I'm working on Linear Algebra homework. I'm having trouble with: $\mathbb{C}\times\mathbb{C}$ is a real vector space. Explain why. Write down a basis for this real vector space.

I'm just confused on what exactly $\mathbb{C}\times\mathbb{C}$ is and why it is a real vector space. It seems it can be defined for vector addition and scalar multiplication. I've seen $+\colon\mathbb{C}\times\mathbb{C}$ and $*\colon\mathbb{C}\times\mathbb{C}$ and I'm confused because I thought it was a cross product?

I know once I figure out how to define $\mathbb{C}\times\mathbb{C}$, I have to show that the vector space properties apply.

Thanks in advance for any help.

  • $\begingroup$ I realize that CxC is the cross of two complex numbers. I'm just not sure how to work out the "crossing" $\endgroup$ – mathmom Feb 19 '15 at 17:47
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    $\begingroup$ $C\times C$ is the set of all ordered pairs $(z,w)$ where $z$ and $w$ are both in $C$. $\endgroup$ – WillO Feb 19 '15 at 17:47
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    $\begingroup$ Hint: what's a basis for $\mathbb C$ as a vector space over $\mathbb R$? $\endgroup$ – Simon S Feb 19 '15 at 17:50
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    $\begingroup$ @mathmom I noticed you wrote $[(a+ib),(c+id)]$ to denote an ordered pair. This is incorrect. I'm not sure why they teach that parentheses, brackets, and braces should be alternated in grade school, but it's wrong. The correct way to write it is $((a+ib),(c+id))$ or simply $(a+ib,c+id)$. $\endgroup$ – Matt Samuel Feb 19 '15 at 18:45
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    $\begingroup$ old habits die hard - thank you. Nomenclature is important. $\endgroup$ – mathmom Feb 19 '15 at 18:51

A vector space is defined as a quadruple $(\mathbf{V},\mathbb{K},\oplus,\odot)$ where $\mathbf{V}$ is a set of elements called vectors, $\mathbb{K}$ is a field $(\mathbb{K},+,\cdot)$ , $\oplus$ is a binary operation (called sum) on $\mathbf{V}$ such that $(\mathbf{V},\oplus)$ is a commutative group and $a\odot\mathbf{v}:\mathbb{K}\times\mathbf{V} \rightarrow \mathbf{V}$ is a scalar multiplication such that, $\forall a,b \in \mathbb{K}$ and $\forall \mathbf{u,v} \in \mathbf{V}$ we have: $$ a\odot(b\star\mathbf{v})=(a\cdot b)\odot\mathbf{v} $$ $$ 1\odot\mathbf{v}=\mathbf{v} $$ $$ a \odot (\mathbf{u}\oplus\mathbf{v})=a \odot\mathbf{u}\oplus a\odot \mathbf{v} $$ $$ (a+b)\odot \mathbf{v}=a\odot \mathbf{v}\oplus b\odot \mathbf{v} $$ Note that $(+,\cdot)$ are the operations on $\mathbb{K}$ and are different from the operations $(\oplus, \odot)$.

In your case $\mathbf{V}= \mathbb{C}\times \mathbb{C}$ i.e the vectors are couple $(\alpha_1,\alpha_2)=\mathbf{a}$ with $\alpha_1,\alpha_2 \in \mathbb{C}$.

Usually we define the operations $\oplus$ as: $$ \mathbf{a}\oplus\mathbf{b}=(\alpha_1,\alpha_2)\oplus(\beta_1,\beta_2)=(\alpha_1+\beta_1,\alpha_2+\beta_2) $$ Where the $+$ operation is the usual sum in $\mathbb{C}$ and, since $(\mathbb{C},+)$ is a commutative group, we can easily see that $(\mathbb{C}\times\mathbb{C},\oplus)$ is a commutative group, with neutral element $(0,0)$ and opposite of $(\alpha_1,\alpha_2)$ the element $(-\alpha_1,-\alpha_2)$.

Now we want $\mathbb{K}=\mathbb{R}$ and we define the scalar multiplication as: $$ r\odot \mathbf{a}=r\odot(\alpha_1,\alpha_2)=(r\cdot\alpha_1,r\cdot \alpha_2) $$ Where $ r\cdot \alpha_i$ is the product in $\mathbb{C}$ and is well defined also for $r\in \mathbb{R}$ and $\alpha_i \in \mathbb{C}$ since $\mathbb{R}$ is a subfield of $\mathbb{C}$.

Now, using the properties of $(\mathbb{C},+,\cdot)$ as a field it is not difficult to see that all the axioms for a scalar multiplication are verified.

As an example we have: $$ r\odot(\mathbf{a}\oplus\mathbf{b})=r\odot\left( (\alpha_1,\alpha_2)\oplus(\beta_1,\beta_2)\right)= r\odot\left( (\alpha_1+\beta_1),(\alpha_2+\beta_2)\right)= $$ $$ = \left( r\cdot(\alpha_1+\beta_1),r\cdot(\alpha_2+\beta_2)\right) = \left( r\cdot\alpha_1+r\cdot\beta_1,r\cdot\alpha_2+r\cdot\beta_2 \right)= $$ $$ =r\odot(\alpha_1,\alpha_2)\oplus r\odot(\beta_1,\beta_2)= r\odot\mathbf{a}\oplus r\odot\mathbf{b} $$

  • $\begingroup$ Sorry may talking italian leads to this confusion... I edit $\endgroup$ – Emilio Novati Feb 19 '15 at 21:21
  • $\begingroup$ My english is primitive... scalar multiplication is better? or is used another name? $\endgroup$ – Emilio Novati Feb 19 '15 at 21:25
  • $\begingroup$ Thank you very much - this clarifies my confusion! $\endgroup$ – mathmom Feb 20 '15 at 18:22
  • $\begingroup$ Welcome! Don't forget to accept if you like :-) $\endgroup$ – Emilio Novati Feb 20 '15 at 19:30

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