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I was given this question in an assignment asking if $u=(0,2,-2), v=(1,-2,1)$ and $w=(4,2,3)$ will span a line, a plane or all of $\mathbb{R}^3$.

What I have done so far is determined that these vectors are linearly independent by row reducing to calculate a diagonal product that is not equal to $0$, proving that it is not linearly dependent. This means that it is a basis for $\mathbb{R}^3$.

What I am confused about is how do I know whether this will span a plane, a line or $\mathbb{R}^3$.

Does a basis of $\mathbb{R}^3$ imply it spans $\mathbb{R}^3$?

How do I know if a vector spans a plane or a line or $\mathbb{R}^3$?

Ryan Shesler
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Peter
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  • Welcome to Mathematics Stack Exchange. If three vectors are linearly independent, they will span a $3$-dimensional space – J. W. Tanner Apr 27 '20 at 18:31
  • What is the definition of the span of a set of vectors? (Note that the question is not about the span of a single vector!) – Mees de Vries Apr 27 '20 at 18:32
  • A (linear) span of a set of vectors is the set of vectors that can be obtained by linear combinations of the vectors from the original set – J. W. Tanner Apr 27 '20 at 18:33
  • @J.W.Tanner thanks! So to clarify, that means this will span all of R3, rather than a line or a plane? What would the process be if this was a plane or a line? – Peter Apr 27 '20 at 18:35
  • If you have three vectors, they could span a space of dimension $0$, $1$, $2$, or $3$; $3$ if they are linearly independent, otherwise less than $3$ – J. W. Tanner Apr 27 '20 at 18:37
  • The dimension of the span is the rank of the matrix of the vectors. – Berci Apr 27 '20 at 18:38

2 Answers2

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If you have $3$ linearly independent vectors, they will span a $3$-dimensional space.

If you have $3$ vectors that are linearly dependent, they will span a space of $0$, $1$, or $2$ dimensions.

Your vectors are independent, so they span $\mathbb R^3$.

Here is an example of three vectors that span a plane: $(1,0,0), (0,1,0), (1,1,0)$.

Here is an example of three vectors that span a line: $(1, 1, 0), (2,2,0), (3,3,0)$.

Here is an example of three vectors that span a zero-dimensional space:

$(0,0,0), (0,0,0), (0,0,0)$.

J. W. Tanner
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  • That clarifies everything. Thank you! So to my understanding, the vector set of (u,v,w) will span R3 because they are 3 linearly independent vectors. For a set of 3 vectors to span a plane, you need a missing pivot, and for it to span a line, the vectors will be multiples. – Peter Apr 27 '20 at 18:42
  • Yes, though I would write $\color{blue}{{}u,v,w\color{blue}{}}$ for a set of vectors – J. W. Tanner Apr 27 '20 at 18:46
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Since we have that given $n$ vectors $v_{1}\cdots v_{n}$ $ \in \mathbb{K}^{n}$ these are linearly independant if and only if det$(v_{1} | \cdots | v_{n}) \ne 0$

(You can find some reference here)

We have that since $$\mbox{det}\begin{pmatrix}0 & 1 & 4 \\ 2 & -2 & 2 \\ -2 & 1 & 3\end{pmatrix} \ne 0$$

They are linearly independant, which means they span a subspace of dimension 3, but since the space is $\mathbb{R}^{3}$, span($u,v,w$) = $\mathbb{R}^{3}$.

Note that is determinant equals to $0$ it means that the vector are not linearly dependant, so they can span a plane (which is the case when just one of them is linearly depedent from the other two), or a line (when two of them linearly depend from a one of them).

jacopoburelli
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