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Let T: V->W be a linear transformation between vector spaces over F and let $v_1,v_2...,v_n$ elements of V

if ${Tv_1, Tv_2, ..., Tv_n}$ is linearly independent, prove that ${v_1,v_2,...,v_n}$ linearly independent as well

So far I have,

$a_1Tv_1 + ... + a_nTv_n = 0$ where $a_1,...,a_n = 0$ then, $T(a_1v_1 + ...+ a_nv_n) = 0$

is it enough now to say that $a_1v_1 + ... + a_nv_n = 0$ and since ${a_1,...,a_n}=0$

${v_1,v_2,...,v_n}$ linearly independent

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This is wrong. You can't assume that the $a_n$'s are $0$. You are supposed to prove that!

Take scalars $a_1,\ldots,a_n$ such that $a_1v_1+\cdots+a_nv_n=0$; you want to prove that each $a_i$ is $0$. But\begin{align}a_1v_1+\cdots+a_nv_n=0\implies&T(a_1v_1+\cdots+a_nv_n)=0\\\iff&a_1T(v_1)+\cdots+a_nT(v_n)=0\\\implies&a_1=\cdots=a_n=0,\end{align}since $T(v_1),\ldots,T(v_n)$ are linearly independent.

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  • $\begingroup$ I thought you could assume an's are 0 since we know that T(v1),...,T(vn) are linearly independent, then work from there $\endgroup$ – Madnobleman Mar 5 '18 at 22:36
  • $\begingroup$ @Madnobleman If all $a_n$'s were $0$, then $a_1v_1+\cdots+a_nv_n$ with be $0$ automatically. $\endgroup$ – José Carlos Santos Mar 5 '18 at 22:38
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We want to prove that $v_1,\dots,v_n$ are independent, so start out from their combination instead: Suppose $$a_1v_1+\dots+a_nv_n=0$$

Now apply $T$ on both sides, use its linearity, and then use the condition, in order to conclude that all $a_i=0$.

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Actually, you wrote things backward. Why did you take $a_1Tv_1+\dots+a_nTv_n=0$ with $a_1,\dots,a_n=0$? Actually, I can really take a linearly dependent family of vectors $v_1=v_2=\dots=v_n=v$ and say "take $a_1v_1+\dots+a_nv_n=0$ with $a_1=\dots=a_n=0$. Then the family is independent because $a_i=0$".

The casual way you prove $P\implies Q$ is that you assume $P$ is true and prove $Q$.

You want to prove that

$Tv_1,\dots,Tv_n$ are linearly independent $\implies$ $v_1,\dots,v_n$ are linearly independent.

So, let's go. Assume that $Tv_1,\dots,Tv_n$ are linearly independent. You tried to translate what it means, which is a good thing, but used it in a quite wrong way. Your assumption means that:

If $a_1,\dots,a_n$ are real numbers such that $a_1Tv_1+\dots+a_nTv_n=0$, then $a_1=\dots=a_n=0$.

This is your assumption. This is a hypothesis. You have nothing to use it at for the moment.

Now what's your goal? You want to show that $v_1,\dots,v_n$ are linearly independent. Translate this:

If $a_1,\dots,a_n$ are real numbers such that $a_1v_1+\dots+a_nv_n=0$, then $a_1=\dots=a_n=0$.

So let's prove this. Let $a_1,\dots,a_n$ be real numbers such that $a_1v_1+\dots+a_nv_n=0$. Why is $a_1=\dots=a_n=0$? I'll let you find out.

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  • $\begingroup$ I understand how showing the proof starting with $a_1v_1+⋯+a_nv_n=0$ is better and then you can conclude that $a_1, ..., a_n = 0$. My confusion lies with why can't you say that $a_1,...a_n = 0$ just from the fact that $Tv_1,Tv_2,...,Tv_n$ is linearly independent when you have $a_1Tv_1+⋯+a_nTv_n=0$ $\endgroup$ – Madnobleman Mar 5 '18 at 22:49
  • $\begingroup$ "why can't you say that $a_1,\dots, a_n=0$ from the fact that $Tv_1,\dots,Tv_n$ is linearly independent and $a_1Tv_1+\dots+a_nTv_n=0$". You CAN say that. This is true. But when in your proof you say "$a_1Tv_1 + ... + a_nTv_n = 0$ where $a_1,...,a_n = 0$ ", where did the $a_1,\dots,a_n$ come from? It's just like you really take them to be $0$ and then put them into $a_1v_1+\dots a_nv_n=0$. All you showed is that if you take $a_1=\dots=a_n=0$, then you will have $a_1v_1+\dots+a_nv_n=0$ and $a_1=\dots=a_n=0$. $\endgroup$ – Scientifica Mar 6 '18 at 17:21
  • $\begingroup$ In other words, you showed that $a_1=\dots=a_n=0$ $\implies$ ($a_1v_1+\dots+a_nv_n=0$ and $a_1=\dots=a_n=0$). But this is not what "$v_1,\dots,v_n$ are linearly independent" means. It means that whenever you take $a_1,\dots,a_n$ such that $a_1v_1+\dots+a_nv_n=0$, you must have $a_1=\dots=a_n=0$. It's like if you take ANY $a_i$ such that $\sum a_iv_i=0$ and then find out that $a_i=0$. $\endgroup$ – Scientifica Mar 6 '18 at 17:23
  • $\begingroup$ Think of this like proving that a function $f:D\to\mathbb{R}$ is injective. You don't say "let $x,y\in D$ such that $\exp{x}=\exp{y}$. Since $\exp$ is injective, $x=y$. To reach injectivity of $f$, it is enough to now say that $f(x)=f(y)$, and since $x=y$, we get the desired result". What you start with to prove $P\implies Q$ is $P$ (unless you're doing a proof by contraposition or contradiction...), so you say "$x,y\in D$ such that $f(x)=f(y)$". That's what you say first. Let's take an example: $D=\mathbb{R}^+$, $f=\exp\circ\ln$, then you say "I have $f(x)=f(y)$. $\endgroup$ – Scientifica Mar 6 '18 at 17:27
  • $\begingroup$ ... this means $\exp(\ln(x))=\exp(\ln(y))$ and since $\exp$ is injective we get $\ln(x)=\ln(y)$ and since $\ln$ is injective we get $x=y$". (yeah $\exp\circ\ln=\text{id}_{\mathbb{R}^+}$ so that's kinda trivial but that's an example). $\endgroup$ – Scientifica Mar 6 '18 at 17:29
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No. You want $a_1\cdot v_1+\dots+a_n\cdot v_n=0 \implies a_1=a_2=\dots=a_n=0$.

Now, $T(0)=0$, since $T$ is linear. Also by linearity, $a_1T(v_1)+\dots+a_nT(v_n)=0$.

Now conclude $a_1=a_2=\dots =a_n=0$ (by linear independence of $T(v_1),\dots ,T(v_n)$...)

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