Question

Show that $\left\{i,1+2i,1+3i\right\}$spans C over R.

Short answer

Answer:

$\left\{i,1+2i,1+3i\right\}$is spans C over R.

Step-by-step solution

Definition of linear span.

The linear span also called as just span of a set of vector in a vector space is the intersection of all subspace containing the set.

Show {i,1+2i, 1+3i}is spans C over R

It is enough to show that the given set is linearly independent and any element in C can be written as linear combination of element of the given set.

Let ${\mathit{C}}_{\mathbf{1}}\mathbf{,}{\mathit{C}}_{\mathbf{2}}\mathbf{,}{\mathit{C}}_{\mathbf{3}}\mathbf{\in }\mathit{R}$such that ${\mathit{C}}_{\mathbf{1}}\left(i\right)\mathbf{+}{\mathit{C}}_{\mathbf{2}}\left(1+2i\right)\mathbf{+}{\mathit{C}}_{\mathbf{3}}\left(1+3i\right)\mathbf{=}\mathbf{0}$

But $\left(1+3i\right)$can be written as linear combination of other two.

Hence, it can be dropped out from basic element.

So, put $\left(1+3i\right)$

Then

$$\begin{gathered} {\mathit{C}}_{\mathbf{1}}\left(i\right)\mathbf{+}{\mathit{C}}_{\mathbf{2}}\left(1+2i\right)\mathbf{=}\mathbf{0}\mathbf{+}\mathbf{0}\mathit{i} \\ \left(C_1+2C_2\right)\mathit{i}\mathbf{+}{\mathit{C}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{+}\mathbf{0}\mathit{i} \end{gathered}$$

This gives

$$\begin{gathered} {\mathit{C}}_{\mathbf{1}}\mathbf{+}\mathbf{2}{\mathit{C}}_{\mathbf{2}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{C}}_{\mathbf{2}}\mathbf{=}\mathbf{0} \end{gathered}$$

${\mathit{C}}_{\mathbf{2}}\mathbf{=}{\mathit{C}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$

The set $\left\{i,1+2i,1+3i\right\}$is lineally independent.