For example,

let S = {$(x_1, x_1)| x_1 \in \mathbb R$} be a subspace of $\mathbb R^2$.

By definition, dim(S) = 1, and dim($\mathbb R^2$) = 2.

Then the set {(1, 1)} only has one vector, so is it linearly independent in S, but is linearly dependent in $\mathbb R^2$?

I know this doesn't make any sense, but we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.

Since (1,1) is in both S and $\mathbb R^2$, but the set {(1, 1)} only spans S, how come it is not only linearly independent in S and linearly dependent in $\mathbb R^2$ (since {(1,1)} does not span $\mathbb R^2$).

Sorry for this stupid question

  • 1
    Vectors are linearly independent of other vectors in the same space so it doesn't make sense to discuss them in different spaces. – CyclotomicField Sep 21 at 22:43
  • 3
    If the set $\{v_1,\dots,v_n\}$ doesn't span $V$, then it need not be linearly dependent, unless $n=\dim V$. In your case the set consists of one vector and $\mathbb{R}^2$ has dimension $2$. – egreg Sep 21 at 22:50
  • 5
    we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in You must have misunderstood something. Two non-parallel vectors are linearly independent in $\,\Bbb R^3\,$, for example, but they certainly don't span $\,\Bbb R^3\,$. – dxiv Sep 21 at 22:51
  • 1
    $\{ 1 \}$ is linearly independent in $\mathbb{R}$ with the usual vector $\mathbb{R}$-vector space structure; but if instead you use the vector space structure $x \oplus y = x + y - 1$, $\lambda \odot x = \lambda x - \lambda + 1$ then $\{ 1 \}$ is linearly dependent with respect to this structure. – Daniel Schepler Sep 21 at 22:59
  • 1
    Another possible example: if you consider $\mathbb{C}$ to be a vector space over $\mathbb{R}$, then $\{ 1, i \}$ is linearly independent. However, if you consider $\mathbb{C}$ to be a vector space over $\mathbb{C}$, then $\{ 1, i \}$ is linearly dependent. – Daniel Schepler Sep 21 at 23:02

You said you “learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.” This isn’t correct, unless “the vector space that it is in” means “the smallest vector space that it is in,” which would not be a typical reader’s understanding. But even with that understanding, the statement is not useful, because the fact is not special to linearly independent sets. Every set of vectors spans the smallest vector space that contains them: a set of vectors spans its span. (That’s practically the definition of span.)

I suspect what you were supposed to learn in class was that “A set of vectors in a vector space $V$ can only be a basis for $V$ if the set spans $V$.” (In addition, it must be a linearly independent set of vectors.)

In particular, the set $\{(1,1)\}$ is a linearly independent set, whether it is considered as a set of vectors in $\mathbb R^2$ or as a set of vectors in what you call $S$. The fact that $\{(1,1)\}$ does not span $\mathbb R^2$ does not tell you anything about the linear independence of the set. (And by the way, any set containing only one vector is a linearly independent set of vectors so long as that one vector is not zero.)

  • Slight correction in first paragraph: A set is not linearly independent if it spans the smallest space that it's contained in... – Chris Custer Sep 21 at 23:39
  • Chris: The OP’s statement was that a space can ONLY be linearly independent if it spans the vector space that it is in. The word “only” reverses the implication. “P only if Q” means “if P then Q.” It is true that a set of vectors is linearly independent ONLY if it spans its span, but as I note, that is not a useful fact. – Steve Kass Sep 22 at 0:37
  • 1
    Ok. I should have read it more carefully. It gives an implication whose conclusion is always true. Now I see. – Chris Custer Sep 22 at 1:01

Suppose $\{v_1,v_2,\dots,v_n\}$ is a basis of $V$. Then $$ \{v_2,\dots,v_n\} $$ doesn't span $V$, but it is linearly independent.

A set of vectors may fail to span $V$, but it can still be linearly independent.

The only case when you can infer linear dependence from the fact that a set fails to span $V$ is when the set has the same number of elements as the dimension of $V$.

In your case this condition is not satisfied, because $\{(1,1)\}$ consists of one element, but the dimension of $\mathbb{R}^2$ is two.

As egreg explained, your example is wrong.

Suppose $V$ and $W$ are both subspaces of a vector space $X$. Then any set of vectors $\{v_1, \ldots, v_k\}$ in the intersection $V \cap W$ that is linearly dependent as a subset of vector space $V$ is also linearly dependent as a subset of $W$. This is because both linear dependence statements are equivalent to the existence of scalars $c_1, \ldots, c_k$, not all $0$, such that $c_1 v_1 + \ldots + c_k v_k = 0$.

Firstly, the definition

... that a set of vectors can only be linearly independent if it spans the vector space that it is in.

is not correct.

A set $S$ of vectors is linearly dependent if there is some $v \in S$ that can be written as a linear combination of other vectors from $S$. It is linearly independent otherwise.

More formally, a set of vectors $S = \{v_1,\ldots,v_k\}$ is linearly independent if $$\sum_{i=1}^{k}\alpha_iv_i = 0 \quad \Rightarrow \quad \alpha_1= \cdots = \alpha_k=0$$ That is, the only way we can make zero from vectors in $S$ is by multiplying every vector by zero.

It is easy to see that if we remove vectors from linearly independent $S$, it stays linearly independent. Conversely, if we add a vector to $S$, it may become dependent. The maximum cardinality (number of elements) of a linearly independent set of vectors in a $n$-dimensional vectors space $V$ is $n$ and this set is then called a basis for $V$. It has the property that it spans entire $V$, that is - every $v \in V$ can be represented as a linear combination of vectors from basis.

Note that the cardinality of a set of vectors tells us nothing precise about its linear (in)dependence.

example 1: $\{(1,1),(2,2)\}$ is linearly dependent in $\mathbb{R}^2$ because $2 \cdot (1,1) = (2,2)$.
example 2: $\{(1,1)\}$ is linearly independent in $\mathbb{R}^2$ but not a basis for $\mathbb{R}^2$ because $|S| = 1 \neq 2 = dim \mathbb{R}^2$..
example 3: $\{(1,1),(2,3)\}$ is linearly independent in $\mathbb{R}^2$ and also a basis for $\mathbb{R}^2$ because $|S| = 2 = dim \mathbb{R}^2$.
example 4: $\{(1,1),(2,3),v\}$ must be linearly dependent in $\mathbb{R}^2$, for any $v \in \mathbb{R}^2$.
example 5: $\{(1,1)\}$ is linearly independent in your space $S := \{(x_1,x_1)| x_1 \in \mathbb{R}\}$ and a basis for it.

I think it's safe to say that if a set of vectors is linearly independent in one space, then it will be in any space that contains it.

This seems clear for any superspace, $W\supset V$...

Also, when a different space contains our space... For instance, the spaces $V=x$-axis, $W=xy$-plane and $U=xz$-plane. We simply get that $\{(1,0)\}$ is linearly independent in (actually) all three spaces.

The reason is that the defining condition, namely that only the trivial linear combination is zero, is independent of the surrounding space...

One could say, for instance, that a set is linearly independent $\iff$ it is a basis for the space it spans. Again, notice there is no mention of any (larger) surrounding space.

As pointed out by @egreg, if there are $n$ vectors, and if $\operatorname{dim} V=n$, then it is true (that the set is linearly independent iff it spans $V$).

In light of @Daniel Schepler's comments, we should note that changing the base field, or the vector space "structure", we can get some different results. This shows your question may be a little better than you originally thought (though it was an error that led you to it)...

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.