# Prove that an isomorphism produces a basis of W

• Sep 30th 2010, 08:26 AM
Runty
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.
• Sep 30th 2010, 09:39 AM
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.
• Sep 30th 2010, 10:23 AM
Runty
Quote:

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.
• Oct 1st 2010, 04:05 AM
HallsofIvy
Quote:

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.

Quote:

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.