$\mathbb{C}$ is a one-dimensional complex vector space. What is its dimension when regarded as a vector space over $\mathbb{R}$?
I don't understand how $\mathbb{C}$ is one-dimensional. Please help me understand that.
Also, I'm pretty sure that when the field is reals we have $\dim(\mathbb{C})=2$. Since when $\alpha$ is real and $z=a+bi$ is complex we have $\alpha z=\alpha a+\alpha bi=\alpha a(1,0)+\alpha b(0,i)$.
How does this look? Any solutions or help is greatly appreciated.
$\endgroup$ 93 Answers
$\begingroup$$\mathbb C$ as an $\mathbb C$-vector space is one dimensional. Any field as a vector space over itself is one dimensional. It has a basis of $\{ 1 \}$.
To see this, consider an element $f \in \mathbb F$ in a field $\mathbb F$. Write $\vec f$ when we are considering $f$ as a member of the vector space. Then as we can write $\vec f = f \vec 1$ for all $\vec f \in \mathbb F$, the multiplicative identity $\vec 1$ is in itself a perfectly good basis for $\mathbb F$.
Using that notation for $\mathbb C$ as an $\mathbb R$-vector space, we can write any vector $\vec z \in \mathbb C$ is
$$\vec z = a\vec 1 + b \vec i \quad \text{ with } a, b \in \mathbb R$$
Hence $\{ \vec 1, \vec i \}$ is a basis.
As for why the dimension of $\mathbb F^n$ is $n$, as you ask in the comments: that's because there is a basis $\{ (1,0,\cdots, 0), (0, 1, \cdots, 0), \cdots, (0,0 , \cdots, 0, 1) \}$ with $n$ elements. Hence by definition of dimension
$$\dim(\mathbb F^n) = n$$
$\endgroup$ 2 $\begingroup$To answer your question of why the dimension of $\Bbb C$ as a vector space over itself is $1$:
Do you agree that the dimension of a vector space is the same as the number of basis elements? So to find the dimension of a vector space over a scalar field, you just have to find a basis for the vector space and then the dimension will be the number of elements in the basis.
Well, if $S$ is a vector space and $F$ is our scalar field, then a basis of $S$ (over $F$) is a set of elements $\{s_{1}, s_{2}, \dots, s_{n} \} \subseteq S$ such that the set is linearly independent, and for each $s \in S$, we can find scalars $f_{1}, f_{2}, \dots, f_{n} \in F$ such that $s = \sum \limits_{i = 1}^{n} f_{i}s_{i}$.
Now that we have that out of the way, replace the above stuff with $S = \Bbb C$ and $F = \Bbb C$. The dimension of $S$ over $F$ is a collection of elements in $S$ such that for each $s \in S$, we can write it as a linear combination of the collection of elements, and the collection of elements is linearly independent.
Here is my candidate for a basis of $\Bbb C$ over $\Bbb C$. It is $\{ 1 \}$. Don't believe me that this is a basis? Check the two conditions:
1) Is it a linearly independent set? Well, for which scalars $c \in \Bbb C$ do we have $c \cdot 1 = 0$? Only $c = 0$, so yes, it is a linearly independent set.
2) Does it span $\Bbb C$ (over the field of scalars $\Bbb C$)? Well, let $\alpha \in C$. Can we find a scalar $c \in \Bbb C$ such that $\alpha = c \cdot 1$? Yes, just let $c = \alpha$. Obviously, $\alpha \cdot 1 = \alpha$. So $\{1 \}$ spans $\Bbb C$ as a vector space over itself.
Since $\{ 1 \}$ satisfies the criteria to be a basis of $\Bbb C$ over itself, and since the set only has one element, then $\Bbb C$ has dimension $1$ as a vector space over itself.
$\endgroup$ 1 $\begingroup$The complex numbers as a vector space over the field of real numbers is of dimension $2$. The two vectors $1$ and $i$ form a basis and any complex vector $a+ib$ is a linear combination of the two vectors $1$ and $i$, multiplied by real scalars $a$ and $b$ and added.
The complex numbers as a vector space over the field of complex numbers is of dimension $1$. Chose any vector in this space. You can multiply it with a scalar $a+bi$ (scalar because the field is the field of complex numbers) to get the arbitrary vector $a+bi$ (now $a+bi$ is any vector in the complex vector space, got by multiplying a unit vector $1$ with the scalar $a+bi$).
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