Not a subspace of the complex numbers?

[diogomgf]

Why isn't the set [MATH]F = \left \{ \begin{bmatrix}a & b\\ -\bar{b} & \bar{a}\end{bmatrix}: \space a,b \in \mathbb{C} \right \}[/MATH] a subspace of [MATH]M_{2\times 2}(\mathbb{C})[/MATH], if [MATH]M_{2\times 2}(\mathbb{C})[/MATH] is a vector space over [MATH]\mathbb{C}[/MATH]?

The matrix has weird formatting. The element with index 2,1 is [MATH]- \bar{b}.[/MATH]

 

[Steven G]

Can you show, using the definition of a vector space, that F is a subspace of M_2x2(C)?

We will go from there.

 

[Steven G]

BTW, on my screen I see three bars. One of which is not over anything. Can you please state the 4 elements in a clearer fashion. Thanks.

 

[diogomgf]

Can you show, using the definition of a vector space, that F is a subspace of M_2x2(C)?

We will go from there.

Meanwhile I think I understood why it is not a subspace.
If it was a subspace, we could have [MATH]\forall \alpha \in \mathbb{C}; \space \alpha \boxdot F_{1} \in F[/MATH] Where [MATH]F_{1} \in F[/MATH].
[MATH]a=\alpha+\beta i; \space \bar{a}=\alpha - \beta i[/MATH] .
So [MATH] i \cdot a=-\beta + \alpha i; \space i\cdot \bar{a}=\beta + \alpha i[/MATH] .
This shows that if we multiply any matrix in [MATH]F[/MATH] by [MATH]i[/MATH] we obtain a matrix that is not in [MATH]F[/MATH].

 

[diogomgf]

BTW, on my screen I see three bars. One of which is not over anything. Can you please state the 4 elements in a clearer fashion. Thanks.

The matrix has weird formatting. The element with index 2,1 is [MATH]- \bar{b}.[/MATH]

 

[Steven G]

bar = '
I thought that if A = r + si, then A' = r-si. You have the equivalent of A' = -r+si in your work.

 

[diogomgf]

bar = '
I thought that if A = r + si, then A' = r-si. You have the equivalent of A' = -r+si in your work.

Typo, fixed.

 

[Steven G]

a=α+βi; ¯a=α−βi . Perfect
So i⋅a=−β+αi; i⋅¯a=β+αi . 2nd equation is wrong (same mistake as before.

 

[diogomgf]

a=α+βi; ¯a=α−βi . Perfect
So i⋅a=−β+αi; i⋅¯a=β+αi . 2nd equation is wrong (same mistake as before.

[MATH]ia = i(\alpha + \beta i) = \alpha i + \beta i^{2} = \alpha i - \beta[/MATH] i think its the same...

 

[Steven G]

[MATH]ia = i(\alpha + \beta i) = \alpha i + \beta i^{2} = \alpha i - \beta[/MATH] i think its the same...

Yes, it is correct. The problem is that is the 1st equation you wrote on that line. I suggested that the 2nd equation is not correct.

 

[diogomgf]

Yes, it is correct. The problem is that is the 1st equation you wrote on that line. I suggested that the 2nd equation is not correct.

[MATH]i \bar{a} = i(\alpha - \beta i) = \alpha i - \beta i^{2} = \alpha i + \beta[/MATH] its the same as I have...

 

[Steven G]

[MATH]i \bar{a} = i(\alpha - \beta i) = \alpha i - \beta i^{2} = \alpha i + \beta[/MATH] its the same as I have...

My eyes are deceiving me tonight. You are correct. Sorry about the time spent on this.

 

[diogomgf]

My eyes are deceiving me tonight. You are correct. Sorry about the time spent on this.

No problem, I've made you spend more time than the other way around

 

[pka]

I have stayed away form this thread because I think all who have given replies are not fully acquainted with the notation of complex numbers.
The complex numbers form a vector space over the field of ordered paired of real numbers.
At the most fundamental the complex numbers is simply an enlargement of the real numbers by one element \({\bf{\mathcal{i}}}~\).
The new number \({\bf{\mathcal{i}}}~\) is the solution of the equation \(x^2+1=0\) Well of course so is \({\bf{\mathcal{-i}}}~\).
We are accustomed to thinking that saying \(z=a+b{\bf{\mathcal{i}}}~\) implies that \(a~\&~b\) are real numbers.
Now here is a standard exercise in any undergraduate course in complex analysis. PROVE
1) \(\overline{z+w}=\overline{z}+\overline{w}\)
2) \(\overline{z\cdot w}=\overline{z}\cdot\overline{w}\)
3)\( \overline{\left(\dfrac{z}{w}\right)}=\dfrac{\overline{z}}{\overline{w}}\)

Now the posted question does not appear to conform to any standard format.
Please correct me if otherwise.

 

[Dr.Peterson]

Now the posted question does not appear to conform to any standard format.
Please correct me if otherwise.

I think you may be missing the nature of the problem because the title is wrong.

The title says, "Not a subspace of the complex numbers?"

The problem should be described as, "Not a subspace of the space of 2x2 matrices over the complex numbers?"

I think post #4 gives a good answer: multiplying a matrix of that form by i yields a matrix not of that form.

The question is not about the complex numbers as a vector space over the real numbers, but about the vector space of 2x2 matrices over the complex numbers.

 

[diogomgf]

I think you may be missing the nature of the problem because the title is wrong.

The title says, "Not a subspace of the complex numbers?"

The problem should be described as, "Not a subspace of the space of 2x2 matrices over the complex numbers?"

I think post #4 gives a good answer: multiplying a matrix of that form by i yields a matrix not of that form.

The question is not about the complex numbers as a vector space over the real numbers, but about the vector space of 2x2 matrices over the complex numbers.

You are right...