# Thread: Prove that an isomorphism produces a basis of W

1. ## Prove that an isomorphism produces a basis of W

I think I may have the answer for this one, but I could use a second opinion.

Suppose $\displaystyle V$ and $\displaystyle W$ are vector spaces over a field $\displaystyle F$.

Now suppose $\displaystyle \psi:V\rightarrow W$ is an isomorphism and $\displaystyle \{ v_1,...,v_n\}$ is a basis of $\displaystyle V$. Prove that $\displaystyle \{\psi(v_1),...,\psi(v_n)\}$ is a basis of $\displaystyle W$.

Here is what I have so far:

Let $\displaystyle \{w_1,...,w_n\}$ be a basis for $\displaystyle W$ where
$\displaystyle w_1=\psi(v_1),...,w_n=\psi(v_n), \forall n>0$.
Then we consider the map
$\displaystyle \psi(x)=x_1w_1+x_2w_2+...+x_nw_n$ where $\displaystyle x=x_1v_1+x_2v_2+...+x_nv_n\in V$.
Since the bases are inverse to each other, we can conclude that $\displaystyle \{\psi(v_1),...,\psi(v_n)\}$ is a basis of $\displaystyle W$.

I don't know if I have any major holes in my proof, so if you see any, please let me know.

2. The last part in particular sounds strange.

What does it mean that the bases are inverse to each other, and why does it imply that $\displaystyle \{\psi(v_1),\ldots,\psi(v_n)\}$ is a basis for $\displaystyle W$?

I would do like this: We need to prove that $\displaystyle \{\psi(v_1),\ldots,\psi(v_n)\}$ is a basis for $\displaystyle W$. So we need to check two things, namely that the vectors in the proposed basis are linearly independent and that they span all of $\displaystyle W$.

Let $\displaystyle x_1\psi(v_1)+\ldots+x_n\psi(v_n)=0$ for some scalars $\displaystyle x_i$. Apply linearity of $\displaystyle \psi$ to get

$\displaystyle \psi(x_1v_1+\ldots+x_nv_n)=0.$

Remember that $\displaystyle \psi$ is an isomorphism, so what do you know about vectors that $\displaystyle \psi$ map to 0?

To see that the basis spans all of $\displaystyle W$, take some vector $\displaystyle w\in W$, and choose a vector $\displaystyle v\in V$, such that $\displaystyle \psi(v)=w$ (why does this vector $\displaystyle v$ exist?).

You know that $\displaystyle \{v_1,\ldots,v_n\}$ is a basis for $\displaystyle V$, so you can write $\displaystyle v$ as a linear combination of these basis vectors. Try applying $\displaystyle \psi$ to this relation and see what happens.

3. Originally Posted by HappyJoe
The last part in particular sounds strange.

What does it mean that the bases are inverse to each other, and why does it imply that $\displaystyle \{\psi(v_1),\ldots,\psi(v_n)\}$ is a basis for $\displaystyle W$?

I would do like this: We need to prove that $\displaystyle \{\psi(v_1),\ldots,\psi(v_n)\}$ is a basis for $\displaystyle W$. So we need to check two things, namely that the vectors in the proposed basis are linearly independent and that they span all of $\displaystyle W$.

Let $\displaystyle x_1\psi(v_1)+\ldots+x_n\psi(v_n)=0$ for some scalars $\displaystyle x_i$. Apply linearity of $\displaystyle \psi$ to get

$\displaystyle \psi(x_1v_1+\ldots+x_nv_n)=0.$

Remember that $\displaystyle \psi$ is an isomorphism, so what do you know about vectors that $\displaystyle \psi$ map to 0?

To see that the basis spans all of $\displaystyle W$, take some vector $\displaystyle w\in W$, and choose a vector $\displaystyle v\in V$, such that $\displaystyle \psi(v)=w$ (why does this vector $\displaystyle v$ exist?).

You know that $\displaystyle \{v_1,\ldots,v_n\}$ is a basis for $\displaystyle V$, so you can write $\displaystyle v$ as a linear combination of these basis vectors. Try applying $\displaystyle \psi$ to this relation and see what happens.
Thanks a lot for the help. That "bases are inverse to each other" part, I actually got off Wikipedia. It applied to ordered bases, though, so that could've made a difference.

Either way, I've gotten it now so, again, thanks a bunch.

4. Originally Posted by Runty
I think I may have the answer for this one, but I could use a second opinion.

Suppose $\displaystyle V$ and $\displaystyle W$ are vector spaces over a field $\displaystyle F$.

Now suppose $\displaystyle \psi:V\rightarrow W$ is an isomorphism and $\displaystyle \{ v_1,...,v_n\}$ is a basis of $\displaystyle V$. Prove that $\displaystyle \{\psi(v_1),...,\psi(v_n)\}$ is a basis of $\displaystyle W$.

Here is what I have so far:

Let $\displaystyle \{w_1,...,w_n\}$ be a basis for $\displaystyle W$ where
$\displaystyle w_1=\psi(v_1),...,w_n=\psi(v_n), \forall n>0$.
You appear to be assuming what you want to prove! That is that $\displaystyle w_1=\psi(v_1),...,w_n=\psi(v_n)$ is a basis.

Instead you need to show that this set (1) is independent and (2) spans the space.

To show independence: Supppose that [tex]a\psi(v_1)+ b\psi(v_2)+ \cdot\cdot\cdot+ z\psi(v_n)= 0[/itex]
Since $\displaystyle \psi$ is a linear transformation, that is the same as $\displaystyle \psi(av_1+ bv_2+ \cdot\cdot\cdot+ zv_n)= 0$

Since $\displaystyle \psi$ is an isomorphism, its kernel is the 0 vector only. Therefore $\displaystyle av_1+ bv_2+ \cdot\cdot\cdot+ av_n= 0$. What does that tell you?

To show that $\displaystyle \{\psi(v_1), \psi(v_n), \cdot\cdot\cdot, \pis(v_n)\}$ span W, let w be any vector in W. Since $\displaystyle \psi$ is an isomorphism, there exist v in V such that $\displaystyle \psi(v)= w$. Since [tex]\{v_1, v_2, \cdot\cdot\cdot, v_n} is a basis for V, there exist scalars a, b, ..., z such that $\displaystyle av_1+ bv_2+ \cdot\cdot\cdot+ zv_n= v$. Apply $\displaystyle \psi$ to both sides of that equation.

Then we consider the map
$\displaystyle \psi(x)=x_1w_1+x_2w_2+...+x_nw_n$ where $\displaystyle x=x_1v_1+x_2v_2+...+x_nv_n\in V$.
Since the bases are inverse to each other, we can conclude that $\displaystyle \{\psi(v_1),...,\psi(v_n)\}$ is a basis of $\displaystyle W$.

I don't know if I have any major holes in my proof, so if you see any, please let me know.