So I am stuck on 51 here:

51. Determine whether the sets $S_1$ and $S_2$ span the same subspace of $\mathbb{R}^3$: $$\begin{align*} S_1 &= \Bigl\{ (1,2,-1),\ (0,1,1),\ (2,5,-1)\Bigr\}\\ S_2 &= \Bigl\{ (-2,-6,0),\ (1,1,-2)\Bigr\} \end{align*}$$

What I did to solve it was to multiply each vector in set $S_1$ by $C$, add the vectors together and set them to zero. Once I reduced the resulting matrix I got the following result: $$\begin{align*} c_1&=-2t\\ c_2&=-t\\ c_3&=t \end{align*}$$

So this result would be linearly dependent. Then I did the same thing for the second set and I got that $c_1=c_2=0$ which means that it is linearly independent. In conclusion, I said that set $S_1$ and $S_2$ does not span the same subspace of $\mathbb{R}^3$. The book says they do. Could anyone point out where I went wrong?


  • 2
    $\begingroup$ You're looking at the wrong thing. To see why, look at a simpler example: $S_1 = \{(1,0,0)\}$ and $S_2 = \{(1,0,0),(2,0,0)\}$. Clearly $S_1$ is linearly independent and $S_2$ is linearly dependent, so by your reasoning they shouldn't span the same subspace of $R^3$, but you shouldn't have any trouble seeing that they both span the $x$-axis. To show that your $S_1$ and $S_2$ span the same space, you need to show that every linear combination of vectors in $S_1$ is also a linear combination of vectors in $S_2$ and vice versa. $\endgroup$ – Brian M. Scott Jun 25 '11 at 18:27

Two sets can span the same subspace even if one is dependent and the other is not.

To solve such problems you can try, for example, seeing if every element of $S_2$ is a linear combination of elements from $S_1$ and vice verse. If they are, the sets span the same space.


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 infering these properties without actually showing them directly (such as dimension arguments liike 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.)


For this kind of problem, finding the reduced row echelon form (rref) seems to be a rather straightforward method.

For the set $S_1$, you form the matrix $$ \begin{pmatrix} 1 & 2 &-1 \\ 0 & 1 & 1 \\ 2 & 5 &-1 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 &-1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 &-1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 &-3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix} $$ For the set $S_2$, you form the matrix $$ \begin{pmatrix} -2 &-6 & 0 \\ 1 & 1 &-2 \end{pmatrix}\sim \begin{pmatrix} 1 & 3 & 0 \\ 1 & 1 &-2 \end{pmatrix}\sim \begin{pmatrix} 1 & 3 & 0 \\ 0 &-2 &-2 \end{pmatrix}\sim \begin{pmatrix} 1 & 3 & 0 \\ 0 & 1 & 1 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 &-3 \\ 0 & 1 & 1 \end{pmatrix} $$

Since we know that elementary row operations do not change row space, we see that $$\newcommand{\span}{\operatorname{span}}\span(S_1) = \span\{(1,0,-3),(0,1,1)\} = \span(S_2).$$

This methods works in general. If the non-zero rows in the rrefs are different, that means $\span(S_1) \ne \span(S_2)$.

To formulate this more clearly, we know that:

Theorem. Let $A$ and $B$ be two $m\times n$ matrices over the same field. The following conditions are equivalent:

  • The matrices $A$ and $B$ are row equivalent.
  • The matrices $A$ and $B$ have the same row space.
  • The matrices $A$ and $B$ have the same reduced row echelon form.
  • $\begingroup$ Thanks for the edit @GaryTam. I somehow missed that $S_{1,2}$ in the question denotes the a of vectors and not a subspace. I still changed $span$ to $\operatorname{span}$ (using \operatorname). (But I'd say that the notation $[(1,0,-3),(0,1,1)]$ is quite common, too.) $\endgroup$ – Martin Sleziak Dec 31 '16 at 14:57

The beginnings of your strategy were very reasonable. If it had turned out that the collection $S_1$ was linearly independent, then you would have known that the space spanned by $S_1$ has dimension $3$, i.e. is the full space. The space spanned by $S_2$ must have dimension $\le 2$, actually exactly $2$. So if $S_1$ had turned out to be a linearly independent set, you could have immediately concluded that the subspaces spanned are different, since they would have different dimension.

Unfortunately, it turned out that the set $S_1$ is linearly dependent, so a simple dimension argument will not solve the problem.

But it is now easy to see that both spaces are $2$-dimensional, they are planes through the origin. If you can show that the two vectors in $S_2$ are each in the span of $S_1$, that will mean the two spaces are the same. If you can show that at least one of the vectors in $S_2$ is not in $S_1$, you will know the spaces are not the same.

Checking this just a matter of looking at some linear equations, but you may be able to eyeball things. For example, the second vector in $S_2$ is the difference between the first two vectors in $S_1$. And the first vector of $S_2$ is (apart from scaling) the sum of the first two vectors in $S_1$. So the spaces are the same.

Since we are in $3$-space, you could alternately, by calculating cross-products, show that the two planes are perpendicular to the same line, and, since they both go through the origin, they are the same plane.


There are many ways to approach this sort of question, based on how much one depends on mechanical approaches vs. intuitive approaches.

Perhaps the most mechanical way of approaching this question is to quite literally Gram-Schmidt these guys to death (as this removes redundant vectors, i.e. linearly dependent vectors, i.e. vectors that are already spanned in the vectors already considered). For example, after one performs G-S on the first set of vectors, we are only left with 2 basis vectors (the third is obliterated). Continuing this, we obliterate the second set of vectors with these two basis vectors. (We note that the second set of vectors, considered separately, are linearly independent - and so also span a space of 2 dimensions).

But what does this accomplish? It's just a mechanical way of explicitly writing each vector in terms of other vectors. So we literally show that anything written in one basis can be written in the other. That's not so exciting.

Suppose we wanted to do it differently still: we can quickly see that the dimension of the space spanned by the first set is 2, and the dimension of the space spanned by the second set is 2. These are so called linear spaces, which is handy. This means that if we find 2 linearly independent vectors contained in both spaces, they are the same. While in one respect, we could just choose the two first vectors and see if they're in the second space, this also means that we could choose any linear combination of the first two vectors as well. So it's slightly more general.

Differently still: each linear subspace is 2 dimensional, i.e. a plane through the origin (it's a linear space, so it contains the origin). You could find the equation of the plane generated by the first two vectors and compare it to the plane generated by the two vectors in the second set. They are the same (multiples of each other). How does one do this? Taking cross products! (if you know them - it's very very simple).

Differently still: after realizing that each subspace is 2 dimensional, throw all 5 vectors in a matrix and row reduce. If 2 are left, they're the same. If there are 3 or more, then they are not the same. Along the same lines, one could (though should not, to be honest) proceed with determinants. Form a matrix whose first and second rows are the vectors from the first set, and whose third row is the first vector of the second set. Form another whose third row is the second vector of the second set. Taking determinants, we will see that both determinants are zero! This means that the 3-dimensional volume of these two parallelopipeds is zero, i.e. that they lie flat! If they lie flat, their sides must be linearly dependent, and since both vectors of the second set are dependent in the first set, they span the same subspace.

Differently still: find a vector not spanned in the first set, find the component orthogonal to the first subspace, and dot this orthogonal component with each vector in the second set. You will get 0 both times, meaning that the two subspaces have the same orthogonal complement, and therefore they are the same. Alternatively, take the cross product of the two vectors in the first set and dot the result with each vector in the second set. You will get 0 again, and this does not bear the burden of finding orthogonal complements in any witty or projective fashion.

Differently still: do it heuristically! By rolling a fair 24-sided die (called a deltoidal icositetrahedron) repeatedly, generate a random set of about 9 different points in 3 space. Find the line of best fit and project onto these two spaces. You will get the same projections! After doing this a couple of times, you can expect this to always work! Afterwords, show that as these 9 points are always on lattice points, at least 1 of the 36 segments joining these 9 points contain a lattice point as well. It will sharpen your skills with pigeonholing ideas.

Differently still, and finally: guess. When I TA'd linear algebra, my students would throw everything they could into matrices and row-reduce them, write a few things about rank and nullity, and solve a linear system of equations (whether asked for or not) on every question. Literally, even this one. When they realized that's not what I asked for, they would write a few illegible lines of work, draw a big $\Longrightarrow$, and say something like "Clearly, they do not span the same space" or "Obviously, they span the same space." For some other TAs, they'd give partial credit. It at least gave me a laugh.

Seriously though, if you would like to explain any of these further, let me know. They all really do work. And yes, it was quite a yarn. But I've been gone for a couple of weeks, and I had to say hello somehow!

  • $\begingroup$ I had similar experience, that some students simply put vectors into matrix as rows (or columns) and row reduce without really knowing what they are doing. However, I think that this is exactly the type of problem, where finding rref seems to me rather efficient way to solve it. $\endgroup$ – Martin Sleziak Feb 4 '16 at 19:26