Welcome to our community

Be a part of something great, join today!

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

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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 $V$. 🤔

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

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

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
$W$ 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:



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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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)


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

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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)

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

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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. (Nod)
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
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:

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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. :oops:
Country Boy 's example gives the counter example where $\phi$ maps every vector to the null vector.
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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. :oops:
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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)

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)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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)

Instead we can use a proof by contradiction.
Suppose $v_1\,\ldots,v_k$ are not linearly independent, then... 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
The proof is wrong.
The particular step assumes that $\phi(x)=0$ implies that $x=0$, but that is not generally true. (Shake)

Instead we can use a proof by contradiction.
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
Yep. (Nod)