# A true or false about an inner product

• Mar 2nd 2009, 09:11 AM
arbolis
A true or false about an inner product
Hi MHF,
I did badly (failed) in the final exam on linear algebra (4.4/10) and so I must retake the exam in late June or early July. I'm still stuck on some exam questions :
Tell whether this is true or false : Let $\displaystyle V$ and $\displaystyle W$ be vector spaces and $\displaystyle T:V\to W$ an isomorphism. If $\displaystyle W$ has an inner product $\displaystyle \langle , \rangle$, then $\displaystyle (X,Y)=\langle TX, TY \rangle$ defines an inner product in $\displaystyle V$.
I admit that I'm totally lost. I don't know what theorem(s) could be useful...
Although $\displaystyle T$ is an isomorfism, I don't think it imples that $\displaystyle V$ and $\displaystyle W$ have the same dimension... I don't even know what are $\displaystyle X$ and $\displaystyle Y$.
Is $\displaystyle T$ the identity linear transformation? If so, I guess it's almost obvious that the answer is "true".
• Mar 2nd 2009, 03:10 PM
ThePerfectHacker
Quote:

Originally Posted by arbolis
Hi MHF,
I did badly (failed) in the final exam on linear algebra (4.4/10) and so I must retake the exam in late June or early July. I'm still stuck on some exam questions :
Tell whether this is true or false : Let $\displaystyle V$ and $\displaystyle W$ be vector spaces and $\displaystyle T:V\to W$ an isomorphism. If $\displaystyle W$ has an inner product $\displaystyle \langle , \rangle$, then $\displaystyle (X,Y)=\langle TX, TY \rangle$ defines an inner product in $\displaystyle V$.
I admit that I'm totally lost. I don't know what theorem(s) could be useful...
Although $\displaystyle T$ is an isomorfism, I don't think it imples that $\displaystyle V$ and $\displaystyle W$ have the same dimension... I don't even know what are $\displaystyle X$ and $\displaystyle Y$.
Is $\displaystyle T$ the identity linear transformation? If so, I guess it's almost obvious that the answer is "true".

"Isomorphism" means they are really the exact same vector space just the elements are called differently. It is like doing arithmetic but instead of writing "1,2,...,9,0" you wrote "A,B,C,...", basically you are just calling the elements in a different way but all you are doing is identically the same. That is what an Isomorphism is. Therefore, the answer is definitely YES!

• Mar 2nd 2009, 04:16 PM
arbolis
I'm not sure I've understood how to prove the affirmation.
What does $\displaystyle TX$ exactly mean? Is it $\displaystyle T \circ X$ ? for an arbitrary $\displaystyle X \in W$? If so... does $\displaystyle \langle TX, TY \rangle = T \circ \langle X, Y \rangle$?

About isomorphismes... Say I have a 2 dimensional vector space. Isn't it isomorphic to infinitly many other 2-dimensional vector spaces although they can be orthogonal? So they are not the same vector space, but still isomorphic because there exist bijective linear transformations from one vector space to each other. I feel I didn't grasp well concepts.
And for the problem I could check if the 4 properties of any inner product are valid in $\displaystyle V$, but I'm sure there's a fastest way to do it. Like justificating with a theorem especially with this $\displaystyle T$ transformation. A little more help would maybe save me but surely make me happy.
• Mar 2nd 2009, 07:04 PM
ThePerfectHacker
Quote:

Originally Posted by arbolis
I'm not sure I've understood how to prove the affirmation.
What does $\displaystyle TX$ exactly mean? Is it $\displaystyle T \circ X$ ? for an arbitrary $\displaystyle X \in W$? If so... does $\displaystyle \langle TX, TY \rangle = T \circ \langle X, Y \rangle$?

Let me give you an example with numbers again. Think about "numbers modulo three". These are $\displaystyle A=\{0,1,2\}$ and we add then in the following way: $\displaystyle 0+0=0,0+1=1,1+2=0,1+1=2,2+2=1,...$. In other words you find the remainder of the number upon division by three. Now, let $\displaystyle B=\{a,b,c\}$ and we will define $\displaystyle a*a=a,a*b=b,b*b=c,b*c=a,c*c=b,...$ i.e. we define *-addition on $\displaystyle B$ to be as if $\displaystyle a$ was $\displaystyle 0$ as if $\displaystyle b$ was $\displaystyle 1$ as if $\displaystyle c$ was $\displaystyle 2$. Of course, $\displaystyle A$ and $\displaystyle B$ are "isomorphic". Because we can pair the elements together so that their addition still pairs up, for example: $\displaystyle 0$ pairs with $\displaystyle a$ and $\displaystyle 1$ pairs with $\displaystyle b$ furthermore $\displaystyle 0+1$ pairs with $\displaystyle a*b$. . Mathematically, it means we can find a function $\displaystyle f: A\to B$ so that $\displaystyle f$ is a bijection (remember being a bijection means we can pair elements) and $\displaystyle f(x+y) = f(x)*f(y)$.

It is common, however, to write $\displaystyle f(x+y) = f(x)+f(y)$, just be careful! The addition $\displaystyle +$ in $\displaystyle A$ is actually different from the addition $\displaystyle *$ in $\displaystyle B$. When dealing with vector spaces say you have one vector space $\displaystyle V$ and another vector $\displaystyle W$, then for $\displaystyle f$ to be an isomorphism it means $\displaystyle f:V\to W$ is a bijection with $\displaystyle f(v_1+v_2) = f(v_1)+f(v_2)$. Again remember that addition in $\displaystyle V$ may be different than the addition in $\displaystyle W$. We use the same $\displaystyle +$ eventhough they are different depending on the vector space. So just remember that.

Now let $\displaystyle \left< ~ , ~ \right>$ be an inner product on $\displaystyle V$. Since $\displaystyle W$ is isomorphic to $\displaystyle V$, with $\displaystyle f:V\to W$ an isomorphism, we can easily define an inner product on $\displaystyle W$. For $\displaystyle w_1,w_2\in W$ we know that there exists unique elements in $\displaystyle V$ with which they are paired, so say, $\displaystyle w_1=f(v_1)\text{ and }w_2 = f(v_2)$. Now define them as if they are paired together, i.e. define $\displaystyle \left< w_1,w_2\right>$ to be $\displaystyle \left<v_1,v_2\right>$. It should be intuitely obvious why this is an inner product on $\displaystyle W$. However, you should prove it mathematically by going through the list of the properties of inner products.
• Mar 3rd 2009, 04:56 AM
arbolis
Thanks a bunch for the lesson! At one moment you wrote $\displaystyle f:V\to M$ but you meant $\displaystyle f:V\to W$.
I reformulate the problem in a more understandable form to me : Let $\displaystyle X$ and $\displaystyle Y$ be in $\displaystyle W$. We have that $\displaystyle \langle X,Y \rangle$ is an inner product. Is $\displaystyle \langle TX, TY \rangle$ an inner product in $\displaystyle V$?
My answer : as $\displaystyle T$ is bijective, I can pair $\displaystyle TX \in V$ with $\displaystyle X \in W$ and the same applies for $\displaystyle TY$ and $\displaystyle Y$.
As $\displaystyle \langle X,Y \rangle$ is an inner product in $\displaystyle W$, as I paired $\displaystyle X$ with $\displaystyle TX$ and $\displaystyle Y$ with $\displaystyle TY$ and also because $\displaystyle T$ is linear, $\displaystyle \langle TX, TY \rangle$ has the same properties than $\displaystyle \langle X,Y \rangle$ and so it's an inner product.
I don't really know how I can justify that for $\displaystyle X\neq \vec 0$, $\displaystyle \langle X, X \rangle >0 \Rightarrow \langle TX, TX \rangle >0$. Maybe I can justify by saying that $\displaystyle \langle X, X \rangle >0$ works whatever $\displaystyle X$ is. And so it works with $\displaystyle TX$ even if this is an element from another vector space since there are isomorphic and I associated $\displaystyle X$ with $\displaystyle TX$. Ok, I think I get it. The other properties are more obvious.