# Show that it is a subspace

• MHB
Gold Member
MHB
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)

Homework Helper
MHB
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)

Gold Member
MHB
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)

Homework Helper
MHB
There is no inverse, is there?

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

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

But how? (Wondering)

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MHB
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)

Gold Member
MHB
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)

Homework Helper
MHB
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)

Gold Member
MHB
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)

Homework Helper
MHB
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 $v$.

Gold Member
MHB
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)