I know the conditions of being a basis. The vectors in set should be linearly independent and they should span the vector space.

However, how do I find two different sets of basis vectors of the $xy$-plane in the three-dimensional space $\mathbb{R}^3$?

  • $\begingroup$ Do you know at least one set of basis vectors? $\endgroup$ – Zev Chonoles Apr 9 '13 at 7:36
  • $\begingroup$ No, unfortunately I don't :( $\endgroup$ – Leslie Apr 9 '13 at 7:39

If I understand the question, you want a spanning set for the $xy$-plane in ${\mathbb R^3}$. A spanning set for the $xy$-plane is $${\mathcal{B}}=\left\{\begin{bmatrix}a\\b\\0\end{bmatrix},\,\begin{bmatrix}c\\d\\0\end{bmatrix}\right\}, $$ where $ad-bc\neq 0$. Any choice of $a,\,b\,c$ and $d$ which satitfy the condition are a spanning set.

Some examples: $$\begin{align}\mathcal{B}_1&=\left\{\begin{bmatrix}1\\0\\0\end{bmatrix}, \,\begin{bmatrix}0\\1\\0\end{bmatrix}\right\}&\text{[the standard basis]}\\ \mathcal{B}_2&=\left\{\begin{bmatrix}1\\1\\0\end{bmatrix}, \,\begin{bmatrix}1\\-1\\0\end{bmatrix}\right\}\\ \mathcal{B}_3&=\left\{\begin{bmatrix}1\\-4\\0\end{bmatrix}, \,\begin{bmatrix}2\\1\\0\end{bmatrix}\right\} \end{align} $$


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