How can two vectors, with three elements each, form a base of a two dimensional space?

Question

I might have misunderstood something(most likely the case), but there's an example. Assume this matrix:

$\begin{bmatrix} 1& 2&3 \\ 1& 1& 2\\ 1& 2 &3 \end{bmatrix}$

Now, the reduced form is:

$\begin{bmatrix} 1& 2&3 \\ 0& -1& -1\\ 0& 0 &0 \end{bmatrix}$

Which give:

$\begin{bmatrix} 1& 2&3 \\ 0& 1& 1\\ 0& 0 &0 \end{bmatrix}$

Now, the three vectors of the matrix can't form a base in R^3, but because the rank of the matrix is 2, I can choose two independant vectors that will form a base for R^2. And this is where I loose comprehension. Let's say I take these two vectors that are independant:

$\begin{matrix} 1\\ 1\\ 1 \end{matrix}$

and

$\begin{matrix} 2\\ 1\\ 2 \end{matrix}$

These two supposedly form a base in R^2. What bothers me is how come there are three elements in the vectors that form a base of R^2, which as a general rule, has two elements in the vectors of its base? What does the third element tell us? Am I supposed to ignore it?

Answer 1

As answered in a comment above, your vectors do not belong to $R^2$.

However, what is meant is that for $span\{(1,1,1),(2,1,2)\} = K \subset R^3$ it holds that $K \simeq R^2$.

This can be seen by $L : K \to R^2 $;

$L((u_{1},u_{2},u_{3})) = (u_{1},u_{2})$

As this was defined on a basis of $K$, this linear transformation is well defined.

It is clearly onto $R^2$ since the image vectors are linearly independent, and letting $L(v) = 0_{R^2}$ we get that $(0,0,\gamma) = a(1,1,1) +b(2,1,2)$ and a simple calculcation shows that $a,b$ must be zero, showing the kernel is empty and $L$ to be injective and thus an isomorphism.

Answer 2

Look at matrix row by row.

Third vector is same as first - (1,2,3)

So, matrix has two independant vectors which describes plane in $R^3$

Answer 3

The two vectors $$ \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix} \qquad \text{and} \qquad \begin{bmatrix} 2 \\ 1 \\ 2\end{bmatrix} $$ for a basis for their span, which is a subspace of $\mathbb{R}^3$. Such a subspace is a vector space on its own and is isomorphic to $\mathbb{R}^2$ (albeit not in a unique way), not equal to $\mathbb{R}^2$.

In general, a linearly independent set $\{v_1,v_2,\dots,v_m\}$ in a vector space $V$ forms a basis of its span. This subspace of $V$ has dimension $m$, but this not makes it equal to $\mathbb{R}^m$.