[Math] Determine whether the set S spans $\mathbb{R^2}$

linear algebra

Here's the question I'm trying to solve:

Determine whether the set $S$ spans $\mathbb{R^2}$:
$S={(1,2),(-2,-4),(1/2,1)}$

I know that the answer is that it does not span $\mathbb{R^2}$ because when I attempt to solve the resultant matrix, a row of zeros is produced – indicating that the vectors are linearly dependent and therefore negating the possibility of S being a spanning set.

To explain, a spanning set, specifically in this case, would have to satisfy
$(a,b)=c_1(1,2)+c_2(2,-4)+c_3(1/2,1)$.

From here, a matrix can be formed to solve for $c_1,c_2,c_3$.

That is fine. However, the other part of the question is along the lines of, "if S does not span $\mathbb{R^2}$, give a geometric description of the subspace it does define."

To me, it is intuitive that this would describe a line in $\mathbb{R^2}$. This is because the vectors are clearly multiples of each other, which would indicate they can be found on the same line. Even though that's the case, would would be a mathematical process that would reach the same conclusion?

Any help will be appreciated.

Edit: I should add that I feel that it can describe a line if one concludes that the free variable that will be generated describes a line. For example: $c_1+c_2(t)$ where $t=c_3$.

Best Answer

They're not just on the same line --- they're on the same line through the origin. That's the same as saying each one is a scalar multiple of the others, and that gives you a mathematical process to reach the conclusion.

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[Math] Linear Algebra: determine whether the sets span the same subspace

Two sets of vectors in the same vector space, $S_1$ and $S_2$, span the same subspace if and only if:

Each vector in $S_1$ can be written as a linear combination of the vectors in $S_2$; and
Each vector in $S_2$ can be written as a linear combination of the vectors in $S_1$.

There may be ways of inferring these properties without actually showing them directly (such as dimension arguments like user6312 suggests), but it really ends up coming down to showing this holds or that this cannot hold.

Note that these conditions are not intrinsic: it's almost never enough to just look at $S_1$ without thinking about $S_2$, and to then look at $S_2$ without looking at $S_1$; it is only in very extreme circumstances that this suffices (when you can prove that the "sizes" of the spans don't match, a dimension argument) that this can prove to be enough.

So just figuring out if the sets are linearly dependent or independent is not enough in this case.

So: are $(-2,-6,0)$ and $(1,1,-2)$ each linear combinations of $(1,2,-1)$, $(0,1,1)$, and $(2,5,-1)$? Yes: you can try solving the two systems of linear equations: $$\begin{align*} \alpha_1 \left(\begin{array}{r}1\\2\\-1\end{array}\right) + \beta_1\left(\begin{array}{c}0\\1\\1\end{array}\right) + \gamma_1\left(\begin{array}{r}2\\5\\-1\end{array}\right) &= \left(\begin{array}{r}-2\\-6\\0\end{array}\right)\\ \alpha_2 \left(\begin{array}{r}1\\2\\-1\end{array}\right) + \beta_2\left(\begin{array}{c}0\\1\\1\end{array}\right) + \gamma_2\left(\begin{array}{r}2\\5\\-1\end{array}\right) &= \left(\begin{array}{r}1\\1\\-2\end{array}\right) \end{align*}$$ and see if they each have solutions. (It can even be done both at once, by doing row reduction on $$\left(\begin{array}{rrr|rr} 1 & 0 & 2 & -2 & 1\\ 2 & 1 & 5 & -6 & 1\\ -1 & 1 & -1 & 0 & -2 \end{array}\right).$$ If either system has no solutions, then you know that not every vector in $S_2$ is in the span of $S_1$ and you are done; if both systems have solutions, then every vector in $S_2$ is in the span of $S_1$, so $\mathrm{span}(S_2)\subseteq \mathrm{span}(S_1)$.

Then you need to see if the converse inclusion holds: is every vector in $S_1$ a linear combination of the vectors in $S_2$? That is, can we solve the three systems of linear equations? $$\begin{align*} \rho_1\left(\begin{array}{r}-2\\-6\\0\end{array}\right) + \sigma_1 \left(\begin{array}{r}1\\1\\-2\end{array}\right) &= \left(\begin{array}{r}1\\2\\-1\end{array}\right)\\ \rho_2\left(\begin{array}{r}-2\\-6\\0\end{array}\right) + \sigma_2 \left(\begin{array}{r}1\\1\\-2\end{array}\right) &= \left(\begin{array}{r}0\\1\\1\end{array}\right)\\ \rho_3\left(\begin{array}{r}-2\\-6\\0\end{array}\right) + \sigma_3 \left(\begin{array}{r}1\\1\\-2\end{array}\right) &= \left(\begin{array}{r}2\\5\\-1\end{array}\right) \end{align*}$$ If we can solve all, then each vector in $S_1$ is in the span of $S_2$, so $\mathrm{span}(S_1)\subseteq \mathrm{span}(S_2)$; together with the previous inclusion, this would show the spans are equal. If some equation cannot be solved, then not every vector in $S_1$ is in the span of $S_2$, so the spans are different.

(There is one thing that you can rescue from your efforts: since you proved that the set $S_1$ is linearly dependent, you can extract from it a linearly independent set (in this case, for instance, the first two vectors), and replace $S_1$ with that set of two vectors (because the third vector is a linear combination of the first two). That will mean that checking that "every vector in $S_1$ is a linear combination of the vectors in $S_2$" and checking that "every vector in $S_2$ is a linear combination of the vectors in $S_1$" will be simpler: instead of checking five things, you only need to check four.)

Linear Algebra – How to Tell if a Set of Vectors Spans a Space?

There are several things you can do. Here are four:

You can set up a matrix and use Gaussian elimination to figure out the dimension of the space they span. They span $\mathbb{R}^3$ if and only if the rank of the matrix is $3$. For example, you have $$\begin{align*} \left(\begin{array}{ccc} 1 & 1 & 1\\ 3 & 2 & 1\\ 1 & 1 & 0\\ 1 & 0 & 0 \end{array}\right) &\rightarrow \left(\begin{array}{ccc} 1 & 0 & 0\\ 3 & 2 & 1\\ 1 & 1 & 0\\ 1 & 1 & 1 \end{array}\right) &&\rightarrow \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 2 & 1\\ 0 & 1 & 0\\ 0 & 1 & 1 \end{array}\right)\\ &\rightarrow \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 2 & 1\\ 0 & 1 & 1 \end{array}\right) &&\rightarrow \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 1 \end{array}\right). \end{align*}$$ (Sequence of operations: exchanged rows 1 and 4; subtracted first row from other rows to make $0$s in first column; exchanged second and third rows; added multiples of the second row to third and fourth row to make $0$s in the second column).

At this point, it is clear the rank of the matrix is $3$, so the vectors span a subspace of dimension $3$, hence they span $\mathbb{R}^3$.

See if one of your vectors is a linear combination of the others. If so, you can drop it from the set and still get the same span; then you'll have three vectors and you can use the methods you found on the web. For example, you might notice that $(3,2,1) = (1,1,1)+(1,1,0)+(1,0,0)$; that means that $$\mathrm{span}\Bigl\{(1,1,1),\ (3,2,1),\ (1,1,0),\ (1,0,0)\Bigr\} = \mathrm{span}\Bigl\{(1,1,1),\ (1,1,0),\ (1,0,0)\Bigr\}.$$
Determine if the vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ lie in the span (or any other set of three vectors that you already know span). In this case this is easy: $(1,0,0)$ is in your set; $(0,1,0) = (1,1,0)-(1,0,0)$, so $(0,1,0)$ is in the span; and $(0,0,1) = (1,1,1)-(1,1,0)$, so $(0,0,1)$ is also in the span. Since the span contains the standard basis for $\mathbb{R}^3$, it contains all of $\mathbb{R}^3$ (and hence is equal to $\mathbb{R}^3$).
Solve the system of equations $$\alpha\left(\begin{array}{c}1\\1\\1\end{array}\right) + \beta\left(\begin{array}{c}3\\2\\1\end{array}\right) + \gamma\left(\begin{array}{c}1\\1\\0\end{array}\right) + \delta\left(\begin{array}{c}1\\0\\0\end{array}\right) = \left(\begin{array}{c}a\\b\\c\end{array}\right)$$ for arbitrary $a$, $b$, and $c$. If there is always a solution, then the vectors span $\mathbb{R}^3$; if there is a choice of $a,b,c$ for which the system is inconsistent, then the vectors do not span $\mathbb{R}^3$. You can use the same set of elementary row operations I used in 1, with the augmented matrix leaving the last column indicated as expressions of $a$, $b$, and $c$.

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[Math] Linear Algebra: determine whether the sets span the same subspace

Linear Algebra – How to Tell if a Set of Vectors Spans a Space?

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