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 <i>_\beta^{A\beta}</i> 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}
\Rightarrow <i>_\beta^{\beta&#039;} x^\beta=<i>_\beta^{\beta&#039;} \beta^{-1}\mathbf{x}</i></i>

for some other basis \beta&#039; where <i>_\beta^{\beta&#039;} </i> is the change of basis matrix from \beta to \beta&#039; coordinates. so

<i>_\beta^{\beta&#039;} x^\beta=<i>_\beta^{\beta&#039;} \beta^{-1}\mathbf{x}= x^{\beta&#039;}</i></i> --(*)

We also know the coordinates of \mathbf{x} in the \beta&#039; coords using the \beta&#039; basis:

\beta&#039; x^{\beta&#039;}=\mathbf{x}
\Rightarrow x^{\beta&#039;}=\beta&#039;^{-1}\mathbf{x} --(**)

(*) and (**) combine to give

<i>_\beta^{\beta&#039;} \beta^{-1}\mathbf{x}=\beta&#039;^{-1}\mathbf{x}</i>
\Rightarrow <i>_\beta^{\beta&#039;} \beta^{-1}=\beta&#039;^{-1} </i>
\Rightarrow <i>_\beta^{\beta&#039;}=\beta&#039;^{-1}\beta</i>

This seems like a nice neat result to me, but if \beta&#039;=A\beta as it is in the book, we have

<i>_\beta^{\beta&#039;}=\beta&#039;^{-1}\beta</i>
\Rightarrow <i>_\beta^{A \beta}=(A\beta)^{-1}\beta</i>
\Rightarrow <i>_\beta^{A \beta}=\beta^{-1}A^{-1}\beta</i>
\not=A^{-1}

However, if \beta&#039;=\beta A
<i>_\beta^{\beta A}=\beta&#039;^{-1}\beta</i>
\Rightarrow <i>_\beta^{\beta A}=(\beta A)^{-1}\beta</i>
\Rightarrow <i>_\beta^{\beta A}=A^{-1}\beta^{-1}\beta</i>
=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
<br /> \text{conversion matrix from }\beta \text{ to } A\beta=\beta^{-1}A^{-1}\beta<br />
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}<br /> &amp; \beta x=(\beta A)y\\<br /> \Rightarrow &amp; A^{-1}\beta^{-1}\beta x=y\\<br /> \Rightarrow &amp; 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}<br /> \left[I\right]_{\beta}^{A\beta} &amp; =\beta^{-1}A^{-1}\beta\\<br /> &amp; =(A\beta)^{-1}\beta\\<br /> &amp; =(\beta^{-1}A\beta)^{-1}\\<br /> &amp; =(\left[A\right]_{\beta}^{\beta})^{-1}\end{aligned}


Hoorah... What a great feeling.

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