# Does Linear Independence in V Imply the Same in W?

• MHB
• mathmari
In summary: Yes. Because $\text{Lin}((1,w)) \sim \mathbb R$, which is different from $\mathbb{R}^2$.In summary, the conversation discusses the relationship between linear maps and vector spaces. It is shown that if the vector space $V$ is the linear span of $k$ vectors, then the vector space $W$ is also the linear span of the images of those vectors under a linear map $\phi$. However, it is also shown that this does not necessarily hold in the opposite direction. The conversation concludes with a counterexample using a non-bijective linear map.
mathmari
Gold Member
MHB
Hey! :giggle:

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 $V=\text{Lin}(v_1, \ldots , v_k)$ ?

:unsure:

mathmari said:
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. (Shake)

$W$ 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 $V$.

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

Indeed. (Nod)

mathmari said:
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?

Klaas van Aarsen said:
$W$ 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 $\mathbb{R}^2=\text{Lin}((1,0),(0,1))$.

:unsure:
Klaas van Aarsen said:
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 ?

:unsure:

mathmari said:
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. (Nod)
mathmari said:
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".

Klaas van Aarsen said:
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? :unsure:

mathmari said:
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. (Nod)

mathmari said:
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 $\phi$.

Klaas van Aarsen said:
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? :unsure:

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

Klaas van Aarsen said:
Yup.

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? :unsure:

mathmari said:
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 $\mathbb{R}^2$.

Klaas van Aarsen said:
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 $(1,w)\in \mathbb{R}^2$ ? :unsure:

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 $\mathbb{R}^2=\text{Lin}((1,w))$.

:unsure:

Last edited by a moderator:
mathmari said:
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.
mathmari said:
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. (Nod)

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

Last edited:
Country Boy said:
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? :unsure:

mathmari said:
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.

Klaas van Aarsen said:
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

:unsure:

mathmari said:
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.

Klaas van Aarsen said:
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? :unsure:

mathmari said:
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. (Nod)

mathmari said:
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 $\phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=0\Rightarrow \alpha_1v_1+ \ldots + \alpha_kv_k=0$. (Shake)

Klaas van Aarsen said:
I'm afraid that we cannot conclude that $\phi (\alpha_1v_1+ \ldots + \alpha_kv_k)=0\Rightarrow \alpha_1v_1+ \ldots + \alpha_kv_k=0$. (Shake)

Ok... So is the proof wrong or does the statement not hold? :unsure:

mathmari said:
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. (Shake)

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

Klaas van Aarsen said:
The proof is wrong.
The particular step assumes that $\phi(x)=0$ implies that $x=0$, but that is not generally true. (Shake)

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.

(Malthe)

Yep. (Nod)

Klaas van Aarsen said:
Yep. (Nod)

Great! Thank you! (Sun)

## 1. Does the statement hold true for any vector space V and linear transformation φ?

Yes, the statement holds true for any vector space V and linear transformation φ. This is because the definition of linear span (Lin) and linear transformation remain consistent regardless of the specific vector space or transformation being used.

## 2. Can the vectors v1, ..., vk be any set of vectors in V?

Yes, the vectors v1, ..., vk can be any set of vectors in V. As long as they span the vector space V, the statement will hold true.

## 3. Is it possible for W to be a proper subset of Lin(φ(v1), ..., φ(vk))?

No, it is not possible for W to be a proper subset of Lin(φ(v1), ..., φ(vk)). This is because the linear transformation φ will map all vectors in V to W, meaning that W must contain all possible linear combinations of the vectors φ(v1), ..., φ(vk).

## 4. Does the statement hold true for infinite-dimensional vector spaces?

Yes, the statement holds true for infinite-dimensional vector spaces. As long as the vectors v1, ..., vk span the vector space V, the statement will hold true regardless of the dimensionality of V.

## 5. Can the statement be proven using a counterexample?

No, the statement cannot be proven using a counterexample. This is because the statement is a logical implication, and in order to disprove it, a counterexample must be found for all possible cases. However, since the statement holds true for all vector spaces and linear transformations, it cannot be disproven using a single counterexample.

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