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 $T(c_1\mathbf{v}+c_2\mathbf{u})=c_1 T(\mathbf{v})+c_2 T(\mathbf{u})=\mathbf{0}.$ ?
$\endgroup$ 44 Answers
$\begingroup$Why? A non-trivial solution means $x \ne 0$. You maybe confused with the concepts.
$\endgroup$ 3 $\begingroup$No $c_1\mathbf{u}+c_2\mathbf{v}$ is not $0$ because $\mathbf{u}$ and $\mathbf{v}$ are linearly independant
$\endgroup$ 2 $\begingroup$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 $$T(av+bu)=aT(v)+bT(u)=0.$$
$\endgroup$ 2 $\begingroup$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.
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