# Finding the dimension of a vector space over different fields?

We were given this question for my linear algebra module:

We view $\Bbb C ^2$ as a vector space over $\Bbb C$,$\Bbb R$ and $\Bbb Q$. Let $$\mathbf x_1 := \begin{pmatrix} i \\ 0 \end{pmatrix}, \mathbf x_2 := \begin{pmatrix} \sqrt 2 \\ \sqrt 5 \end{pmatrix}, \mathbf x_3 := \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \mathbf x_4 := \begin{pmatrix} i\sqrt 3 \\ \sqrt 3 \end{pmatrix}, \mathbf x_5 := \begin{pmatrix} 1 \\ 3 \end{pmatrix} \in \Bbb C ^2$$

Find dim$_F$(Span$_F$($\mathbf x_1,\mathbf x_2,\mathbf x_3,\mathbf x_4,\mathbf x_5$)) for F=$\Bbb C$,$\Bbb R$ and $\Bbb Q$.

So I have found that for F=$\Bbb C$, the dimension is 2, but I'm struggling with the other fields. Whats been confusing me is say I want to find a basis for Span$_F$($\mathbf x_1,\mathbf x_2,\mathbf x_3,\mathbf x_4,\mathbf x_5$) over $\Bbb R$. If I want to apply Gaussian elimination to obtain the minimal spanning set, which scalars am I allowed to used say when scaling the rows when applying row ops. Do the scalars always have to the field elements, say in this case just $\Bbb R$?

Sorry if this seems trivial but couldn't really find any good sources online and lecturer didn't cover it in lectures.

Thanks a lot!

• The scalars are from the field you are over. – Yunus Syed Mar 28 '17 at 22:43

Hint:

For the dimension over $\mathbf R$, consider these vectors in $\mathbf C^2$ as vectors in $\mathbf R^4$: $$\mathbf x_1=\begin{pmatrix}0\\1\\0\\0 \end{pmatrix},\enspace\mathbf x_2=\begin{pmatrix} \sqrt2\\0\\\sqrt5\\0\end{pmatrix},\enspace\mathbf x_3=\begin{pmatrix}0\\0\\1\\0 \end{pmatrix}, \enspace\mathbf x_4=\begin{pmatrix} 0\\\sqrt3\\\sqrt3\\0\end{pmatrix},\enspace\mathbf x_5=\begin{pmatrix}1\\0\\3\\0 \end{pmatrix}$$

Can there be non-trivial relations between these vectors with rationalcoefficients?

• Hello, we haven't covered anything as such in our lectures, how would that help me in this case? Thanks! – The Mysterious Pigeon Mar 28 '17 at 23:46
• You don't have to know much on field extensions – just the degree of $F/K=[F:K]$, which is the dimension of $F$ as a vector space over $K$ (if $K=\mathbf K$, $F=\mathbf C$, it's $2$). If $\;(\omega_1,\dots,\omega_n)$ is a basis of $F$ over $K$, prove that a basis of $V$ over $K$ is just a basis of $V$ over $F$, multiplied by $\omega_1,\dots,\omega_n$. – Bernard Mar 28 '17 at 23:58
• In the OP's situation, the set $V$ depends on the field $F$ (i.e., is the "$F$-span" of the set of vectors). Consequently, all the OP's dimensions are finite. :) – Andrew D. Hwang Mar 29 '17 at 0:22
• @Andrew D. Hwang: Al dimensions oren't finite on $\mathbf Q$… Explicitly, he/she must extract a basis from the given space of vectors, then multiply by a basis of $\mathbf C$ over $\mathbf R$, then a Hamel basis of $\mathbf R$ over $\mathbf Q$ (albeit the word ‘explicit’ is a little too much here…) – Bernard Mar 29 '17 at 0:30
• Oh! I see. I misunderstood the question. I've changed my answer. – Bernard Mar 29 '17 at 1:31


There's a recursive algorithm for extracting a basis from a finite ordered spanning set $(\Vx_{i})_{i=1}^{n}$ over an arbitrary field $K$, which may be easier in your situation than Gaussian elimination:

Start with $T_{0} = \varnothing$. For each $i = 1, \dots, n$, inductively define $$T_{i} = \begin{cases} T_{i-1} & \Vx_{i} \in \Span_{K}(T_{i-1}), \\ T_{i-1} \cup \{\Vx_{i}\} & \Vx_{i} \not\in \Span_{K}(T_{i-1}). \end{cases}$$ It's straightforward to check that the final set $T_{n}$ is a basis for $\Span_{K}(\Vx_{i})_{i=1}^{n}$.

In your case, $T_{3} = (\Vx_{i})_{i=1}^{3}$ is linearly independent over $\Reals$, hence over $\Ratls$.

Since $\Vx_{4} = \sqrt{3}(\Vx_{1} + \Vx_{3})$ and $\Vx_{5} \in \Span_{\Reals}(\Vx_{2}, \Vx_{3})$, the set $T_{3}$ is a basis, so $\dim_{\Reals}\bigl(\Span_{\Reals}(\Vx_{i})_{i=1}^{5}\bigr) = 3$.

The calculation over the rationals is similar:

$\dim_{\Ratls}\bigl(\Span_{\Ratls}(\Vx_{i})_{i=1}^{5}\bigr) = 5$.