# Thread: Llinear transformations: Help with proving linear independency

1. ## Llinear transformations: Help with proving linear independency

Hi guys.

I had the following statement to prove:

Let $\displaystyle \varphi :V\rightarrow W$ be a linear transformation, and $\displaystyle v_1, v_2,...v_n\in V$ a set of vectors.
If $\displaystyle v_1, v_2,...v_n$ are linear independent, and $\displaystyle \varphi$ is one-to-one, then $\displaystyle \varphi(v_1), \varphi(v_2),...,\varphi(v_n)$ are also linear independent.

that's how I proved it:
$\displaystyle v_1, v_2,...v_n$ are linear independent, so if:

$\displaystyle \alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n=0$

then: $\displaystyle \alpha_1=\alpha_2=...=\alpha_n=0$

now, since $\displaystyle \varphi$ is a linear transformation, it holds that $\displaystyle \varphi(\vec{0}_V)=\vec{0}_W$, therefore:

$\displaystyle \varphi(\vec{0}_V)=\varphi(\alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n)=$
$\displaystyle \alpha_1\varphi(v_1)+\alpha_2\varphi(v_2)+...+ \alpha_n\varphi(v_n)=\vec{0}_W$

and since we know that $\displaystyle \alpha_1=\alpha_2=...=\alpha_n=0$, then $\displaystyle \varphi(v_1), \varphi(v_2),...,\varphi(v_n)$ are linear independent.

my question here is: why should I care that $\displaystyle \varphi$ is one-to-one?

any help would be greatly appreciated.

2. ## Re: Llinear transformations: Help with proving linear independency

Originally Posted by Stormey
Let $\displaystyle \varphi :V\rightarrow W$ be a linear transformation, and $\displaystyle v_1, v_2,...v_n\in V$ a set of vectors.
If $\displaystyle v_1, v_2,...v_n$ are linear independent, and $\displaystyle \varphi$ is one-to-one, then $\displaystyle \varphi(v_1), \varphi(v_2),...,\varphi(v_n)$ are also linear independent.

now, since $\displaystyle \varphi$ is a linear transformation, it holds that $\displaystyle \varphi(\vec{0}_V)=\vec{0}_W$, therefore:

$\displaystyle \varphi(\vec{0}_V)=\varphi(\alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n)=$
$\displaystyle \alpha_1\varphi(v_1)+\alpha_2\varphi(v_2)+...+ \alpha_n\varphi(v_n)=\vec{0}_W$

and since we know that $\displaystyle \alpha_1=\alpha_2=...=\alpha_n=0$, then $\displaystyle \varphi(v_1), \varphi(v_2),...,\varphi(v_n)$ are linear independent.

my question here is: why should I care that $\displaystyle \varphi$ is one-to-one?

any help would be greatly appreciated.
What you did is not a proof.
You should start with, suppose that $\displaystyle \alpha_1\varphi(v_1)+\alpha_2\varphi(v_2)+...+ \alpha_n\varphi(v_n)=\vec{0}_W$
From which you must show that each $\displaystyle \alpha_k$ must equal zero.

Now as you said $\displaystyle \varphi(\vec{0}_V)=\varphi(\alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n)$ is also that sum.

But that means $\displaystyle \vec{0}_V=\alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n$ because $\displaystyle \varphi$ is one-to-one.

Are you done? Why?

3. ## Re: Llinear transformations: Help with proving linear independency

Hi Plato, and thanks for the help.

Yeah, we're done, if $\displaystyle \vec{0}_V=\alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n$, then $\displaystyle \alpha_1=\alpha_2=...=\alpha_n=0$
because $\displaystyle v_1, v_2,...v_n$ are linear independent, and therefore $\displaystyle \varphi(v_1), \varphi(v_2),...,\varphi(v_n)$ are linear independent.

what was wrong with mine?
why doesn't it considered a valid proof?

4. ## Re: Llinear transformations: Help with proving linear independency

Originally Posted by Stormey
what was wrong with mine?
why doesn't it considered a valid proof?
You started out with what you wanted to prove. So you went nowhere.
That is why you never had to use one-to-one.
You are to show that $\displaystyle \left\{ {\varphi ({v_k})} \right\}_{k = 1}^n$ is an independent set.
So you start with a linear combination equaling the zero vector. Not the other way round.