-1
$\begingroup$

Can I say if the vector $B$ is defined as follows that: $$B = \begin{bmatrix} a - b \\ 2a + b \end{bmatrix}, a, b \in \mathbb{R} $$

$B = \mathbb{R}^2$

Or do we require something like

$$B = \begin{bmatrix} c \\ d \end{bmatrix}, c, d \in \mathbb{R} $$

Where $c$ and $d$ are independent of each other?

by the definition of euclidean space $\mathbb{R}^2$?

  • 0
    More than "a vector", $\;B\;$ seems to be a poor intent to describe a subspace of $\;\Bbb R^2\;$, and then the question would, probably, be whether $\;B=\Bbb R^2\;$ ... – 2017-01-08
  • 0
    @DonAntonio, yes, so does $B$ satisfy the requirements? Because $a$ and $b$ are dependent so I wasnt sure – 2017-01-08
  • 0
    @Gaandmit independence of $a$ and $b$ is not what you should be aiming for as it has nothing to do with "set" $B$ being equal to $\mathbb{R}^2$. Instead see any vector in $B$ can be written as a sum of two "independent" vectors $\begin{bmatrix}1\\2\end{bmatrix}$ and $\begin{bmatrix}-1\\1\end{bmatrix}$. – 2017-01-08

1 Answers 1

1

Assuming the question actually meant to ask whether

$$B:=\left\{\,\binom{a-b}{2a+b}\in\Bbb R^2\;/\;a,b\in\Bbb R\,\right\}=\Bbb R^2\;\,?$$

The answer is yes, because $\;\dim B=2\;$, as for example $\;\binom 12\,,\,\,\binom03\;$ belong to $\;B\;$ and are linearly independent (why?) . Finish now the argument.

  • 0
    What is dim here? – 2017-01-08
  • 0
    I showed that $(a-b, 2a + b) = (x, y)$ for any real $x,y$ you can find a solution for $(a, b)$ proving that every vector in $R^2$ can be written as a combination of $(a- b, 2a + b)$ is that not enough? – 2017-01-08
  • 0
    If you don't know what dimension, one of the most important and basic things in linear algebra, is then this answer is not for you. Perhaps someone else can come up with something more basic. – 2017-01-08