0
$\begingroup$

This question in my book

Let $T:\mathbb{R}^n\to \mathbb{R}^m$ be a linear transformation. Suppose $\{\mathbf{u}, \mathbf{v}\}$ is a linearly independent set, but $\{T(\mathbf u), T(\mathbf v)\}$ is a linearly dependent set. Show that $T(\mathbf x)$ has a nontrivial solution. [Hint: Use the fact that $c_1\,T(\mathbf u) + c_2\,T(\mathbf v) = \mathbf 0$ for some weights $c_1$ and $c_2$, not both zero.]

This answer in the solution manual is

Suppose that $\{\mathbf u,\mathbf v\}$ is a linearly independent set in $\mathbb\{R\}^n$ and yet $T(\mathbf u)$ and $T(\mathbf v)$ are linearly dependent. Then there exist weights $c_1, c_2$ not both zero, such that $c_1 \, T(\mathbf u) + c_2 \, T(\mathbf v) = \mathbf 0$. Because $T$ is linear, $T(c_1\mathbf u + c_2\mathbf v) = \mathbf 0$. That is, the vector $\mathbf x = c_1\mathbf u + c_2\mathbf v$ satisfies $T(\mathbf x) = \mathbf 0$. Furthermore, $\mathbf x$ cannot be the zero vector, since that would mean that a nontrivial linear combination of $\mathbf u$ and $\mathbf v$ is zero, which is impossible because $\mathbf u$ and $\mathbf v$ are linearly independent. Thus, the equation $T(\mathbf x) = \mathbf 0$ has a nontrivial.

Now I'm confused.

If $c_1\mathbf{u}+c_2\mathbf{v} \ne \mathbf{0}$ how can $T(c_1\mathbf{v}+c_2\mathbf{u})=c_1 T(\mathbf{v})+c_2 T(\mathbf{u})=\mathbf{0}.$ ?

$\endgroup$
  • $\begingroup$ Another thing: in my textbook a linear transformation specifically means a linear mapping from a linear space $V$ to itself. So I don't think it proper to call $T$ a linear transformation in your context. Just linear mapping is ok. $\endgroup$ – Vim Jan 23 '15 at 9:47
  • $\begingroup$ $x=0$ is the trivial solution. You have proven it has a nontrivial solution. $\endgroup$ – Winther Jan 23 '15 at 9:59
  • $\begingroup$ $Tx = 0$ can have lots of nontrivial solutions. take $T = 0$ to see an extreme case of this. $\endgroup$ – abel Jan 23 '15 at 11:04
  • $\begingroup$ Why not? Have you learned solving linear systems like $Ax=b (b\ne 0)$ and $Ax=0$? For the latter, which is called a homogeneous equation, if and only if $rank(A)<n$ where $n$ denotes the number of variables in the system, say, the number of columns of $A$, then it has non-trivial solutions. $\endgroup$ – Vim Jan 23 '15 at 16:25
2
$\begingroup$

Why? A non-trivial solution means $x \ne 0$. You maybe confused with the concepts.

$\endgroup$
  • $\begingroup$ That could definitely be the case. In my world a non-trivial solution means that the weights are not both zero for the equation $c_1\mathbf u + c_2\mathbf v = \mathbf 0$. I do not know what it means when the equation $\ne 0$ ? $\endgroup$ – Erikzzz Jan 23 '15 at 16:09
  • $\begingroup$ Well that is simply a solution but has nothing to do with trivial or not. In fact I think we call 0 the trivial solution just because $A 0=0$ is too "plain" and even doesn't depend on $A$ at all, but that's different for a non-zero solution, which is perhaps therefore called "non-trivial" $\endgroup$ – Vim Jan 23 '15 at 16:15
  • $\begingroup$ What do you mean by "the equation $\ne 0$"? An equation like $Ax=b$ where $b \ne 0$? It is literally called a non-homogeneous equation. $\endgroup$ – Vim Jan 23 '15 at 16:20
2
$\begingroup$

No $c_1\mathbf{u}+c_2\mathbf{v}$ is not $0$ because $\mathbf{u}$ and $\mathbf{v}$ are linearly independant

$\endgroup$
  • $\begingroup$ if $c_1\mathbf{u}+c_2\mathbf{v} \ne \mathbf{0}$ how can $T(c_1\mathbf{v}+c_2\mathbf{u})=c_1 T(\mathbf{v})+c_2 T(\mathbf{u})=\mathbf{0}.$ ? $\endgroup$ – Erikzzz Jan 23 '15 at 16:13
  • $\begingroup$ The starting point is because $T(\mathbf{u})$ and $T(\mathbf{v})$ are linearly dependant one can find $c_1\neq 0$ and $c_2\neq 0$ such that $c_1 T(\mathbf{u})+c_2 T(\mathbf{v})=0$ $\endgroup$ – marwalix Jan 23 '15 at 16:41
0
$\begingroup$

Since $u,v$ are not lineary dependent $av+bu= 0$ if and only if $a=b=0$. Since, $T(v),T(u)$ are lineary dependent there exist $a,b$ not both of them zero such that $aT(v)+bT(u)=0$. Therefore $av+bu\neq 0$ and $$T(av+bu)=aT(v)+bT(u)=0.$$

$\endgroup$
  • $\begingroup$ This is the contradiction I was talking about. 1) I understand that because u,v are independent there should exist weights in a linear combination of these vectors not both being zero 2) I understand that by the same in-dependency relationship none of the two vectors can be the zero vector 3) I understand that this implies that the outcome of the linear combination of weights and the two vectors can never be zero $\endgroup$ – Erikzzz Jan 23 '15 at 16:03
  • $\begingroup$ What I don't understand is when a linear transformation is made from that same linear combination the outcome can all of a sudden be equal to zero with a non trivial solution, I don't understand how this is possible and how it is shown that this is true, looks like some rules are being applied back and forth but not "real" proof $\endgroup$ – Erikzzz Jan 23 '15 at 16:04
0
$\begingroup$

We can use rank nullity theorem for proof. Only we have to show that dim of nullity is not zero. Suppose $T:\mathbb{R}^n \rightarrow \mathbb{R}^n.$ $n=dim (Im T)+ nullity.$ Now if $\{T(u), T(v)\}$ are dependent then $dim(Im T)<n.$ Then $nullity \neq 0.$ This proved the result.

$\endgroup$
  • $\begingroup$ I'm sorry I do not know this way of making a proof. also I do not know the rank nullity theorem $\endgroup$ – Erikzzz Jan 23 '15 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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