# Show that $T(\mathbf x)=\mathbf 0$ has a nontrivial solution

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

• 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. – Vim Jan 23 '15 at 9:47
• $x=0$ is the trivial solution. You have proven it has a nontrivial solution. – Winther Jan 23 '15 at 9:59
• $Tx = 0$ can have lots of nontrivial solutions. take $T = 0$ to see an extreme case of this. – abel Jan 23 '15 at 11:04
• 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. – Vim Jan 23 '15 at 16:25

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

• 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 $\ne 0$ ? – Erikzzz Jan 23 '15 at 16:09
• 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" – Vim Jan 23 '15 at 16:15
• 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. – Vim Jan 23 '15 at 16:20

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

• if $c_1\mathbf{u}+c_2\mathbf{v} \ne \mathbf{0}$ how can $T(c_1\mathbf{v}+c_2\mathbf{u})=c_1 T(\mathbf{v})+c_2 T(\mathbf{u})=\mathbf{0}.$ ? – Erikzzz Jan 23 '15 at 16:13
• 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 $c_1 T(\mathbf{u})+c_2 T(\mathbf{v})=0$ – marwalix Jan 23 '15 at 16:41

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 $$T(av+bu)=aT(v)+bT(u)=0.$$

• 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 – Erikzzz Jan 23 '15 at 16:03
• 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 – Erikzzz Jan 23 '15 at 16:04

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.

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