Show that it is a subspace

[mathmari]

Hey!

Let $1\leq m, n\in \mathbb{N}$, let $\phi :\mathbb{R}^n\rightarrow \mathbb{R}^m$ a linear map and let $U\leq_{\mathbb{R}}\mathbb{R}^n$, $W\leq_{\mathbb{R}}\mathbb{R}^m$ be subspaces.
I want to show that:

1. $\phi (U)$ is subspace of $\mathbb{R}^m$.
2. $\phi^{-1} (W)$ is subspace of $\mathbb{R}^n$.

I have done the following:

1. We have that $\phi (U)=\{\phi (u) \mid u\in U\}$.
   
   Since $U$ is a subspace we have that $0\in U$. Therefore $\phi (0)\in \phi (U)$. Since $\phi$ is a linear map we have that $\phi (0)=0$ and so we get that $0\in \phi (U)$.
   
   Let $\phi (u_1), \phi (u_2)\in \phi (U)$. Then we have that $\phi (u_1)+\phi (u_2)=\phi (u_1+u_2)$, since $\phi$ is linear.
   Since $U$ is a subspace we have that since $u_1, u_2\in U$ then $u_1+u_2\in U$. Therefore we get that $\phi (u_1+u_2)\in \phi (U)$ and so we have that $\phi (u_1)+\phi (u_2)\in \phi (U)$.
   
   Let $\lambda\in \mathbb{R}$ and $\phi (u_1)\in \phi (U)$. Then we have that $\lambda \phi (u_1)=\phi (\lambda u_1)$, since $\phi$ is linear.
   Since $U$ is a subspace we have that since $\lambda\in \mathbb{R}$ and $u_1\in U$ then $\lambda u_1\in U$. Therefore we get that $\phi (\lambda u_1)\in \phi (U)$ and so we have that $\lambda \phi (u_1)\in \phi (U)$.
   
   
   
   That means that $\phi (U)$ is subspace of $\mathbb{R}^m$.
   
   
   
   Is everything correct? (Wondering)
   
   
2. We have that $\phi$ is linear. Does it follow then that $\phi^{-1}$ is also linear? (Wondering)

 

[I like Serena]

Is everything correct?

Hey mathmari!

Yep. (Nod)

2. We have that $\phi$ is linear. Does it follow then that $\phi^{-1}$ is also linear?

Let's see... suppose we pick $\phi: u \mapsto 0$.
That is a linear map isn't it?
What is $\phi^{-1}$? Is it a linear map? Is it a function for that matter? (Wondering)

 

[mathmari]

Let's see... suppose we pick $\phi: u \mapsto 0$.
That is a linear map isn't it?
What is $\phi^{-1}$? Is it a linear map? Is it a function for that matter? (Wondering)

There is no inverse, is there? (Wondering)

 

[I like Serena]

There is no inverse, is there?

Indeed. So we'll have to solve the problem differently. (Thinking)

 

[mathmari]

Indeed. So we'll have to solve the problem differently. (Thinking)

But how? (Wondering)

 

[I like Serena]

But how? (Wondering)

As a linear map, $\phi(0)=0$, so $0\in\phi^{-1}(\{0\})$.
Since W is a subspace, it must contain 0.
So we must have that $0\in \phi^{-1}(W)$, don't we? (Wondering)

Suppose $u,v\in \phi^{-1}(W)$, then what can we find out about $u+v$? (Wondering)

 

[mathmari]

Suppose $u,v\in \phi^{-1}(W)$, then what can we find out about $u+v$? (Wondering)

We have that $\phi(u), \phi(v) \in W$. Since $W$ is a subspace we have that $\phi(u) +\phi (v) \in W$. Since $\phi$ is linear we get that $\phi (u+v) \in W$. We apply the inverse and we get $u+v\in \phi^{-1}(W)$.

Is everything correct? (Wondering)

 

[I like Serena]

We have that $\phi(u), \phi(v) \in W$. Since $W$ is a subspace we have that $\phi(u) +\phi (v) \in W$. Since $\phi$ is linear we get that $\phi (u+v) \in W$. We apply the inverse and we get $u+v\in \phi^{-1}(W)$.

Is everything correct? (Wondering)

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

I thought about that again and now I got stuck. I said that $\phi (u), \phi (v)\in W$. Is this correct? Does this hold because $W$ is a subspace of $\mathbb{R}^m$ and the image of $\phi$ is $\mathbb{R}^m$ ? (Wondering)

 

[I like Serena]

I thought about that again and now I got stuck. I said that $\phi (u), \phi (v)\in W$. Is this correct? Does this hold because $W$ is a subspace of $\mathbb{R}^m$ and the image of $\phi$ is $\mathbb{R}^m$ ?

We assumed that $u\in\phi^{-1}(W)$. Doesn't this mean by definition that $\phi(u)\in W$?
That is, isn't $\phi^{-1}(W)\overset{\text{def}}{=}\{x : \phi(x) \in W\}$? (Wondering)
The same applies to $v$.

 

[mathmari]

We assumed that $u\in\phi^{-1}(W)$. Doesn't this mean by definition that $\phi(u)\in W$?
That is, isn't $\phi^{-1}(W)\overset{\text{def}}{=}\{x : \phi(x) \in W\}$? (Wondering)
The same applies to $v$.

Ahh yes! (Blush)

Thank you very much! (Yes)