Let $B = \{(1,0,1),(0,1,3)\} \subset \mathbb{R}^3$ and $V = span(B)$. Let $C = \{\vec{c_1} = (1,2,7), \vec{c_2} = (3,4,15)\}$ (Assume that $B$ and $C$ are linearly independent).

Show that that $span(C) = V$.

I don't really have an elegant way of explaining this, but this is my best attempt:

The span of a set is the set of all linear combinations of that set. So for set $B$...

If $a_1,a_2,a_3 \in \mathbb{R}$ \begin{equation*} \begin{bmatrix} 1&0\\ 0&1\\ 1&3 \end{bmatrix} \begin{bmatrix} a_1\\ a_2\\ a_3 \end{bmatrix} = \begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix} = span(B) \end{equation*}

If $\vec{b} = (b_1,b_2,b_3)$ and I set $\vec{b} = \vec{0}$ and then augment and RREF...

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

The same RREF matrix is produced if I was to go through the same process with $C$

Hence $span(C) = V$?

I would love it if someone was able to further explain if my reasoning is correct and if it is, WHY it is.

  • $\begingroup$ You should be multiplying the $3 \times 2$ matrix by a vector in $\Bbb{R}^2$, not a vector from $\Bbb{R}^3$. $\endgroup$ Mar 26 '16 at 20:33
  • $\begingroup$ @NobleMushtak So I should instead by multiplying by $\begin{bmatrix} a_1\\ a_2\\ \end{bmatrix} $? Why do we lose a dimension? $\endgroup$
    – RedShift
    Mar 26 '16 at 20:36
  • $\begingroup$ You only have two columns, so the vector you multiply should only have two entries. Meanwhile, you have three rows, so the vector you get as a result should have three entries. $\endgroup$ Mar 26 '16 at 20:36

Unfortunately, your reasoning is not correct. What you have shown is that $B$ and $C$ both span two-dimensional subspaces of $\mathbb R^3$. You have not shown that they both span the same subspace. But this is a good step Once you know that the spans of $C$ have the same dimensions, it suffices to show that you can write both of the elements of $C$ as a linear combination of the elements of $B$. For example, $(1,2,7)=1\times(1,0,1)+2\times(0,1,3)$. Can you figure out the other one?

  • $\begingroup$ Oh okay, so $(3,4,15) = 3 \times (1,0,1) + 4 \times (0,1,3) $ Just trying to understand... Since the spans of $B$ and $C$ both have the same dimensions, and $C$ can be written as a linear combination of the elements from $B$, this implies that $B$ and $C$ both span the same subspace? $\endgroup$
    – RedShift
    Mar 26 '16 at 20:51
  • $\begingroup$ Yes. Since we can write everything in $C$ as linear combinations of things in $B$, it means that $B$ also spans the span of $C$, so $\text{span}\,C\subset\text{span}\,B=V$. But since the dimension of $\text{span}\,C$ is the same as the dimension of $V$, they are equal. $\endgroup$
    – Alex S
    Mar 26 '16 at 20:58

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