal) every time I hear the requirement emphasized, a little part of me dies :( Complaining about it is theraputic for me, and I hope it doesn't bother you. – rschwieb Mar 08 '13 at 21:37
  • 1
    @rschwieb Sorry to hear that. I did not mean to hurt anybody here. Are there lots of definitions that you can't stand hearing? Just curious. No, there are not so many things that bother me in general. – Julien Mar 08 '13 at 21:39
  • 1
    @julien No, I just dislike the ones phrase in nitpicky ways. Requiring things to be nonempty just seems so obtuse compared to keeping in mind that all vector spaces have 0, and subspaces are vectorspaces. – rschwieb Mar 08 '13 at 21:42
  • 1
    @rschwieb Do you mean: $0$ is in our set is better than our set is non-empty? Sure, I agree. But I find that you get upset easily... – Julien Mar 08 '13 at 21:57
  • @julien Upset?! Thta's completely wrong: I'm feeling sympathy for students with teachers who tie their students' shoelaces together with such details. Thanks for the therapy though, it has been really helpful. – rschwieb Mar 08 '13 at 22:00
  • 1
    @rschwieb Upset was maybe not the appropriate word. English is not my native language. But I checked in the dictionary and sarcastic applies to your last comment. And that was not useful. Especially when there is no reason. – Julien Mar 08 '13 at 22:07
  • @julien Aw man... people are taking what I've said too seriously :) I've only meant it all in the silliest sense. I said at the beginning: your position is absolutely sound! There is no attack here, just friendly banter. – rschwieb Mar 09 '13 at 01:50
  • 1
    I think Andreas' proof is slick and clean, but it requires muzzlator's exercise, the statement of which she is not yet familiar with (though by now she might be!). I would suggest she just stick with her original proof, adding that the space has a zero element. – Rohit Chatterjee Jun 20 '13 at 04:37

1 Answers1

6

Note $$ \begin{pmatrix} x && x+y \\ x-y && y \end{pmatrix} = x \begin{pmatrix} 1 && 1 \\ 1 && 0 \end{pmatrix} + y \begin{pmatrix} 0 && 1 \\ -1 && 1 \end{pmatrix} $$ so it is the subspace generated by the two matrices $$ \begin{pmatrix} 1 && 1 \\ 1 && 0 \end{pmatrix}, \qquad \begin{pmatrix} 0 && 1 \\ -1 && 1 \end{pmatrix}. $$

Andreas Caranti
  • 68,873
  • Thank you! I have just one more question, I clearly understand your example, so, given that 2 $M_{2x2}$ matrices generate the example matrix, then it is automatically proved that it's also a subspace (the generated one)? – Susana Mar 08 '13 at 21:28
  • 1
    Yes, your set is the set of all linear combinations of those two matrices, and thus a subspace. – Andreas Caranti Mar 08 '13 at 21:29
  • 2
    Exercise: Prove that the space of all linear combinations (also called the span) of any subset of a vector space forms a subspace. – muzzlator Mar 08 '13 at 21:33