# Thread: another basis and dimension question

1. ## another basis and dimension question

Let the subset $\displaystyle U \subset \mathbb{R}^4$ be defined by

$\displaystyle U = \left \{ v | v = \left [\begin{matrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{matrix} \right ] , v_1 - v_2 + v_3 = 0 , v_2 + 2v_3 - v_4 = 0 \right \}$

1. Show that $\displaystyle U$ is a sub-space of $\displaystyle \mathbb{R}^4$

Because $\displaystyle \mathbb{R}^4$ is a vector space and $\displaystyle U \subset \mathbb{R}^4$ U inherits the same operations as $\displaystyle \mathbb{R}^4$. If $\displaystyle U$ is a sub-space then

a) U is non-empty, eg. $\displaystyle \left [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right ] \in U$

b) $\displaystyle p \in U$ and $\displaystyle q \in U$, then $\displaystyle p + q \in U$

$\displaystyle \left [\begin{matrix} p_4 - 3p_3 \\ p_4 - 2p_3 \\ p_3 \\ p_4 \end{matrix} \right ] + \left [\begin{matrix} q_4 - 3q_3 \\ q_4 - 2q_3 \\ q_3 \\ q_4 \end{matrix} \right ] = \left [\begin{matrix} (p_4 + q_4) - 3(p_3 + q_3) \\ (p_4 + q_4) - 2(p_3 + q_3) \\ (p_3 + q_3) \\ (p_4 + q_3) \end{matrix} \right ]$

c) $\displaystyle \alpha \left [\begin{matrix} p_4 - 3p_3 \\ p_4 - 2p_3 \\ p_3 \\ p_4 \end{matrix} \right ] \in U$

2. Find a basis and dimension of $\displaystyle U$

$\displaystyle U = \left \{ v_4 \left [\begin{matrix} 1 \\ 1 \\ 0 \\ 1 \end{matrix} \right ] + v_3 \left [\begin{matrix} -1 \\ -2 \\ 1 \\ 0 \end{matrix} \right ] , v_3, v_4 \in \mathbb{R} \right \}$

pretty simple, just the two column vectors above and so, dim(U) is 2.

3. How do I change the basis to a basis over $\displaystyle \mathbb{R}^4$ ?

2. Hi bmp05.

Express $\displaystyle v_3$ and $\displaystyle v_4$ in terms of $\displaystyle v_1$ and $\displaystyle v_2.$

$\displaystyle v_3\ =\ v_2-v_1$

$\displaystyle v_4\ =\ 2v_3+v_2\ =\ 3v_2-2v_1$

Hence each vector in $\displaystyle U$ is of the form

$\displaystyle \begin{bmatrix}v_1\\v_2\\v_2-v_1\\3v_2-2v_1\end{bmatrix}\ =\ v_1\begin{bmatrix}1\\0\\-1\\-2\end{bmatrix}\,+\,v_2\begin{bmatrix}0\\1\\1\\3\en d{bmatrix}$

This shows that $\displaystyle \left\{\begin{bmatrix}1\\0\\-1\\-2\end{bmatrix},\,\begin{bmatrix}0\\1\\1\\3\end{bma trix}\right\}$ is a basis for $\displaystyle U,$ which therefore has dimension 2. (Note that it also shows that $\displaystyle U$ is a subspace of $\displaystyle \mathbb R^4.)$

3. I believe the final question was how to extend the given basis, $\displaystyle U = \left \{ v_4 \left [\begin{matrix} 1 \\ 1 \\ 0 \\ 1 \end{matrix} \right ] + v_3 \left [\begin{matrix} -1 \\ -2 \\ 1 \\ 0 \end{matrix} \right ] , v_3, v_4 \in \mathbb{R} \right \}$ to a basis for $\displaystyle R^4$. Just find two more independent vectors. That will give a set of 4 independent vectors which will form a basis for $\displaystyle R^4$. Looks to me like $\displaystyle \left[\begin{matrix}1 \\ 0 \\ 0\\ 0\end{matrix}\right]$ and $\displaystyle \left[\begin{matrix}0 \\ 1 \\ 0 \\ 0\end{matrix}\right]$ will do.