7
$\begingroup$

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 $M_{2\times 2}$

$\endgroup$
14
  • 2
    $\begingroup$ It also needs to be non empty. But this is easy. $\endgroup$
    – Julien
    Commented Mar 8, 2013 at 21:24
  • 1
    $\begingroup$ @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) $\endgroup$
    – rschwieb
    Commented Mar 8, 2013 at 21:28
  • 1
    $\begingroup$ @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. $\endgroup$
    – Julien
    Commented Mar 8, 2013 at 21:34
  • 1
    $\begingroup$ @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. $\endgroup$
    – Julien
    Commented Mar 8, 2013 at 21:39
  • 1
    $\begingroup$ @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. $\endgroup$
    – rschwieb
    Commented Mar 8, 2013 at 21:42

1 Answer 1

6
$\begingroup$

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 $$ \begin{pmatrix} 1 && 1 \\ 1 && 0 \end{pmatrix}, \qquad \begin{pmatrix} 0 && 1 \\ -1 && 1 \end{pmatrix}. $$

$\endgroup$
3
  • $\begingroup$ 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)? $\endgroup$
    – Susana
    Commented Mar 8, 2013 at 21:28
  • 1
    $\begingroup$ Yes, your set is the set of all linear combinations of those two matrices, and thus a subspace. $\endgroup$ Commented Mar 8, 2013 at 21:29
  • 2
    $\begingroup$ Exercise: Prove that the space of all linear combinations (also called the span) of any subset of a vector space forms a subspace. $\endgroup$
    – muzzlator
    Commented Mar 8, 2013 at 21:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .