[SOLVED]Show that it is a subspace

mathmari

Well-known member
MHB Site Helper
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?

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

Klaas van Aarsen

MHB Seeker
Staff member
Is everything correct?
Hey mathmari !!

Yep.

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?

mathmari

Well-known member
MHB Site Helper
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?
There is no inverse, is there?

Klaas van Aarsen

MHB Seeker
Staff member
There is no inverse, is there?
Indeed. So we'll have to solve the problem differently.

MHB Site Helper

Klaas van Aarsen

MHB Seeker
Staff member
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?

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

mathmari

Well-known member
MHB Site Helper
Suppose $u,v\in \phi^{-1}(W)$, then what can we find out about $u+v$?
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?

Klaas van Aarsen

MHB Seeker
Staff member
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?
Yep.

mathmari

Well-known member
MHB Site Helper
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 $\mathbb{R}^m$ ?

Klaas van Aarsen

MHB Seeker
Staff member
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 $\phi^{-1}(W)\overset{\text{def}}{=}\{x : \phi(x) \in W\}$?
The same applies to $v$.

mathmari

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

Thank you very much!!