0
$\begingroup$

I'm having a bit of trouble showing that the following subsets of $\Bbb{R}^n$ are subspaces of $\Bbb{R}^n (n>2)$. I know that I need to show that they are closed under addition and multiplication, but I do I need to show it for the $nn$ and the $nm$ case? If so, how do I show it for the $nm$ case? Furthermore, is part d the set of $n$-tuples?

a) The symmetric matrices

b) The nonsingular matrices

c) The diagonal matrices

d) $\{x \mid \sum_{j=1}^{n}x_j =0\}$

$\endgroup$
1
  • $\begingroup$ As for the first three (a,b,c), those matrices aren't even defined in the $nm$ case (if n is not equal to m). Part d is the set of n-tuples whose components sum to zero (at least i think..it doesn't really make sense written as it is) $\endgroup$
    – Franco
    Commented Apr 19, 2014 at 0:38

1 Answer 1

0
$\begingroup$

First, what is a symmetric matrix?, its definition already implies it has to be a nxn matrix. Second,nonsingular, same argument, it is nonsingular if and only if it has nonzero determinant, and only square matrices have nonzero determinant. Same for diagonal,only square matrices have diagonals,and same for d).

$\endgroup$
4
  • $\begingroup$ So how do I know d) is also nxn? $\endgroup$ Commented Apr 19, 2014 at 0:46
  • $\begingroup$ If I understand it correctly, $x_j$ are column vectors and there are n of them, also each of them is n-dimensional meaning $x_j=(x_{j,1},\dots x_{j,n})$. $\endgroup$
    – L.G
    Commented Apr 19, 2014 at 0:50
  • $\begingroup$ Okay. So the fact that it says they are equal to zero implies they pass the origin correct? $\endgroup$ Commented Apr 19, 2014 at 0:54
  • $\begingroup$ No that means they are linearly dependent therefore parallel to each other. $\endgroup$
    – L.G
    Commented Apr 19, 2014 at 1:07

You must log in to answer this question.

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