# Is the image $T(S)$ of a linearly dependent set $S$ under a linear transformation $T$ linearly dependent?

My attempt: $\exists c_1,\cdots,c_k\in F,~\exists v_1,\cdots,v_k\in V,~c_1v_1+\cdots+c_kv_k=0$, where $c_i$ are not all zero. Then $c_1T(v_1)+\cdots+c_kT(v_k)=0$. However, we can not deduce that $T(v_1),\cdots,T(v_k)$ are linearly dependent, since there may be $i,j$ such that $T(v_i)=T(v_j)$, so the coefficients $c_i$ and $c_j$ have the chance to sum up to zero!

For example, even though we can get that $2T(v_1)+0T(v_2)-2T(v_3)=0$, we cannot say that $T(v_1),T(v_2),T(v_3)$ is linearly dependent, since $T(v_1)$ and $T(v_3)$ could probably be the same!

• The condition is that the $c_i$ are not (individually) all zero; whether the sum of two or more or the $c_i$ are zero is irrelevant. – Marc van Leeuwen Sep 2 '17 at 12:00
• @MarcvanLeeuwen I don't understand what you mean. Can you give some more detailed comments or a complete proof? :) – Eric Sep 2 '17 at 12:01
• For linear dependence( independence) , you should think of families of vectors. Repeats are allowed. – orangeskid Sep 2 '17 at 14:23

You are confused with the definition of linear dependence. A $k$-tuple of vectors (like the $T(v_i)$) is linearly dependent as soon as there exists some non-trivial linear combination of them that gives zero. You've got such a combination here, the "non-trivial" is true because the $c_i$ were assumed from the outset to not all be zero. End of story.

To cut short on the comments, the statement as in your title, about linearly dependent sets is just not true. Mapping $T:\Bbb R^3\to\Bbb R^2$ by dropping the final coordinate is certainly linear, and the image under $T$ of the linearly dependent set $\{\,(1,0,z)\mid z\in\Bbb R\,\}$ is the linearly independent singleton set $\{(1,0)\}$. (Here $S$ is infinite, but taking any $3$ or more distinct vectors from it gives a linearly dependent finite set with the same property). The statement is true however when stated for $k$-tuples of vectors (linearly dependent $k$-tuple gives another linearly dependent $k$-tuple), and that is how it should have been stated.

The problem is that by laziness one often says "set" instead of "$k$-tuple" (and moreover, one much more often considers linearly independent vectors than dependent ones, in which case the laziness is relatively harmless, since being independent implies being distinct).

• For example, even though we can get that $2T(v_1)+0T(v_2)-2T(v_3)=0$, we cannot say that $T(v_1),T(v_2),T(v_3)$ is linearly dependent, since $T(v_1)$ and $T(v_3)$ could probably be the same, it is to say, what we have shown is $2T(v_1)-2T(v_1)+0T(v_2)=0$, this does not mean that $T(v_1)$ and $T(v_3)$ are linearly dependent right? – Eric Sep 2 '17 at 12:07
• If a set contains two identical vectors, it's linearly dependent, for $1 a + (-1)b$ is a nontrivial linear combination of the two (identical) vectors. – John Hughes Sep 2 '17 at 12:08
• The fact that $2T(v_1)+0T(v_2)-2T(v_3)=0$ most definitely implies that $T(v_1), T(v_2), T(v_3)$ are linearly dependent. That $T(v_1)$ and $T(v_3)$ could be the same (as in fact they will) does not hurt, it only makes the triple more obviously linearly dependent (because two among them are already linearly dependent, worse: equal). – Marc van Leeuwen Sep 2 '17 at 12:10
• In set theory, $\{a,a\}$ is the same as $\{a\}$. So do you mean $\{(1,2)\}$ is linearly depedent? (Since it is the same as $\{(1,2),(1,2)\}$). The definition of linearly dependency of a set $S$, as I seen in many textbooks, don't mentioned the duplication-situation as you said. – Eric Sep 2 '17 at 12:12
• I'm not saying you cannot define linear dependence of sets (which is useful, mostly to define linearly independent sets, possibly infinite), it is just not the most basic useful notion. That basic notion is "linearly dependent $k$-tuples of vectors". One must be able to say for instance, "if $v=w$ then $v,w$ are linearly dependent" (meaning the $2$-tuple $[v,w]$ is dependent), or "let $u,v,w$ be linearly independent vectors", which is not the same as "let $\{u,v,w\}$ be a set of linearly independent vectors" (because that does not imply the mentioned set has three elements). – Marc van Leeuwen Sep 2 '17 at 12:36

The problem is simple, and your proof is actually a correct one, but your confusion is justified, because the definition of "linear combination" given in most books is a little ambiguous.

They usually say "a linear combination is an expression of the form $a_1 v_1 + \ldots + a_k v_k$", and that it's nontrivial if at least one of the $a_i$ is nonzero. Tthe problem is that if $k = 2$, and $v_1 = -v_2$, then $$1 v_1 + 1 v_2$$ appears to be a nontrivial linear combination, but a little algebra reduces it to be $$0 v_1$$ which is a trivial linear combination, so which is it?

The real problem here is the notion of "an expression of the form XYZ" isn't really well-defined, or at least is a meta-idea rather than one about the values represented by expressions.

The right solution is to give an unambiguous definition of the notion. Here goes:

"A linear combination of the sequence of vectors $v_1, \ldots, v_k$ is a sequence of coefficients $c_1, \ldots, c_k$ in the underlying field of the vector space.

"We say that the linear combination is trivial if all $c_i$ are zero; we say that a nontrivial linear combination "is zero" if the sum $$c_1 v_1 + \ldots + c_k v_k = 0$$

"In general, in referring to linear combinations, we are sloppy in distinguishing between the combination itself (i.e., the sequence of coefficients), and what might be called the "value" of the combination, the vector $$c_1 v_1 + \ldots + c_k v_k = 0$$ which is often what's needed in practice."

With that definition, the combination $(1, 1)$ in my example is clearly nontrivbial, but the associated value, $1 v_1 + 1 v_2$, turns out to be zero.

I've never seen a book that made this kind of distinction, alas.

• Suppose $v_1=-v_2$, then isn't $1 v_1 - 1 v_2=v_1+(-v_2)=v_1+v_1=2v_1$? – Eric Sep 2 '17 at 13:10
• Fixed, thanks. I was thinking two things at once. ("I need two cancelling vectors", "I need coefficients that'll make things cancel") and both won. :( – John Hughes Sep 2 '17 at 14:02
• OK. Well, I haven't seen books metioning "nontrivial" or "trivial" linear combinations. So I don't concern about how "nontrivial" or "trivial" linear combinations are ambiguous, because they don't appear in my textbooks; my texbooks defined linear (in)dependence without using nontrivial or trivial combination. – Eric Sep 2 '17 at 14:13
• "Let $v_1,\cdots,v_k$ be distinct vectors. If $\forall c_1,\cdots,c_k\in F,(c_1v_1+\cdots+c_kv_k=0\to c_1=\cdots=c_k=0)$, then we called $v_1,\cdots,v_k$ linearly independent." Is this definition ambiguous? Or set version: "Let $S\subseteq V$. If there exists finitely distinct vectors $v_1,\cdots,v_k\in S$ and $c_1,\cdots,c_k\in F$ not all zero such that $c_1v_1+\cdots+c_kv_k=0$, then we called $S$ L.D. otherwise L.I." Is this definition ambiguous or having any defect? – Eric Sep 2 '17 at 14:18
• (ps: the set version I just metioned minutes ago is exactly refering to uncountably infinite set version. Some authors didn't define it, but some did.) Anyway, it is soooo weird that many textbooks didn't pay attention on this issue. And I think it is important, maybe not for engineers, but for students in mathematics deparments. Anyway, thanks for your patient replying. – Eric Sep 3 '17 at 18:03

(My post is too long to fit into a comment.)

I guess you need to think through your example. If the characteristic of the field is not 2, and $2v_1+0v_2-2v_3=0$, then $v_1=v_3$, so $S=\{v_1, v_2\}$.

The right way to reason in linear algebra is to reason with collections of vectors that preserve multiplicities. These collections may be:

1. bags,
2. tuples (also known as lists),
3. indexed families.

A family of vectors in $V$ indexed by $I$ is just a function of type $I\to V$. “Family” is a fancy term which is often used. I guess that families are the most convenient.

The linear combination of a family of vectors $a$ with coefficients $c$ is $\sum_{i\in I}c(i)\cdot a(i)$. $c$ is a function of type $I\to F$. $I$ is a finite set. Linear combination per se is a function that takes a family of vectors, a family of scalars, and returns a scalar. There is the Euclidean vector space over $I\to F$, so linear combination is linear in its last argument. $a$ is linearly independent iff $\forall c(\sum_{i\in I}c(i)\cdot a(i)=0 \to \forall i(c(i)=0))$, equivalently, the kernel of linear combination of $a$ is $\{0\}$.

The definition of linearly independent family of vectors $a:I\to V$ when $a$ is infinite is as follows, adapted from the Wikipedia. $a$ is linearly independent iff for every finite subset $J$ of $I$, $a\upharpoonright J$ ($a$ restricted to $J$) is linearly independent.

So, some textbooks are sloppy. We can't fix textbooks. I recommended you “Linear Algebra Done Right” by Sheldon Axler. The author pays attention to formalities. This is why I am studying that textbook. If you are bound to a specific textbook, replace “set” with “collection” in your mind and prove modified theorems by yourself. IMHO, this situation is typical, actually. Textbooks are not ideal; they contain typos, mistakes, and ugly proofs. Maybe some day you will write your own better textbook. ☺

OK, let's get back to the original question: Is the image $T(S)$ of a linearly dependent set $S$ under a linear transformation $T$ linearly dependent?

We must to notice that, there are many possible definitions of span, linear combination and linearly dependence/independence, each of them again have several versions, one talking on $v_1,v_2,\cdots,v_k$(this vector can be required to be totally distinct, or not necessarily distinct, which means there are again two possible choices one may choose), one talking on a set, especially having the ability to talk on infinite set of vectors, such as $\{1,x,x^2,x^3,x^4,\cdots\}$.

First, suppose we adopt the version of span, linear combination and linearly dependence/independence that talking about $v_1,\cdots,v_k$(not set) which are not necessarily distinct. Then in this case, for example, we may say $v,v$ is linearly dependent. Then it is true that for any $v_1,\cdots,v_k\in V$ not necessarily distinct, if they are linearly dependent, then $T(v_1),\cdots,T(v_k)$(possible duplicated here) are linearly dependent.

However, the important point is that, the original proposition in the title, as one might not expect, is false. This is because the set language has an essence that repeating the same object in a set doesn't behave differently than the original set. I just thought and found an unbeatable counterexample.

Let $T:\Bbb R^3\to\Bbb R^3$ defined by $T(\mathbb{x})=\begin{bmatrix}3&0&0\\0&2&0\\0&0&0\end{bmatrix}\mathbb{x}$.

Let $S=\{(1,2,3),(1,2,1),(1,3,7),(1,3,5)\}$. Then since $S\subseteq \Bbb R^3$, then $S$ must be linearly dependent.

Next, by the definition of image of a set under a function, we see that $T(S)=\{(3,4,0),(3,6,0)\}$. Notice that this result followed by the definition of image of a set under a function, which has nothing to do with linear combination or span or linearly (in)dependence. So it doesn't depend on what kind of defintion you wish to choose! Then, as you can see, $T(S)$ is literally and undoubtfully linearly independent. So the original proposition in the title is FALSE.