MHB No prefix added: How do I find a basis of a $\mathbb{R}$-vector space?

[mathmari]

Hey!

I want to prove that $$V=\left \{\begin{pmatrix}a & b\\ c & d\end{pmatrix} \mid a,b,c,d\in \mathbb{C} \text{ and } a+d\in \mathbb{R}\right \}$$ is a $\mathbb{R}$-vector space.
I want to find also a basis of $V$ as a $\mathbb{R}$-vector space. We have the following:
Let $K$ be a field. A vector space over $K$ (or $K$-vector space) is a set $V$ with an addition $V \times V \rightarrow V : (x, y) \mapsto x + y$ and a scalar multiplication $K \times V \rightarrow V : (\lambda , x) \mapsto \lambda \cdot x$, so that the following holds:

• (V1) : $(V,+)$ is an abelian group, with the neutral element $0$.
• (V2) : $\forall a, b \in K, \forall x \in V : (a + b) \cdot x = a \cdot x + b \cdot x$
• (V3) : $\forall a \in K, \forall x, y \in V : a \cdot (x + y) = a \cdot x + a \cdot y$
• (V4) : $\forall a, b \in K, \forall x \in V : (ab) \cdot x = a \cdot (b \cdot x)$
• (V5) : $\forall x \in V : 1 \cdot x = x$ ( $1 = 1_K$ is the identity in $K$).

We have that it is closed under addition and multiplication, right? (Wondering)

The properties 2-5 are also satisfied, or not? How can we check the property 1? (Wondering) Could you give me a hint how to find a basis? (Wondering)

 

[HallsofIvy]

\begin{pmatrix}a & b \\ c & d \end{pmatrix}= \begin{pmatrix}a & 0 \\ 0 & 0 \end{pmatrix}+ \begin{pmatrix}0 & b \\ 0 & 0 \end{pmatrix}+ \begin{pmatrix}0 & 0 \\ c & 0 \end{pmatrix}+ \begin{pmatrix}0 & 0 \\ 0 & d \end{pmatrix}
= a\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}+ b\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}+ c\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}+ d\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}

 

[mathmari]

\begin{pmatrix}a & b \\ c & d \end{pmatrix}= \begin{pmatrix}a & 0 \\ 0 & 0 \end{pmatrix}+ \begin{pmatrix}0 & b \\ 0 & 0 \end{pmatrix}+ \begin{pmatrix}0 & 0 \\ c & 0 \end{pmatrix}+ \begin{pmatrix}0 & 0 \\ 0 & d \end{pmatrix}
= a\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}+ b\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}+ c\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}+ d\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}

Ah! And so, the basis is $$\{\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}\}$$ right? (Wondering)

 

[I like Serena]

I think we need a couple more elements in the basis, since the entries in the matrix are complex, but the scalars we multiply by are real.
Furthermore, we have the restriction that the imaginary part of the trace must be zero.
I think we need 7 elements in the basis. (Thinking)

We have that it is closed under addition and multiplication, right? (Wondering)

The properties 2-5 are also satisfied, or not? How can we check the property 1? (Wondering)

Yes.
For property 1, we already know that regular matrix addition of complex matrices is an abelian group.
So it's sufficient to verify it's a subgroup, meaning we have to verify if the neutral element is an element of the group, and if all additive inverses are elements of the group. (Thinking)

 

[mathmari]

I think we need a couple more elements in the basis, since the entries in the matrix are complex, but the scalars we multiply by are real.
Furthermore, we have the restriction that the imaginary part of the trace must be zero.
I think we need 7 elements in the basis. (Thinking)

Ah ok... (Thinking)

Is the basis the followig?
$$\left \{\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, \begin{pmatrix} 0 & i \\ 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 \\ i & 0 \end{pmatrix} \right \}$$

(Wondering)

 

[I like Serena]

Yes (Mmm)

 

[mathmari]

For property 1, we already know that regular matrix addition of complex matrices is an abelian group.
So it's sufficient to verify it's a subgroup, meaning we have to verify if the neutral element is an element of the group, and if all additive inverses are elements of the group. (Thinking)

We have that $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\in V$ since $0\in\mathbb{R}\subset \mathbb{C}$.

When $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\in V$ then $\begin{pmatrix} -a & -b \\ -c & -d \end{pmatrix}\in V$, since $-a,-b,-c,-d\in \mathbb{C}$ and $-a-d=-(a+d)\in \mathbb{R}$, since $a+d\in \mathbb{R}$.

Is this correct? (Wondering)

 

[I like Serena]

Yup. (Mmm)

 

[mathmari]

We have the following:

\begin{align*}\begin{pmatrix}a & b \\ c & d \end{pmatrix}& = \begin{pmatrix}a_1+a_2i & b_1+b_2i \\ c_1+c_2i & d_1+d_2i \end{pmatrix} = \begin{pmatrix}a_1 & b_1 \\ c_1 & d_1 \end{pmatrix}+\begin{pmatrix}a_2i & b_2i \\ c_2i & d_2i \end{pmatrix} \\ &=
\left (\begin{pmatrix}a_1 & 0 \\ 0 & 0 \end{pmatrix}+ \begin{pmatrix}0 & b_1 \\ 0 & 0 \end{pmatrix}+ \begin{pmatrix}0 & 0 \\ c_1 & 0 \end{pmatrix}+ \begin{pmatrix}0 & 0 \\ 0 & d_1 \end{pmatrix}\right )+ \left (\begin{pmatrix}a_2i & 0 \\ 0 & d_2i \end{pmatrix}+ \begin{pmatrix}0 & b_2i \\ 0 & 0 \end{pmatrix}+ \begin{pmatrix}0 & 0 \\ c_2i & 0 \end{pmatrix} \right ) \\ &=
\left (a_1\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}+ b_1\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}+ c_1\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}+ d_1\begin{pmatrix}0 & 0 \\ 0 &1 \end{pmatrix}\right )+ \left (\begin{pmatrix}a_2i & 0 \\ 0 & d_2i \end{pmatrix}+ b_2\begin{pmatrix}0 & i \\ 0 & 0 \end{pmatrix}+ c_2\begin{pmatrix}0 & 0 \\ i & 0 \end{pmatrix} \right )\end{align*}

How do we get from $\begin{pmatrix}a_2i & 0 \\ 0 & d_2i \end{pmatrix}$ the $\begin{pmatrix}i & 0 \\ 0 & -i \end{pmatrix}$ ? (Wondering)

 

[I like Serena]

The trace of the matrix must be real, so $\text{Im}(a+b) = 0 \quad\Rightarrow\quad d_2 = -a_2$. (Thinking)

 

[mathmari]

The trace of the matrix must be real, so $\text{Im}(a+b) = 0 \quad\Rightarrow\quad d_2 = -a_2$. (Thinking)

Ah ok... I see! To show that $V$ is a real vector space, could we maybe show that it is a subvector space of the real vector space of the complex matrices? (Wondering)

We have that $I_2=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}$ is an element of $V$ since $1,0\in \mathbb{C}$ and $1+1=2\in \mathbb{R}$.

So, the set is non-empty. We have the followig:
$$\begin{pmatrix}a_1 & b_1\\ c_1 & d_1\end{pmatrix}+\begin{pmatrix}a_2 & b_2\\ c_2 & d_2\end{pmatrix}=\begin{pmatrix}a_1+a_2 & b_1+b_2\\ c_1+c_2 & d_1+d_2\end{pmatrix}\in V$$
Since $(a_1+a_2)+(d_1+d_2)=(a_1+d_1)+(a_2+d_2)\in \mathbb{R}$, since $(a_1+d_1)\in \mathbb{R}$ and $(a_2+d_2)\in \mathbb{R}$.

We have also that $$r\begin{pmatrix}a & b\\ c & d\end{pmatrix}=\begin{pmatrix}ra & rb\\ rc & rd\end{pmatrix}\in V$$
since $ra+rd=r(a+d)\in \mathbb{R}$, since $a+d\in \mathbb{R}$ and $r\in \mathbb{R}$. Or is it not known that the set of complex matrices is a real vector space? (Wondering)

 

[I like Serena]

Ah ok... I see! To show that $V$ is a real vector space, could we maybe show that it is a subvector space of the real vector space of the complex matrices? (Wondering)

What's a "real" vector space? (Wondering)

The real vector space is the space of the real matrices with real scalar multiplication.
The complex vector space is the space of the complex matrices with complex scalar multiplication, which is indeed a known vector space.
So yes, it suffices if we can show that $V$ is a sub vector space of the complex vector space.

We have that $I_2=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}$ is an element of $V$ since $1,0\in \mathbb{C}$ and $1+1=2\in \mathbb{R}$.

So, the set is non-empty. We have the followig:
$$\begin{pmatrix}a_1 & b_1\\ c_1 & d_1\end{pmatrix}+\begin{pmatrix}a_2 & b_2\\ c_2 & d_2\end{pmatrix}=\begin{pmatrix}a_1+a_2 & b_1+b_2\\ c_1+c_2 & d_1+d_2\end{pmatrix}\in V$$
Since $(a_1+a_2)+(d_1+d_2)=(a_1+d_1)+(a_2+d_2)\in \mathbb{R}$, since $(a_1+d_1)\in \mathbb{R}$ and $(a_2+d_2)\in \mathbb{R}$.

We have also that $$r\begin{pmatrix}a & b\\ c & d\end{pmatrix}=\begin{pmatrix}ra & rb\\ rc & rd\end{pmatrix}\in V$$
since $ra+rd=r(a+d)\in \mathbb{R}$, since $a+d\in \mathbb{R}$ and $r\in \mathbb{R}$.

The identity matrix is not necessarily part of our vector space, since we're not multiplying matrices.

The zero matrix does have to be an element (implying it's not empty), just like every additive inverse. (Thinking)

And you have already shown closure for addition and closure for scalar multiplication.

 

[mathmari]

The identity matrix is not necessarily part of our vector space, since we're not multiplying matrices.

The zero matrix does have to be an element (implying it's not empty), just like every additive inverse. (Thinking)

And you have already shown closure for addition and closure for scalar multiplication.

Do we know that the zero matrix exists because for each $u,v\in V$ it must hold that $u+v\in V$, and also for $u=-v$ ? (Wondering)

 

[I like Serena]

Do we know that the zero matrix exists because for each $u,v\in V$ it must hold that $u+v\in V$, and also for $u=-v$ ? (Wondering)

That's one way yes.
With the proof that all additive inverses are in $V$, it follows that the zero matrix is in there as well.
Alternatively we can also tell because its trace, which is $0+0=0$, is a real number. (Thinking)

 

[mathmari]

That's one way yes.
With the proof that all additive inverses are in $V$, it follows that the zero matrix is in there as well.
Alternatively we can also tell because its trace, which is $0+0=0$, is a real number. (Thinking)

I see... Thank you very much! (Mmm)