Given that $V = \mathbb{C}^{3}$ how does one determine if a set of vectors $A \subset V$ is a basis? Am I allowed to assume that the the field is now $\mathbb{C}$ and so I am allowed to use all $\lambda \in \mathbb{C}$ as scalars? If so, $E = \{ (1,0,0), (0,1,0), (0,0,1)\}$ is still the usual basis for $\mathbb{C}^{3}$, right?

Basically my approach right now would be to immediately rule out any sets of vectors that were not exactly $3$ elements. Then I would put the three vectors in row echelon form and if all rows were non-zero I would conclude that the given set were a basis (exactly as I would if $V= \mathbb{R}^{3}$), is this OK?

And then in general, I am now thinking that any basis of $\mathbb{R}^{n}$ would also be a basis for $\mathbb{C}^{n}$, but not the other way around, is that right?

Assuming that I am not already way off at this point, is there a big difference between $\mathbb{C}^{n}$ and $\mathbb{R}^{n}$ when it comes to bases?

  • 1
    $\begingroup$ Everything you say is correct. $\endgroup$ – Phira Apr 23 '11 at 20:13

When you talk about a vector space, you are implicitly saying over what field. When one mentions $\mathbb{C}^3$ (or more generally, $\mathbf{F}^n$ for any field $\mathbb{F})$, if nothing else is said then the standard convention is that you are considering it as a vector space over $\mathbb{C}$ (or in the more general case, over $\mathbf{F}$). In that situation, yes, the "standard basis" which consists of the vectors $\mathbf{e}_i$ that have a $1$ in the $i$th coordinate and $0$s elsewhere form a basis.

The notion of dimension works in any vector space over any field, so it is still true that every basis for $\mathbb{C}^3$ (as a vector space over $\mathbb{C}$) must have $3$ elements, etc.

You need to be a bit careful when you say that "any basis of $\mathbb{R}^n$ would also be a basis for $\mathbb{C}^n$ but not the other way around" because the sets of vectors are not exactly the same.

However, that said, yes: if $\mathbf{v}_1,\ldots,\mathbf{v}_n$ are $n$ vectors in $\mathbb{R}^n$ that are linearly independent over $\mathbb{R}$, then they are also linearly independent over $\mathbb{C}$: to see this, take $\lambda_j\in\mathbb{C}$, and write it as $\lambda_j = \alpha_j + i\beta_j$ with $\alpha_j,\beta_j\in\mathbb{C}$. Then if $$\lambda_1\mathbf{v}_1 + \cdots + \lambda_n\mathbf{v}_n = \left(\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n\right) + i\left(\beta_1\mathbf{v}_1+\cdots+\beta_n\mathbf{v}_n\right) = \mathbf{0},$$ it follows that all $\alpha_j$ have to be zero and all $\beta_j$ have to be zero by the linear independence over $\mathbb{R}$. In particular, the vectors, when considered as elements of $\mathbb{C}^n$, are a basis.

Conversely, if $\mathbf{v}_1,\ldots,\mathbf{v}_n$ are vectors in $\mathbb{C}^n$, then if they are linearly independent over $\mathbb{C}$ (allowing complex scalars), then they are also linearly independent over $\mathbb{R}$. So if they are in fact contained in $\mathbb{R}^n$ (entries in $\mathbb{R}$), and they were linearly independent over $\mathbb{C}$, they are also linearly independent over $\mathbb{R}$. So a set of vectors of $\mathbb{R}^n$ that are basis for $\mathbb{C}^n$ are also a basis for $\mathbb{R}^n$.

The "big difference" between $\mathbb{C}^n$ and $\mathbb{R}^n$ is that vectors in bases for $\mathbb{C}^n$ are allowed to have complex entries. This becomes important when you consider linear transformations (and especially linear operators). If you restrict yourself to bases in which all vectors have all coordinates real, you may not be able to find vectors for which linear transformations are "nice"; but once you allow yourself complex entries, a lot of things that are impossible to do in general over $\mathbb{R}$ are always doable over $\mathbb{C}$. (If you already know about linear transformations, it may be impossible to diagonalize over $\mathbb{R}$ and yet be able to do it over $\mathbb{C}$; and Jordan Canonical Forms always exist over $\mathbb{C}$, but not always over $\mathbb{R}$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.