# If V=Lin(v_1, ... , v_k) then does it follow that W=Lin(φ(v_1), ... , φ(v_k)) ?

#### mathmari

##### Well-known member
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Hey!!

Let $V,W$ be $\mathbb{R}$-vector spaces and let $\phi:V\rightarrow W$ be a linear map. Let $1\leq k\in \mathbb{N}$ and let $v_1, \ldots , v_k\in V$.

If $V=\text{Lin}(v_1, \ldots , v_k)$ then does it follow that $W=\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$ ?

I have shown that $v_1, \ldots , v_k$ are linearly independent iff $\phi (v_1), \ldots , \phi (v_k)$ are linearly independent.

We have that $\phi (V)\subseteq W$ and since $\phi$ is linear we get that $\text{Lin}(\phi (v_1), \ldots , \phi (v_k))\subseteq W$, right?

The dimension of $V=\text{Lin}(v_1, \ldots , v_k)$ is at most $k$, then the dimension of $\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$ is also at most $k$.

But the dimesion of $W$ can be greater. Is that correct?

So we get $\subseteq$ and not necessarily $=$, or not?

What if $W=\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$ then does it follow that $V=\text{Lin}(v_1, \ldots , v_k)$ ?

#### Klaas van Aarsen

##### MHB Seeker
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Let $V,W$ be $\mathbb{R}$-vector spaces and let $\phi:V\rightarrow W$ be a linear map. Let $1\leq k\in \mathbb{N}$ and let $v_1, \ldots , v_k\in V$.

If $V=\text{Lin}(v_1, \ldots , v_k)$ then does it follow that $W=\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$ ?
Hey mathmari !!

Nope.

$W$ could have an independent vector that is not in the image.
That is, $\phi$ is not necessarily surjective.
It already suffices if $W$ is of higher dimension than $V$.

So we get $\subseteq$ and not necessarily $=$, or not?
Indeed.

What if $W=\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$ then does it follow that $V=\text{Lin}(v_1, \ldots , v_k)$ ?
Not necessarily I think.
If we drop the requirement that $V$ is the linear span of the $k$ vectors, then there could be an independent $v_{k+1}$ that still maps to $W$, couldn't it?

#### mathmari

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$W$ could have an independent vector that is not in the image.
That is, $\phi$ is not necessarily surjective.
It already suffices if $W$ is of higher dimension than $V$.
So a counterexample could be then the map $\phi:\mathbb{R}\rightarrow \mathbb{R}^2$ with $v\mapsto (v,0)$.
We have that $\mathbb{R}=\text{Lin}(1)$. Then $\phi (1)=(1,0)$ but $\mathbb{R}^2=\text{Lin}((1,0),(0,1))$.

Not necessarily I think.
If we drop the requirement that $V$ is the linear span of the $k$ vectors, then there could be an independent $v_{k+1}$ that still maps to $W$, couldn't it?
I haven't really understood that part.
So we suppose that $V$ is not the linear span of the $k$ vectors. That means that there must be also an independent $v_{k+1}$ in the set ?

#### Klaas van Aarsen

##### MHB Seeker
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So a counterexample could be then the map $\phi:\mathbb{R}\rightarrow \mathbb{R}^2$ with $v\mapsto (v,0)$.
We have that $\mathbb{R}=\text{Lin}(1)$. Then $\phi (1)=(1,0)$ but $\mathbb{R}^2=\text{Lin}((1,0),(0,1))$.
Yep.

I haven't really understood that part.
So we suppose that $V$ is not the linear span of the $k$ vectors. That means that there must be also an independent $v_{k+1}$ in the set ?
It's not that there "must be", but that there "could be".

#### mathmari

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It's not that there "must be", but that there "could be".
Suppose that $V\neq \text{Lin}(v_1, \ldots , v_k)$ that means that either $V\subset \text{Lin}(v_1, \ldots , v_k)$ or $\text{Lin}(v_1, \ldots , v_k)\subset V$.

It cannot hold that $V\subset \text{Lin}(v_1, \ldots , v_k)$ since $v_1, \ldots , v_k\in V$.

Therefore it must be $\text{Lin}(v_1, \ldots , v_k)\subset V$, so there is an element in $V$ that is not in $\text{Lin}(v_1, \ldots , v_k)$.
Let $v\in V$ with $v\notin \text{Lin}(v_1, \ldots , v_k)$.
Then $\phi (v)\in W=\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$. That means that $\phi (v)=c_1\phi (v_1)+\ldots +c_k \phi (v_k) \Rightarrow \phi (v)=\phi (c_1 v_1+\ldots +c_k v_k) \Rightarrow \phi (v-(c_1 v_1+\ldots +c_k v_k))=0$.

For a linear map it holds that $\phi(x)=0\iff x=0$, or not?

#### Klaas van Aarsen

##### MHB Seeker
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Suppose that $V\neq \text{Lin}(v_1, \ldots , v_k)$ that means that either $V\subset \text{Lin}(v_1, \ldots , v_k)$ or $\text{Lin}(v_1, \ldots , v_k)\subset V$.

It cannot hold that $V\subset \text{Lin}(v_1, \ldots , v_k)$ since $v_1, \ldots , v_k\in V$.

Therefore it must be $\text{Lin}(v_1, \ldots , v_k)\subset V$, so there is an element in $V$ that is not in $\text{Lin}(v_1, \ldots , v_k)$.
Let $v\in V$ with $v\notin \text{Lin}(v_1, \ldots , v_k)$.
Then $\phi (v)\in W=\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$. That means that $\phi (v)=c_1\phi (v_1)+\ldots +c_k \phi (v_k) \Rightarrow \phi (v)=\phi (c_1 v_1+\ldots +c_k v_k) \Rightarrow \phi (v-(c_1 v_1+\ldots +c_k v_k))=0$.
Yes.

For a linear map it holds that $\phi(x)=0\iff x=0$, or not?
Not necessarily. Only if $\phi$ is bijective, but that is not given.
If $\phi(x)=0$ that merely means that $x$ is in the kernel (aka null space) of $\phi$.

#### mathmari

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Not necessarily. Only if $\phi$ is bijective, but that is not given.
If $\phi(x)=0$ that merely means that $x$ is in the kernel (aka null space) of $\phi$.
So, if we find a non-bijective map we get a counterexample, or not?

#### Klaas van Aarsen

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So, if we find a non-bijective map we get a counterexample, or not?
Yup.

#### mathmari

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Let's consider $\phi:\mathbb{R}^2\rightarrow \mathbb{R}$ with $(v,w)\mapsto v$.

We have that $\mathbb{R}=\text{Lin}(1)=\text{Lin}(\phi (1,w))$, for each $w\in \mathbb{R}$. So it must be $\mathbb{R}^2=\text{Lin}((1,w))$.

Is that enough to get a contradiction?

#### Klaas van Aarsen

##### MHB Seeker
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Let's consider $\phi:\mathbb{R}^2\rightarrow \mathbb{R}$ with $(v,w)\mapsto v$.

We have that $\mathbb{R}=\text{Lin}(1)=\text{Lin}(\phi (1,w))$, for each $w\in \mathbb{R}$. So it must be $\mathbb{R}^2=\text{Lin}((1,w))$.

Is that enough to get a contradiction?
Yes. Because $\text{Lin}((1,w)) \sim \mathbb R$, which is different from $\mathbb{R}^2$.

#### mathmari

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Yes. Because $\text{Lin}((1,w)) \sim \mathbb R$, which is different from $\mathbb{R}^2$.
Why does it hold that $\text{Lin}((1,w)) \sim \mathbb R$ ? Isn't it $(1,w)\in \mathbb{R}^2$ ?

#### mathmari

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Do we use here the dimensions?

Or do we justify that as follows?

It must hold $\mathbb{R}^2=\text{Lin}((1,w))$. But $(0,1)\in \mathbb{R}^2$ cannot be written as a linear combination of $(1,w)$. Thereforeit doesn't hold that $\mathbb{R}^2=\text{Lin}((1,w))$.

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#### Klaas van Aarsen

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Why does it hold that $\text{Lin}((1,w)) \sim \mathbb R$ ? Isn't it $(1,w)\in \mathbb{R}^2$ ?

Do we use here the dimensions?
I meant that $\text{Lin}((1,w))$ is isomorphic with $\mathbb R$.
We can look at the dimension as well, which is 1.
Or do we justify that as follows?

It must hold $\mathbb{R}^2=\text{Lin}((1,w))$. But $(0,1)\in \mathbb{R}^2$ cannot be written as a linear combination of $(1,w)$. Thereforeit doesn't hold that $\mathbb{R}^2=\text{Lin}((1,w))$.
Yep. That also works.

#### Country Boy

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For a linear map it holds that $\phi(x)=0\iff x=0$, or not?

No! Absolutely not. The function mapping EVERY vector to 0 is a linear map:
f(u+ v)= 0=0+ 0= f(u)+ f(v) and f(av)= 0= a(0)= af(v).

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#### mathmari

##### Well-known member
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No! Absolutely not. The function mapping EVERY vector to 0 is a linear map:
f(u+ v)= 0=0+ 0= f(u)+ f(v) and f(av)= 0= a(0)= af(v).
I have shown the following, but now I am not sure if that is correct :

Suppose that $v_1, \ldots , v_k$ are linearly independent. That means that $\alpha_1 v_1+ \ldots + \alpha_kv_k=0 \Rightarrow \alpha_1=\ldots =\alpha_k=0$.
Let $\phi$ be a linear map, so we have that $\phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=\phi (0)\Rightarrow \phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=0 \Rightarrow \alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0$.
So we have that $\alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0\Rightarrow \alpha_1=\ldots =\alpha_k=0$, which means that $\phi (v_1), \ldots , \phi(v_k)$ are linearly independent.

And for the other direction:

Suppose that $\phi (v_1), \ldots , \phi (v_k)$ are linearly independent. That means that $\alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0 \Rightarrow \alpha_1=\ldots =\alpha_k=0$.
Since $\phi$ is a linear map we have that $\alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0\Rightarrow \phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=0\Rightarrow \alpha_1v_1+ \ldots + \alpha_kv_k=0$.
So we get $\alpha_1v_1+ \ldots + \alpha_kv_k=0\Rightarrow \alpha_1=\ldots =\alpha_k=0$, which means that $v_1, \ldots , v_k$ are linearly independent.

According to your previous note the second part, must be wrong, or not?

#### Klaas van Aarsen

##### MHB Seeker
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Suppose that $\phi (v_1), \ldots , \phi (v_k)$ are linearly independent.
We cannot assume this.
If $\phi$ is "merely" a linear map, it does not follow that the images of independent vectors are independent as well.
Country Boy 's example gives the counter example where $\phi$ maps every vector to the null vector.

#### mathmari

##### Well-known member
MHB Site Helper
We cannot assume this.
If $\phi$ is "merely" a linear map, it does not follow that the images of independent vectors are independent as well.
Country Boy 's example gives the counter example where $\phi$ maps every vector to the null vector.
So are both directions wrong?

So if $\phi:V\rightarrow W$ is a linear map and $V, W$ are vector spaces, and $v_1, \ldots , v_k\in V$ then both following implications are wrong, aren't they?
1. $v_1, \ldots , v_k$ linearly independent $\Rightarrow$ $\phi (v_1), \ldots , \phi (v_k)$ linearly independent
2. $\phi (v_1), \ldots , \phi (v_k)$ linearly independent $\Rightarrow$ $v_1, \ldots , v_k$ linearly independent

#### Klaas van Aarsen

##### MHB Seeker
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So are both directions wrong?

So if $\phi:V\rightarrow W$ is a linear map and $V, W$ are vector spaces, and $v_1, \ldots , v_k\in V$ then both following implications are wrong, aren't they?
1. $v_1, \ldots , v_k$ linearly independent $\Rightarrow$ $\phi (v_1), \ldots , \phi (v_k)$ linearly independent
2. $\phi (v_1), \ldots , \phi (v_k)$ linearly independent $\Rightarrow$ $v_1, \ldots , v_k$ linearly independent
Number 1 is incorrect for "just a linear map", but 2 is correct.

#### mathmari

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Number 1 is incorrect for "just a linear map", but 2 is correct.
Statement number 1:
Suppose that $v_1, \ldots , v_k$ are linearly independent. That means that $\alpha_1 v_1+ \ldots + \alpha_kv_k=0 \Rightarrow \alpha_1=\ldots =\alpha_k=0$.
Let $\phi$ be a linear map, so we have that $\phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=\phi (0)\Rightarrow \phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=0 \Rightarrow \alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0$.
But $\phi (v_1), \ldots , \phi(v_k)$ are not necessarily linearly independent, since $\phi( v_i)$ can be equal to $0$.

Counterexample: $\phi : x\mapsto 0$ then we get $\alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0$ also for all coefficients non-zero.

Proof for Statement number 2:
Suppose that $\phi (v_1), \ldots , \phi (v_k)$ are linearly independent. That means that $\alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0 \Rightarrow \alpha_1=\ldots =\alpha_k=0$.
Since $\phi$ is a linear map we have that $\alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0\Rightarrow \phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=0\Rightarrow \alpha_1v_1+ \ldots + \alpha_kv_k=0$.
So we get $\alpha_1v_1+ \ldots + \alpha_kv_k=0\Rightarrow \alpha_1=\ldots =\alpha_k=0$, which means that $v_1, \ldots , v_k$ are linearly independent.

Is everything correct?

#### Klaas van Aarsen

##### MHB Seeker
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Statement number 1:
Suppose that $v_1, \ldots , v_k$ are linearly independent. That means that $\alpha_1 v_1+ \ldots + \alpha_kv_k=0 \Rightarrow \alpha_1=\ldots =\alpha_k=0$.
Let $\phi$ be a linear map, so we have that $\phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=\phi (0)\Rightarrow \phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=0 \Rightarrow \alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0$.
But $\phi (v_1), \ldots , \phi(v_k)$ are not necessarily linearly independent, since $\phi( v_i)$ can be equal to $0$.

Counterexample: $\phi : x\mapsto 0$ then we get $\alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0$ also for all coefficients non-zero.
Correct.

Proof for Statement number 2:
Suppose that $\phi (v_1), \ldots , \phi (v_k)$ are linearly independent. That means that $\alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0 \Rightarrow \alpha_1=\ldots =\alpha_k=0$.
Since $\phi$ is a linear map we have that $\alpha_1\phi (v_1)+ \ldots + \alpha_k\phi (v_k)=0\Rightarrow \phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=0\Rightarrow \alpha_1v_1+ \ldots + \alpha_kv_k=0$.
So we get $\alpha_1v_1+ \ldots + \alpha_kv_k=0\Rightarrow \alpha_1=\ldots =\alpha_k=0$, which means that $v_1, \ldots , v_k$ are linearly independent.
I'm afraid that we cannot conclude that $\phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=0\Rightarrow \alpha_1v_1+ \ldots + \alpha_kv_k=0$.

#### mathmari

##### Well-known member
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I'm afraid that we cannot conclude that $\phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=0\Rightarrow \alpha_1v_1+ \ldots + \alpha_kv_k=0$.
Ok... So is the proof wrong or does the statement not hold?

#### Klaas van Aarsen

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Ok... So is the proof wrong or does the statement not hold?
The proof is wrong.
The particular step assumes that $\phi(x)=0$ implies that $x=0$, but that is not generally true.

Suppose $v_1\,\ldots,v_k$ are not linearly independent, then...

#### mathmari

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The proof is wrong.
The particular step assumes that $\phi(x)=0$ implies that $x=0$, but that is not generally true.

Suppose $v_1\,\ldots,v_k$ are not linearly independent, then...
Ahh ok!

So we have the following:

We suppose that $v_1, \ldots , v_k$ are not linearly independent. This means that $\alpha_1v_1+ \ldots +\alpha_iv_i+\ldots + \alpha_kv_k=0$ and at least one of the coefficients is non-zero, say $\alpha _i$.

Then we have the following:
\begin{align*}\alpha_1v_1+ \ldots +\alpha_iv_i+\ldots + \alpha_kv_k=0 &\Rightarrow \alpha_iv_i=-\alpha_1v_1- \ldots -\alpha_{i-1}v_{i-1}-\alpha_{i+1}v_{i+1}-\ldots - \alpha_kv_k \\ & \Rightarrow v_i=-\frac{\alpha_1}{\alpha_i}v_1- \ldots -\frac{\alpha_{i-1}}{\alpha_i}v_{i-1}-\frac{\alpha_{i+1}}{\alpha_i}v_{i+1}-\ldots - \frac{\alpha_k}{\alpha_i}v_k\end{align*}
We apply the map $\phi$ and we get:
\begin{align*}&\phi ( v_i)=\phi \left (-\frac{\alpha_1}{\alpha_i}v_1- \ldots -\frac{\alpha_{i-1}}{\alpha_i}v_{i-1}-\frac{\alpha_{i+1}}{\alpha_i}v_{i+1}-\ldots - \frac{\alpha_k}{\alpha_i}v_k\right ) \\ & \Rightarrow \phi ( v_i)=-\frac{\alpha_1}{\alpha_i}\phi (v_1)- \ldots -\frac{\alpha_{i-1}}{\alpha_i}\phi (v_{i-1})-\frac{\alpha_{i+1}}{\alpha_i}\phi (v_{i+1})-\ldots - \frac{\alpha_k}{\alpha_i}\phi (v_k) \\ & \Rightarrow \frac{\alpha_1}{\alpha_i}\phi (v_1)+ \ldots +\frac{\alpha_{i-1}}{\alpha_i}\phi (v_{i-1})+\phi ( v_i)+\frac{\alpha_{i+1}}{\alpha_i}\phi (v_{i+1})+\ldots + \frac{\alpha_k}{\alpha_i}\phi (v_k)=0\end{align*}
So we have a linear combination of $\phi (v_1), \ldots , \phi (v_k)$ that is equal to zero and not all coefficients are equal to zero.
This means that $\phi (v_1), \ldots , \phi (v_k)$ are not linearly independent, a contradiction.

So the assumption that $v_1, \ldots , v_k$ are not linearly independent is wrong. That means that $v_1, \ldots , v_k$ are linearly independent.

Staff member
Yep.

#### mathmari

##### Well-known member
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Great!! Thank you!!