Is the Change of Basis Matrix in My Book Wrong?

[Mmmm]

Homework Statement

I have posted this problem on another website (mathhelpforum) but have received no replies. I don't know whether this is because no one knows what I am talking about or if it's just that no one can find a fault with my reasoning. Please please please could you post a reply even if it's just to say "Looks ok to me, but what would I know?" as there is the (remote) possibility that the book is wrong here and it'd really do me good to know this - the more replies saying it looks ok the happier I'll be. It's playing havok with my confidence.

My book (Tensor Geometry - Poston & Dodson) says the following:

If $\beta = (b_1,..., b_n)$ is a basis for X, and $A : X \rightarrow Y$ is an isomorphism, then $A\beta = (Ab_1,..., Ab_n)$ is a basis for Y.
If $\beta$ is a basis for X and $A : X \rightarrow X$ is an isomorphism, the change of basis matrix [itex]_\beta^{A\beta}[/itex] is exactly the matrix $([A]_\beta^\beta)^{-1}$.

I just can't seem to agree with this result!

Homework Equations

The Attempt at a Solution

After hours of tearing my hair out I have come up with the following argument...

For some basis $\beta$, some vector $\mathbf{x}$ and its representation $x^\beta$ in the $\beta$ coords.

$$\beta x^\beta=\mathbf{x}$$
$$\Rightarrow x^\beta=\beta^{-1}\mathbf{x}$$
[tex]\Rightarrow _\beta^{\beta'} x^\beta=_\beta^{\beta'} \beta^{-1}\mathbf{x}[/tex]

for some other basis $\beta'$ where [itex]_\beta^{\beta'} [/itex] is the change of basis matrix from $\beta$ to $\beta'$ coordinates. so

[tex] _\beta^{\beta'} x^\beta=_\beta^{\beta'} \beta^{-1}\mathbf{x}= x^{\beta'}[/tex] --(*)

We also know the coordinates of $\mathbf{x}$ in the $\beta'$ coords using the $\beta'$ basis:

$$\beta' x^{\beta'}=\mathbf{x}$$
$$\Rightarrow x^{\beta'}=\beta'^{-1}\mathbf{x}$$ --(**)

(*) and (**) combine to give

[tex]_\beta^{\beta'} \beta^{-1}\mathbf{x}=\beta'^{-1}\mathbf{x}[/tex]
[tex]\Rightarrow _\beta^{\beta'} \beta^{-1}=\beta'^{-1} [/tex]
[tex]\Rightarrow _\beta^{\beta'}=\beta'^{-1}\beta[/tex]

This seems like a nice neat result to me, but if $\beta'=A\beta$ as it is in the book, we have

[tex] _\beta^{\beta'}=\beta'^{-1}\beta[/tex]
[tex]\Rightarrow _\beta^{A \beta}=(A\beta)^{-1}\beta[/tex]
[tex]\Rightarrow _\beta^{A \beta}=\beta^{-1}A^{-1}\beta[/tex]
$$\not=A^{-1}$$

However, if $\beta'=\beta A$
[tex] _\beta^{\beta A}=\beta'^{-1}\beta[/tex]
[tex]\Rightarrow _\beta^{\beta A}=(\beta A)^{-1}\beta[/tex]
[tex]\Rightarrow _\beta^{\beta A}=A^{-1}\beta^{-1}\beta[/tex]
$$=A^{-1}$$

which is the required result...

I have tried some basic examples with actual numbers and the results support what I have here... Unless I have some fundamental misunderstanding of all this and what it is supposed to mean, which is quite possible...

 

[Outlined]

It id a linear isomorphism isn't it? Seems really basic and straightforward I must say.

 

[Mmmm]

It id a linear isomorphism isn't it? Seems really basic and straightforward I must say.

Yes, it's linear. So you would say I'm right then?
or are you saying it's straightforward to get the required result?

 

[Outlined]

I did not check your #3 point but you can easily check it is a basis by definition:

A basis B of a vector space V over a field F is a linearly independent subset of V that spans (or generates) V.

Would this help you?

 

[Mmmm]

I did not check your #3 point but you can easily check it is a basis by definition:

A basis B of a vector space V over a field F is a linearly independent subset of V that spans (or generates) V.

Would this help you?

I know $A\beta$ is definitely a basis, the question is whether the identity map between the two bases is equal to $A^{-1}$.
ie does $A^{-1}\mathbf{x^{\beta}}=\mathbf{x^{A\beta}}$
My reasoning says it should be $A^{-1}\mathbf{x^{\beta}}=\mathbf{x^{\beta A}}$

notation: $\mathbf{x^{\beta}}$ meaning the vector $\mathbf{x}$ under the $\beta$ basis
$\mathbf{x^{A\beta}}$ meaning the vector $\mathbf{x}$ under the $A\beta$ basis

 

[Outlined]

Th identity map is just a function which leaves the input unchanged.

 

[Mmmm]

Th identity map is just a function which leaves the input unchanged.

In this case I am changing bases, so the identity map leaves the vector unchanged but changes the components...if you see what i mean...
I'm beginning to get the feeling that the terminology used in this book isn't standard!

 

[Outlined]

In that case it comes down to looking at the (easy) equation

[b₁ b₂ ... b_n]x = [Ab₁ Ab₂ ... Ab_n]y = A[b₁ b₂ ... b_n]y

Here [ ] is a matrix with elements inside as columns
By multiplying with A or A^-1 you can get your vector in the coordinate representation you want.

 

[Mmmm]

In that case it comes down to looking at the (easy) equation

[b₁ b₂ ... b_n]x = [Ab₁ Ab₂ ... Ab_n]y = A[b₁ b₂ ... b_n]y

Here [ ] is a matrix with elements inside as columns
By multiplying with A or A^-1 you can get your vector in the coordinate representation you want.

Right! which is what I did and got
$$\text{conversion matrix from }\beta \text{ to } A\beta=\beta^{-1}A^{-1}\beta$$
rather than the $A^{-1}$ which the book claims. Am I right?

 

[Outlined]

y = [b1 b2 ... bn]^-1A^-1[b1 b2 ... bn]x

I think you are right indeed. But maybe the book is talking about [b1 b2 ... bn]x while you are about x, which is a difference. Look carefully at what the book means.

 

[Mmmm]

so using your notation:
[b1 b2 ... bn]x = [Ab1 Ab2 ... Abn]y = A[b1 b2 ... bn]y

where x is in $\beta$ coords and y is in $A \beta$ coords

taking the first and last of these equalities:

$$[b_1 b_2 ... b_n]x = A[b_1 b_2 ... b_n]y$$
$$\Rightarrow A^{-1}[b_1 b_2 ... b_n]x = [b_1 b_2 ... b_n]y$$
$$\Rightarrow [b_1 b_2 ... b_n]^{-1}A^{-1}[b_1 b_2 ... b_n]x = y$$

 

[Outlined]

I edited my post, please read again. I think you as well as the book are right but your are talking about slightly different things.

 

[Mmmm]

y = [b1 b2 ... bn]^-1A^-1[b1 b2 ... bn]x

I think you are right indeed. But maybe the book is talking about [b1 b2 ... bn]x while you are about x, which is a difference. Look carefully at what the book means.

but even then you would have the matrix being [b1 b2 ... bn]^-1A^-1

 

[Mmmm]

In that case it comes down to looking at the (easy) equation

[b₁ b₂ ... b_n]x = [Ab₁ Ab₂ ... Ab_n]y = A[b₁ b₂ ... b_n]y

Here [ ] is a matrix with elements inside as columns
By multiplying with A or A^-1 you can get your vector in the coordinate representation you want.

OK... but
[b₁ b₂ ... b_n]x = A[b₁ b₂ ... b_n]y

multiplying by A^-1 gives A^-1[b₁ b₂ ... b_n]x = [b₁ b₂ ... b_n]y

but [b₁ b₂ ... b_n]y is meaningless right? it's the components in Ab paired with the b basis!

 

[Outlined]

It seems that you fully understand what is going on, I wouldn't worry about a possible mistake in the book and just work on some exercises. May you have a problem with one of those exercises you can always come back here.

btw: From your opening post I see you write something like $$(A^{\beta}_{\beta})^{-1}$$ (quote from the book) so that is something like what you say. Again: most important point is that you do understand what is going on and in that case you are fine.

 

[Mmmm]

It seems that you fully understand what is going on, I wouldn't worry about a possible mistake in the book and just work on some exercises. May you have a problem with one of those exercises you can always come back here.

btw: From your opening post I see you write something like $$(A^{\beta}_{\beta})^{-1}$$ (quote from the book) so that is something like what you say. Again: most important point is that you do understand what is going on and in that case you are fine.

Thanks, that helps a lot.

Did you notice that in my first post I mentioned that if you choose the second basis as $\beta A$ rather than $A\beta$ then you get the desired result?

$$\begin{aligned} & \beta x=(\beta A)y\\ \Rightarrow & A^{-1}\beta^{-1}\beta x=y\\ \Rightarrow & A^{-1}x=y\end{aligned}$$Unfortunately the basis $\beta A$ isn't as nice geometrically as $A\beta$.
$A\beta$ is the image of $\beta$ under A (which turns out to be a basis for the image of A). Whereas what is $\beta A$? The vectors that A represents in the basis $\beta$? (yuk!) (I'm not even sure if this is a basis for the image of A...Probably not...)

I sort of hoped that the nice geometrical object would have the nice Identity map. I suppose I really wanted to be wrong!

 

[Mmmm]

Wooooooooooooohooooooooooooooooooooooooooooo...

I've figured it out...

At last...

My mistake was in thinking that $A=\left[A\right]_{\beta}^{\beta}$.

The map $A:X\rightarrow X$ maps a vector in the vector space X to a new vector in X.

Wheras $\left[A\right]_{\beta}^{\beta}$ maps the components of a vector in the $\beta$ basis to new components in the $\beta$ basis.

The result is the same but the maps are different.

You can do the map $\left[A\right]_{\beta}^{\beta}$ in terms of $A$ by first converting components into a vector ($\beta(x^{\beta})$), then applying A ($A(\beta x^{\beta})$) and then converting back into components ($\beta^{-1}(A\beta x^{\beta})$).

ie

$$\left[A\right]_{\beta}^{\beta}=\beta^{-1}A\beta$$

My result in my first post was $\left[I\right]_{\beta}^{A\beta}=\beta^{-1}A^{-1}\beta$ which completely confused me (I was expecting $\left[I\right]_{\beta}^{A\beta}=A^{-1}$)

But now I can take this further

$$\begin{aligned} \left[I\right]_{\beta}^{A\beta} & =\beta^{-1}A^{-1}\beta\\ & =(A\beta)^{-1}\beta\\ & =(\beta^{-1}A\beta)^{-1}\\ & =(\left[A\right]_{\beta}^{\beta})^{-1}\end{aligned}$$


Hoorah... What a great feeling.

I am finally feeling at one with my book again...