Basis for set of 2x2 complex symmetric matrices

[csnsc14320]

Homework Statement

Give the basis and dimension of the set of all 2x2 complex symmetric matrices.

Homework Equations

The Attempt at a Solution

I know that if the coefficients were real, then I could just have the basis

$$\left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array} \right), \left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right), \left( \begin{array}{cc} 0 & 0\\ 0 & 1 \end{array} \right)$$

but if the entries of the matrix can be in the form of a+bi, where a and b are real numbers, do I need three SEPARATE matrices with "i" coefficients or can i combine them somehow?

i.e., extend the dimension to 6 by:

$$\left( \begin{array}{cc} i & 0\\ 0 & 0 \end{array} \right) \left( \begin{array}{cc} 0 & i\\ i & 0 \end{array} \right) \left( \begin{array}{cc} 0 & 0\\ 0 & i \end{array} \right)$$

and thus my dimension would be 6?

 

[aPhilosopher]

I think that you can combine them with complex scalar multiplication.

e.g.

$$\left( \begin{array}{cc} i & 0\\ 0 & 0 \end{array} \right)=i \left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array} \right)$$

 

[csnsc14320]

I think that you can combine them with complex scalar multiplication.

e.g.

$$\left( \begin{array}{cc} i & 0\\ 0 & 0 \end{array} \right)=i \left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array} \right)$$

but then that wouldn't span the space?

if i wanted the matrix $$\left( \begin{array}{cc} 1 & 3-2i\\ 3-2i & 16-i \end{array} \right)$$

wouldnt i need:
$$\left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array} \right) + 3\left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right) - 2\left( \begin{array}{cc} 0 & i\\ i & 0 \end{array} \right) + 16\left( \begin{array}{cc} 0 & 0\\ 0 & 1 \end{array} \right) -\left( \begin{array}{cc} 0 & 0\\ 0 & i \end{array} \right)$$

I don't see a way of obtaining this matrix without having a 6 dimensional basis where the 1's and i's are separated?

 

[csnsc14320]

but then that wouldn't span the space?

if i wanted the matrix $$\left( \begin{array}{cc} 1 & 3-2i\\ 3-2i & 16-i \end{array} \right)$$

wouldnt i need:
$$\left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array} \right) + 3\left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right) - 2\left( \begin{array}{cc} 0 & i\\ i & 0 \end{array} \right) + 16\left( \begin{array}{cc} 0 & 0\\ 0 & 1 \end{array} \right) -\left( \begin{array}{cc} 0 & 0\\ 0 & i \end{array} \right)$$

I don't see a way of obtaining this matrix without having a 6 dimensional basis where the 1's and i's are separated?

wait a minute... multiplying by "i" is allowed isn't it? in which case I would have:

$$i^3 \left( \begin{array}{cc} i & 0\\ 0 & 0 \end{array} \right) + 3i^3\left( \begin{array}{cc} 0 & i\\ i & 0 \end{array} \right) - 2\left( \begin{array}{cc} 0 & i\\ i & 0 \end{array} \right) + 16i^3\left( \begin{array}{cc} 0 & 0\\ 0 & i \end{array} \right) -\left( \begin{array}{cc} 0 & 0\\ 0 & i \end{array} \right)$$

in which case, my dimension would be 3, or just the three matrices with i's since i can obtain the 1's by a multiplication of i^3?

i think that's w hat you were getting at aPhiliosopher - i don't think i understood what you meant the first time i read it

 

[aPhilosopher]

$$\left( \begin{array}{cc} 1 & 3-2i\\ 3-2i & 16-i \end{array} \right) = \left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array} \right) + (3 - 2i) \left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right) + (16 - i) \left( \begin{array}{cc} 0 & 0\\ 0 & 1 \end{array} \right)$$

 

[csnsc14320]

$$\left( \begin{array}{cc} 1 & 3-2i\\ 3-2i & 16-i \end{array} \right) = \left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array} \right) + (3 - 2i) \left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right) + (16 - i) \left( \begin{array}{cc} 0 & 0\\ 0 & 1 \end{array} \right)$$

i make things too difficult :p

so just the 1's would work then. i forgot about multiplying by i.

thanks!

 

[Phrak]

What is the basis of the vector space over the field of reals?

 

[csnsc14320]

What is the basis of the vector space over the field of reals?

$$\left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array} \right), \left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right), \left( \begin{array}{cc} 0 & 0\\ 0 & 1 \end{array} \right)$$

which should also be the basis over the complex too since i can just multiply by multiples of i as well

 

[aPhilosopher]

Nooooo, that's not a basis over the reals.

You had it right before I got here if you want it over the reals.

I shouldn't have assumed it was over the complex field.

 

[Phrak]

Nooooo, that's not a basis over the reals.

You had it right before I got here if you want it over the reals.

Thanks, aPhilosoper. I wasn't really sure.

 

[csnsc14320]

Nooooo, that's not a basis over the reals.

You had it right before I got here if you want it over the reals.

I shouldn't have assumed it was over the complex field.

i want it over the complex?

 

[aPhilosopher]

Well, you're going to have to tell me. Technically, this is an incompletely specified problem. Is this vector space over the real or complex numbers? I would assume complex but don't really have a clue because it could go either way. At any rate, that question gives you your answer to what scalars are allowed in linear combinations. If you can't use complex numbers, then you have to have the three additional basis vectors that are just i times the first three.Thanks for pointing this out Phrack.

 

[csnsc14320]

Well, you're going to have to tell me. Technically, this is an incompletely specified problem. Is this vector space over the real or complex numbers? I would assume complex but don't really have a clue because it could go either way. At any rate, that question gives you your answer to what scalars are allowed in linear combinations. If you can't use complex numbers, then you have to have the three additional basis vectors that are just i times the first three.


Thanks for pointing this out Phrack.

its a vector space over the complex :D

and can have complex entries