1
$\begingroup$

Find the value of $c$ such that $\operatorname{span}([1,2,-1],[0,1,3])$ includes $[3,-2,c]$

First thing I notice is that the vectors $[1,2,-1],[0,1,3]$, are non-parallel, and so their span will be $\mathbb{R}^3$. Thus, $c$ can attain any value since there are $\mathbb{R}^3$ possible vectors.

However my textbook says $c = -27$. This is how they got $-27$:

Let $S_1, S_2 \in \mathbb{R}$ (scalar multiples).

$S_1[1,2,-1]+S_2[0,1,3] = [3,-2,c]$

$S_1 = 3$ (equation 1)

$2S_1 + S_2 = -2$ (equation 2)

$-S_1 + 3S_2 = c$ (equation 3)

Subbing (equation 1) into (equation 2):

$2(3) + S_2 = -2$

$S_2 = -8$

Subbing $S_2 = -8$ and $S_1 = 3$ into (equation 3):

$-(3) + 3(-8) = c$

$c = -27$.

My question is, am I wrong to say $c$ can be any value? The span of the two vectors contain infinitely many vectors, correct? So why is $c$ a specific value in this case?

  • 0
    Infinitely many vectors, but not *any* vector. Otherwise $\mathbf R^3$ would have dimension at most $2$… – 2017-01-06
  • 0
    Yeah I meant $R^2$ – 2017-01-06
  • 0
    I'm sorry, but *where* did you mean $\mathbf R^2$? – 2017-01-06
  • 0
    Oh sorry i made a mistake in my question – 2017-01-06

4 Answers 4

1

First thing I notice is that the vectors $[1,2,-1],[0,1,3]$, are non-parallel, and so their span will be $\mathbb{R}^3$. Thus, $c$ can attain any value since there are $\mathbb{R}^3$ possible vectors.

This is your first problem. $\text{dim}\mathbb{R}^3 = 3$, so this contradicts that 2 vectors can span $\mathbb{R}^3$.

Based on your question, it seems like you may not know vector spaces in depth. However, it suffices to find simply some $a, b \in \mathbb{R}$ such that $$a\begin{bmatrix}1 \\ 2 \\ -1\end{bmatrix} + b \begin{bmatrix}0 \\ 1 \\ 3\end{bmatrix} = \begin{bmatrix}3 \\ -2 \\ c\end{bmatrix}$$

Row reduce this system and you can find a solution.

1

No, the span of those two vectors is a subspace of dimension $2$, and most importantly it does not contain the vector $(0,0,1)$ as a consequence it cannot contain both $(3,-2,c)$ and $(3,-2,c')$ for $c\neq c'$

1

I'll try to address all of your questions.

$\mathbb R^3$ has dimension $3$ so its base must have $3$ linearly independent vectors. Therefore the span of the $2$ vectors in question can't be $\mathbb R^3$.

I think this together with the $3$ equations in your question clarify your confusion around why $c$ can't be any value and why it must be $-27$.

Lastly, the subspace given by $(a, a, a)$ has infinitely many vectors but obviously doesn't span the entire $\mathbb R^3$, because for example the vector $(1, 2, 3)$ is not part of it.

  • 0
    What do you mean by 'its base must have $3$ linearly independent vectors' – 2017-01-06
  • 0
    $[1,0,0][0,1,0][0,0,1]$ ? – 2017-01-06
  • 0
    @KSplitX exactly! that's an example of a base of $\mathbb R^3$ – 2017-01-06
  • 0
    So then why did you say the span can't be $R^3$? – 2017-01-06
  • 0
    @KSplitX I meant that the vector space spanned by the $2$ vectors in the question can't be $\mathbb R^3$. Because there are only $2$. The other example you gave with $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ is a good example of a base of $\mathbb R^3$. Does that make sense? – 2017-01-06
  • 0
    Yea makes sense , thanks – 2017-01-06
1

The simplest to find this value of $c$ is row reduction. Consider the ‘row-augmented’ matrix $\;\begin{bmatrix}1&2&-1\\0&1&3\\3&-2&c\end{bmatrix}$. If this matrix has rank $2$, the last row vector will be in the span of the first two row vectors. \begin{align} \begin{bmatrix}1&2&-1\\0&1&3\\3&-2&c\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&2&-1\\0&1&3\\0&-8&c+3\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&2&-1\\0&1&3\\0&0&c+27\end{bmatrix} \end{align} Thus the condition is $\;c=-27$.