Skew symetric matries and basis

[robierob12]

Lately I have been been studying basis and demension.

For a more interesting problem I wanted to see if I could find the basis of the vector space of all 3x3 skew symetric matricies.

Usually, I can find a general form for these types of problem. Such as the general form of a symetric matricie. But skew symetric matricies seem to have more than one form

[0 a b]
[-a 0 c]
[-b -c 0]

and

[0 a -b]
[-a 0 -c]
[b c 0]

I proved that this form of a skew symetric matrice is a basis

[0 a b]
[-a 0 c]
[-b -c 0]

but is it true for the vector space of all 3x3 skew symetric matricies of that form or all skew symetric matricies?

 

[matt grime]

So you not see that those two forms you gave describe exactly the same set of matrices?

You proved what was a basis?

 

[robierob12]

I see that they are both skew symetric but because the general form of the two looked different they may not fit all skew semetric matricies.

Iam trying to find a basis for the vector space of all 3x3 symetric matricies.

I used this as my set

0 1 0
-1 0 0
0 0 0

0 0 1
0 0 0
-1 0 0

0 0 0
0 0 1
0 -1 0

 

[matt grime]

So you're trying to find a set of matrices such that every skew symmetric basis is a linear combination of them. Do you not see how to write any of the matrices in your first post in terms of those three things? Remember, you can multiply basis vectors by any scalar, including -a or -b or -c,...

 

[HallsofIvy]

You seem to be concerned that while the set containing
$$\left(\begin{array}{ccc}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$
$$\left(\begin{array}{ccc}0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{array}\right)$$
and
$$\left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array}\right)$$

is a basis, so is the set containing
$$\left(\begin{array}{ccc}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$
$$\left(\begin{array}{ccc}0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right)$$
and
$$\left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{array}\right)$$

There is certainly no problem with that- any vector space has an infinite number of distinct bases!

 

[robierob12]

I can't believe I didnt catch that. Thanks