Finding the dimension of a vector space over different fields?

Question

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!

Answer 1

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?

Answer 2

$\newcommand{\Field}[1]{\mathbf{#1}}\newcommand{\Reals}{\Field{R}}\newcommand{\Cpx}{\Field{C}}\newcommand{\Ratls}{\Field{Q}}\DeclareMathOperator{\Span}{Span}\newcommand{\Vx}{\mathbf{x}}$Briefly, yes: "You can use scalars in the field $K$ when taking spans and bases over $K$."

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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:

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