0
$\begingroup$

Given that $W = \begin{bmatrix} 3s - 2t \\ s + 2t \\ 2s + 3t \\ \end{bmatrix}$ such that $s$ and $t$ are real numbers is a subspace of $\mathbb{R}^3$, I need to write a basis of $W$ and state the dimension of $W$.

I believe that the basis is the set $ \begin{bmatrix} 3\\ 1\\ 2\\ \end{bmatrix}$ , $ \begin{bmatrix} -2\\ 2\\ 3\\ \end{bmatrix}$ since they both contain pivot positions. I am not sure if this is correct though. Also I do not know how to find the dimensino of $W$.

  • 0
    Does this help? $$\begin{bmatrix} 3s - 2t \\ s + 2t \\ 2s + 3t \\ \end{bmatrix} = s\begin{bmatrix} 3 \\ 1 \\ 2\end{bmatrix} + t\begin{bmatrix} -2 \\ 2 \\ 3\end{bmatrix}$$ So $W$ is the set of linear combinations of $\begin{bmatrix} 3 \\ 1 \\ 2\end{bmatrix}$ and $\begin{bmatrix} -2 \\ 2 \\ 3\end{bmatrix}$. – 2017-02-21
  • 0
    I have already established that from a previous part of the question – 2017-02-21
  • 0
    OK. Well this clearly shows that those two vectors form a *spanning set* for $W$. So what's the relationship between a spanning set and a basis? – 2017-02-21
  • 0
    linear independence ? – 2017-02-21
  • 0
    That is not a full sentence (or idea) ... but yes, it involves linear independence. – 2017-02-21

1 Answers 1

0

If you show ⎢312⎤⎦⎥[312] , ⎡⎣⎢−223⎤⎦⎥ are linearly independent then they form a basis for W as they span W which would mean the dimension is 2.

  • 0
    they are linearly independent . correct? – 2017-02-21
  • 0
    yes it can be shown easily – 2017-02-21