# Help with linear transformation / subspace proof

• July 27th 2009, 09:21 PM
Danneedshelp
Help with linear transformation / subspace proof
Q: Let $T:V\rightarrow{W}$ be a linear transformation, and let $U$ be a subspace of $W$. Prove that the set $T^{-1}(U)=\{\vec{v}\in{V} | T(v)\in{U}\}$ is a subspace of $V$. What is $T^{-1}(U)$ if $U=\{\vec{0}\}$ ?

A: I dont have much, so I will show my train of thought...

First, I want to show $T^{-1}(U)$ is nonempty. I can do this by simply noting $U$ is a subspace; therefore, $T^{-1}(U)$ is nonempty. I don't know if I am reading the set correctly. Do I wan't to show the range of $T^{-1}(U)$ is nonempty?

I am confused by what $T^{-1}(U)$ represents.

Does it mean:

$T(\vec{v})=c_{1}T(\vec{v_{1}})+c_{2}T(\vec{v_{2}}) +
...+c_{k}T(\vec{v_{k}})\in{U}$

$T^{-1}(T(\vec{v}))=\vec{v}=c_{1}\vec{v_{1}}+c_{2}\vec{ v_{2}}+
...+c_{k}\vec{v_{k}}\in{V}$

???

Thanks
• July 27th 2009, 10:01 PM
CaptainBlack
Quote:

Originally Posted by Danneedshelp
Q: Let $T:V\rightarrow{W}$ be a linear transformation, and let $U$ be a subspace of $W$. Prove that the set $T^{-1}(U)=\{\vec{v}\in{V} | T(v)\in{V}\}$ is a subspace of $V$. What is $T^{-1}(U)$ if $U=\{\vec{0}\}$ ?

A: I dont have much, so I will show my train of thought...

First, I want to show $T^{-1}(U)$ is nonempty. I can do this by simply noting $U$ is a subspace; therefore, $T^{-1}(U)$ is nonempty. I don't know if I am reading the set correctly. Do I wan't to show the range of $T^{-1}(U)$ is nonempty?

I am confused by what $T^{-1}(U)$ represents.

Does it mean:

$T(\vec{v})=c_{1}T(\vec{v_{1}})+c_{2}T(\vec{v_{2}}) +
...+c_{k}T(\vec{v_{k}})\in{U}$

$T^{-1}(T(\vec{v}))=\vec{v}=c_{1}\vec{v_{1}}+c_{2}\vec{ v_{2}}+
...+c_{k}\vec{v_{k}}\in{V}$

???

Thanks

$T^{-1}U$ is the inverse image of $U$ under $T$. That is it is the set of all elements $x$ of W such that $Tx\in U$.

That $T^{-1}U$ is non-empty is shown as $T0_V=0_W$ that is for any linear transformation the zero vector maps to the zero vector. Then since $0_W\in U$ we have $0_V\in T^{-1}U$

To show that $T^{-1}U$ is a subspace of $V$ it is sufficient to show that for any scalars $\alpha_1$, $\alpha_2$ and $x_1, x_2 \in T^{-1}U$ that:

$
\alpha_1x_1+\alpha_2x_2 \in T^{-1}U
$

CB
• July 28th 2009, 10:04 AM
Danneedshelp
Quote:

Originally Posted by CaptainBlack

To show that $T^{-1}U$ is a subspace of $V$ it is sufficient to show that for any scalars $\alpha_1$, $\alpha_2$ and $x_1, x_2 \in T^{-1}U$ that:

$
\alpha_1x_1+\alpha_2x_2 \in T^{-1}U
$

CB

Is that all that needs to be shown or do I have to further explain this?

Thank you