Expert: Paul Klarreich - 10/3/2009

Question
What is the dimension of the vector space of 2 x 2 symmetric matrices? of skew-symmetric matrices? How can I generalize to 3x3 matrices and then to n x n matrices?

Answer
Questioner: Sombra
Country: United States
Category: Advanced Math
Question: What is the dimension of the vector space of 2 x 2 symmetric matrices? of skew-symmetric matrices? How can I generalize to 3x3 matrices and then to n x n matrices?
....................................................
Hi, Sombra,

It's not 3?  The dimension of a V.S. is the minimum number of elements needed to generate all the elements of the space.  So for the (general) space, whose elements are all the 2X2's, which we could write

((a,b),(c,d))    << is the notation clear?

<< this means
[ab]
[cd]
.....................
you need four elements, which could be

((1,0),(0,0))
((0,1),(0,0))
((0,0),(1,0))
((0,0),(0,1))

and I think you could easily see that no one of these can be made from the others.

BUT the second and third are not symmetric. So replace them with:

((1,0),(0,0))
((0,1),(1,0))
((0,0),(0,1))

because any symmetric matrix has the form ((a,b),(b,d))  and again you can make any from a linear combo of these three.  Obviously the dimension is 3.

.....................
Skew-symmetric matrices are:  ((0,b),(-b,0)) and now the dimension is 3, since only one element can vary.
...........................................
What about 3X3, n X n?

Perhaps it is easiest to do the n X n directly.  We will eliminate the 'lower half', which isn't really a half.

There are  n^2 elements in the entire matrix.
There are  n elements in the diagonal of the matrix.
That leaves n^2-n 'off' the diagonal.
Take away half of those.
The number of independent elements is

n^2 - (n^2 - n)/2 =

n^2 - n^2/2 + n/2 =

n^2/2 + n/2 = (n^2 + n)/2
..............................
Checking:  n = 2 gives 3.

n = 3:

[xxx]
[.xx]
[..x]
or 6 elements.

And (3^2 + 3)/2 = 12/2 = 6.

I think that's it for symmetric ones.

For skew-symmetrics, we have to subtract off the 'n' diagonal elements, because they are forced to be zero.  

Dimension will be  (n^2 + n)/2 - n = (n^2 - n)/2