Basis of $V=\{a\cdot(1,2,3)^T\}$ I've a vector space
$V=\left\{a\left(\begin{array}
 {l}1 \\ 2 \\ 3\end{array}\right)\right\}$ $a$ is any real number.
Can I choose it's basis as $\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$.
I think that it spans the space but then I can also have it's basis the set of these three vectors
$\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)$
$\left(\begin{array}{l}0 \\ 1\\ 0\end{array}\right)$$\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)$.
Something is wrong with my logic. Can anyone please point out my error.
Thank you

Yes, the set $A$ containing only one element, $\begin{pmatrix}1\\2\\3\end{pmatrix}$, is indeed a basis for $V$. It is very easy to show that the set satisfies both conditions that are in the definition of basis:

*

*It is true that the span of $A$ is equal to $V$, i.e. $\mathrm{span}(A)=V$.

*It is true that the set $A$ is linearly independent.


On the other hand, the set $B=\left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}$, containing the three vectors you list is not a basis for $V$, because the first condition fails. The span of $B$ includes the vector $\begin{pmatrix}1\\0\\0\end{pmatrix}$ which is not an element of $V$. This means that the span is not equal to $V$, so $B$ cannot be a basis for $V$.
