# Can a set of 4 vectors with 3 entries each only span R2 if the third row reduces to all zeros?

I'm a bit confused as to how dimension, dimension of span, and dimension of column space all relate with regards to a basis. The question is worded as follows:

Find the dimension of the span of the set v1 = (1,0,2), v2 = (3,1,1), v3=(9,4,-2), v4=(-7,3,1). It gives this hint: "Recall: the dimension of a vector space is the number of elements in a basis."

I know that for a set of vectors to be a basis for a vector space, none of its members can be a linearly combination of the others. To ensure this, I put these 4 columns into an augmented matrix with a column of zeros on the right hand side, because the pivot columns of this matrix, once its in in RREF, should be the columns of the original non-reduced set to form a basis. After getting it into RREF, I found that the bottom row reduces to all zeros, and the first and second columns are pivot columns.

Does this mean that the dimension is two since there are two pivot columns? Can there be no linear combination that leads to an entry in the third dimension of vector space since the third row is all zeros? I'm confused about the implication of this third row of zeros, especially since I started with more columns than rows. I appreciate any and all help.

• In your title, you say "span $\mathbb{R}^2$", but the vectors are in $\mathbb{R}^3$. Commented Jul 13, 2014 at 18:26
• I added a post, but I wanted to make a remark about your comment that there are more columns than rows. If there are more columns than rows, and we say there are $m$ rows and $n$ columns, you can have at most $m$ pivots in your reduced matrix. What I mean to say is, depending on if you have more rows than columns, or more columns than rows, whichever one you have less of, that is the maximum number of pivots you could potentially have. So if you have 4 rows and 2 columns, you can have at most 2 pivots (you would have to reduce the matrix to determine exactly how many you have). Commented Jul 13, 2014 at 18:47

So the question is to find the span of $v_{1} = \begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix}$, $v_{2} = \begin{bmatrix} 3 \\ 1 \\ 1 \\ \end{bmatrix}$, $v_{3} = \begin{bmatrix} 9 \\ 4 \\ -2 \\ \end{bmatrix}$, and $v_{4} = \begin{bmatrix} -7 \\ 3 \\ 1 \\ \end{bmatrix}$.

Before trying to solve this problem, it is important to know what span means. The first thing you need to know is where the vectors live. To figure this out, count the number of components in one of the vectors. In this case, there are 3 components in each vector, so these vectors live in $\mathbb{R}^{3}$. The first component corresponds to the $x$-axis, the second component corresponds to the $y$-axis, and the third component corresponds to the $z$-axis.

Now, if we take two linearly independent vectors, say $v_{1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ and $v_{2} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$, then these vectors clearly live in $\mathbb{R}^{3}$, since they each have 3 components. Since they are linearly independent, and there are 2 of them, they span a plane in $\mathbb{R}^{3}$. This particular plane is isomorphic to $\mathbb{R}^{2}$, but it is not $\mathbb{R}^{2}$, because vectors in $\mathbb{R}^{2}$ all look like $\begin{bmatrix} x \\ y \\ \end{bmatrix}$ (i.e., they have 2 components only). A plane in $\mathbb{R}^{3}$ is clearly 2-dimensional since that is how we define a plane, but we do not describe it as $\mathbb{R}^{2}$. Instead, we simply say it is a plane in $\mathbb{R}^{3}$.

Here is an example of what I mean: span$\left \{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right \}$ is the $XY$-plane in $\mathbb{R}^{3}$, while span$\left \{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right \}$ is the $XZ$-plane in $\mathbb{R}^{3}$, and span$\left \{\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right \}$ is the $YZ$-pane in $\mathbb{R}^{3}$. All three of the spans I just mentioned are isomorphic to $\mathbb{R}^{2}$, but they are distinct planes in $\mathbb{R}^{3}$, which is why we must describe them as planes in $\mathbb{R}^{3}$, rather than just as $\mathbb{R}^{2}$. When described as planes, they can be differentiated from each other, which is good because they are different planes.

Now, on to your question, as you correctly stated, to determine which vectors are linearly independent (in order to determine the dimension of the span), we put the vectors as columns in a matrix and reduce to RREF. So after doing that, we get that

$\begin{bmatrix} 1 & 3 & 9 & -7 \\ 0 & 1 & 4 & 3 \\ 2 & 1 & -2 & 1 \\ \end{bmatrix}$ reduces to $\begin{bmatrix} 1 & 0 & -3 & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

We see that there are three pivots! The number of pivots is the number of linearly independent columns, and the pivots correspond to the linearly independent columns, so the vectors $\left \{ \begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ 1 \\ \end{bmatrix}, \begin{bmatrix} -7 \\ 3 \\ 1 \\ \end{bmatrix} \right \}$ (which are $v_{1}$, $v_{2}$, and $v_{4}$) are the linearly independent ones, and $v_{3}$ is linearly dependent on them. Since there are three linearly independent vectors, the span of all four vectors is equal to the span of the three linearly independent ones. Three linearly independent vectors span a subspace that is 3-dimensional. But these vectors live in $\mathbb{R}^{3}$, which is 3-dimensional itself, so their span must be equal to $\mathbb{R}^{3}$. If these vectors happened to live in $\mathbb{R}^{4}$, then their span would be a 3-dimensional subspace of $\mathbb{R}^{4}$.

So to answer your question, these four vectors could have spanned a 2-dimensional subspace of $\mathbb{R}^{3}$ if only two of the four were linearly independent. But we had three pivots in our matrix, so three linearly independent vectors, which meant the span of the four was equal to the span of the three linearly independent vectors (i.e., span$\{ v_{1}, v_{2}, v_{3}, v_{4} \} =$ span$\{ v_{1}, v_{2}, v_{4} \}$), and they spanned all of $\mathbb{R}^{3}$ in this case because they live in the 3-dimensional space and their span is 3-dimensional.

It is correct that the number of pivots in the RREF is equal to the dimension of the span of the (column) vectors.

Your confusion about the third row of zeros seems to be about this: while row operations do not change the dimension of the span of a set of column vectors, they do change the vectors themselves and hence the span. In other words, the subspace spanned by the columns of the RREF is not necessarily the same as the subspace spanned by the original columns (in fact, you yourself have noticed that each original column fails to be in the span of the columns of the RREF, because of their third entries).

A good way to remember this is to imagine that there's only one column vector. Think about what row operations look like on an $n\times1$ matrix. Do they change the span of the single column vector?