1
$\begingroup$

I'm having a bit of trouble understanding the concept of dimension of a span. I know that the dimension of a set of vectors is the number of vectors in the basis, but for the following questions I am not told that the set is a basis, nor that the set is linearly independent. How can I conclude the following:

  1. Suppose S is a collection of 93 vectors in $\mathbb{R(74)}$. Then, dim(span(S)) is greater than or equal to...

All I know that I can conclude from this question is that S is not a basis of $\mathbb{R(74)}$ since there are 93 vectors in S and therefore it is linearly dependent. I'm not sure if this can help me solve the question, however.

  1. Suppose S is a collection of 44 pairwise distinct vectors in $\mathbb{R(61)}$. Then dim(span(S)) is greater than or equal to...

I'm not sure what exactly it means that the vectors are "pairwise distinct". If anyone can provide a definition that would be greatly appreciated.

Thanks for any help!

$\endgroup$
5
  • $\begingroup$ "I know that the dimension of a set of vectors is the number of vectors in the basis..." You probably need to go re-read the material about which sets have dimensions, and what constitutes a basis, and how basis and dimension are related. (At the very least, you need to rewrite that sentence.) $\endgroup$ Commented Mar 13, 2018 at 19:40
  • $\begingroup$ "I'm not sure what exactly it means that the vectors are pairwise distinct" A collection of vectors, $\{v_1,v_2,v_3,\dots, v_n\}$ are considered "pairwise distinct" iff for every pair $i,j\in \{1,2,\dots,n\}$ if $i\neq j$ then $v_i\neq v_j$. I.e. no vector appears in the collection more than once. (note that in your problem though, simply being unequal does not mean much since $v\neq 2v$ when $v\neq 0$ but clearly $v$ and $2v$ are closely related) $\endgroup$
    – JMoravitz
    Commented Mar 13, 2018 at 19:47
  • $\begingroup$ @JohnHughes I have read the material on the following topics. I know that a set of vectors is a basis of a vector space if that set is linearly independent and the span of the set equals the vector space. As for how basis and dimension are related, my book states that: "The number of vectors in a basis of V is the dimension of V, dim(V)." I can't really see how that differs from the way I wrote it, please let me know how I can make my statement more clear? $\endgroup$
    – Melissa
    Commented Mar 13, 2018 at 19:49
  • $\begingroup$ As for finding specific lower bounds to the dimension of the span for your question... for (1) consider what happens if all 93 vectors are the zero vector. For (2) consider what happens if all 44 vectors are distinct multiples of the same nonzero vector. (Note: this bound is tight since if we go any lower, that would imply all vectors are 0 and thus not pairwise distinct) For specific upper bounds (not asked for, but useful practice), consider the number of vectors and the space they reside in and what happens if as many of the vectors are linearly independent as possible. $\endgroup$
    – JMoravitz
    Commented Mar 13, 2018 at 19:53
  • $\begingroup$ OK. Here's a set of vectors: $\{ (1,0), (2, 0), (3, 0)\}$. According to your sentence, the dimension of this set is the number of vectors in the basis. That leads me to ask "What basis?" and "Even if you gave me a basis, what dimension would you say this particular set has? Notice that the set contains exactly 3 vectors." $\endgroup$ Commented Mar 13, 2018 at 20:05

2 Answers 2

1
$\begingroup$

Let $V$ be a vector space with finite dimension $\dim(V)=n$

Let $U=\{u_1,u_2,u_3,\dots,u_k\}$ be a collection of vectors from $V$

We have the following tight bounds:

$$0\leq \dim(\text{span}(U))\leq \min(n,k)$$

It could be as low as zero, for example in the case where all vectors in $U$ are the zero vector.

In the case that $k\leq n$ it could be as high as $k$ when all $k$ of the vectors are linearly independent and it could be no higher since if all $k$ of the vectors are linearly independent then $U$ would be a basis for $\text{span}(U)$, implying the dimension is exactly $k$.

In the case that $n\leq k$ it could be as high as $n$ when $\text{span}(U)=V$ and it could not be higher since that would imply we could find $n+1$ or more linearly independent vectors in $V$, contradicting that $\dim(V)=n$, i.e. that there is a basis of $n$ vectors in $V$ that spans the whole space.


You ask for a definition of "pairwise distinct." A collection of objects $\{a_1,a_2,a_3,\dots,a_n\}$ are considered pairwise distinct iff for every $i,j\in\{1,2,3\dots,n\}$ we have $i\neq j\implies a_i\neq a_j$, or equivalently $a_i=a_j\implies i=j$. That is to say, no object appears in the collection more than once.


For your specific problems:

In your specific problem, the first example could indeed have all vectors as the zero vector, so $0\leq \dim\text{span}(S)\leq 74$

$~$

In your second problem, we cannot have all be the zero vector since the vectors would no longer be pairwise distinct. We could however have our collection be distinct scalar multiples of the same nonzero vector. So, $1\leq \dim\text{span}(S)\leq 44$

$\endgroup$
0
$\begingroup$

For the first question:

Suppose S is a collection of 93 vectors in R(74). Then, dim(span(S)) is greater than or equal to..

We can say that dimension of the span(S) is somewhere between 0 and 74.

For the second question :

Suppose S is a collection of 44 pairwise distinct vectors in R(61). Then dim(span(S)) is greater than or equal to...

We can say that dimension of the span(S) is somewhere between 1 and 44.

The reason for 1 is that they are all distinct so there are at least one non-zero vector there.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .