I Similarity transformation, basis change and orthogonality

[Azad Koshur]

I've a transformation $T$ represented by an orthogonal matrix $A$ , so $A^TA=I$. This transformation leaves norm unchanged.

I do a basis change using a matrix $B$ which isn't orthogonal , then the form of the transformation changes to $B^{-1}AB$ in the new basis( A similarity transformation).

Since we only changed our representation of the transformation $T$ then transformation $B^{-1}AB$ should also leave norm unchanged which means that $B^{-1}AB$ should be orthogonal.

Therefore $B^{-1}AB.{{[B^{-1}AB}}]^T=I$.

This suggests that $B^TB=I$ which means it is orthogonal, but that is a contradiction.

Can anyone tell me if what I did wrong.
Thank you.

 

[martinbn]

I've a transformation $T$ represented by an orthogonal matrix $A$ , so $A^TA=I$. This transformation leaves norm unchanged.

I do a basis change using a matrix $B$ which isn't orthogonal , then the form of the transformation changes to $B^{-1}AB$ in the new basis( A similarity transformation).

Since we only changed our representation of the transformation $T$ then transformation $B^{-1}AB$ should also leave norm unchanged which means that $B^{-1}AB$ should be orthogonal.

Therefore $B^{-1}AB.{{[B^{-1}AB}}]^T=I$.

This suggests that $B^TB=I$ which means it is orthogonal, but that is a contradiction.

Can anyone tell me if what I did wrong.
Thank you.

You say it suggests $B^TB=I$, but does it imply it?

 

[Azad Koshur]

You say it suggests $B^TB=I$, but does it imply it?

$B^TB=I$ is a solution but I'm not sure it's the only one.

 

[hutchphd]

For instance let $\mathbb A=\mathbb 1$. Then there are no additional requirements upon $\mathbb B$ and your supposition is manifestly incorrect.

 

[PeroK]

$B^TB=I$ is a solution but I'm not sure it's the only one.

Let's look at your equation:
$$B^{-1}AB(B^{-1}AB)^T = I \ \Leftrightarrow \ AB(B^TA^T(B^{-1})^T) = B$$$$\Leftrightarrow \ ABB^TA^T(B^T)^{-1} = B \ \Leftrightarrow \ ABB^TA^T = BB^T$$And we can see that if $A$ is orthogonal, then this equation holds whenever $BB^T$ is invariant under the transformation $X \ \rightarrow \ AXA^T$.

So, perhaps your assumption that $B^{-1}AB$ is orthogonal is false? Did you try to prove it?

 

[Azad Koshur]

Let's look at your equation:
$$B^{-1}AB(B^{-1}AB)^T = I \ \Leftrightarrow \ AB(B^TA^T(B^{-1})^T) = B$$$$\Leftrightarrow \ ABB^TA^T(B^T)^{-1} = B \ \Leftrightarrow \ ABB^TA^T = BB^T$$And we can see that if $A$ is orthogonal, then this equation holds whenever $BB^T$ is invariant under the transformation $X \ \rightarrow \ AXA^T$.

So, perhaps your assumption that $B^{-1}AB$ is orthogonal is false? Did you try to prove it?

Wikipedia says :
"In linear algebra, two n-by-n matrices A and B are called similar if there exists an invertible n-by-n matrix P such that

${\displaystyle B=P^{-1}AP.}$

Similar matrices represent the same linear map under two (possibly) different bases, with P being the change of basis matrix"
In our case A was the transformation and $B^{-1}AB$ was the transformation in another basis. Since the underlying transformation preserves norm then $B^{-1}AB$ has to persevere it as well. But all norm preserving matrices are orthogonal so $B^{-1}AB$ has to be as well.
Where does this argument go wrong?

Link:https://en.m.wikipedia.org/wiki/Matrix_similarity

 

[Azad Koshur]

For instance let $\mathbb A=\mathbb 1$. Then there are no additional requirements upon $\mathbb B$ and your supposition is manifestly incorrect.

But I don't see where the argument went wrong?

 

[hutchphd]

You have assumed the matrix B to not be orthogonal but then show that it can be orthogonal. This is not the same as showing it must be orthogonal and does not serve as negation.

 

[PeroK]

Wikipedia says :
"In linear algebra, two n-by-n matrices A and B are called similar if there exists an invertible n-by-n matrix P such that

${\displaystyle B=P^{-1}AP.}$

Similar matrices represent the same linear map under two (possibly) different bases, with P being the change of basis matrix"
In our case A was the transformation and $B^{-1}AB$ was the transformation in another basis. Since the underlying transformation preserves norm then $B^{-1}AB$ has to persevere it as well. But all norm preserving matrices are orthogonal so $B^{-1}AB$ has to be as well.
Where does this argument go wrong?

Link:https://en.m.wikipedia.org/wiki/Matrix_similarity

An orthogonal transformation may not be represented by an orthogonal matrix in a general basis - although always in an orthonormal basis. Your transformation may be norm-preserving, but the matrix representation in a general basis may not be orthogonal.

 

[PeroK]

In finite-dimensional spaces, the matrix representation (with respect to an orthonormal basis) of an orthogonal transformation is an orthogonal matrix.

https://en.wikipedia.org/wiki/Orthogonal_transformation

 

[wrobel]

Since we only changed our representation of the transformation then transformation should also leave norm unchanged

yes but how is this norm presented in this new basis?

In general if $G$ is a matrix of the inner product then the orthogonality of a matrix $A$ means that $A^TGA=I$

 

[Azad Koshur]

An orthogonal transformation may not be represented by an orthogonal matrix in a general basis - although always in an orthonormal basis. Your transformation may be norm-preserving, but the matrix representation in a general basis may not be orthogonal.

An orthogonal transformation may not be represented by an orthogonal matrix in a general basis - although always in an orthonormal basis. Your transformation may be norm-preserving, but the matrix representation in a general basis may not be orthogonal.

Yes got it now.

 

[Office_Shredder]

I think the really interesting question here is, given two vectors $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ in the new basis representation, how do you decide if they are orthogonal?

 

[Azad Koshur]

I think the really interesting question here is, given two vectors $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ in the new basis representation, how do you decide if they are orthogonal?

I think represent those into orthogonal basis coordinates and then calculate the norm using the usual way

 

[Office_Shredder]

I think represent those into orthogonal basis coordinates and then calculate the norm using the usual way

That's right. So when you say a matrix is orthogonal, you are making a claim about the rows and columns being orthogonal (notice matrix multiplication of the matrix with its transpose is doing the dot product of the rows and the columns). But if you have the matrix represented in the new basis, you would first have to transform the matrix back to being in the old basis before computing all those dot products to decide if it's orthogonal.

 

[PeroK]

I think the really interesting question here is, given two vectors $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ in the new basis representation, how do you decide if they are orthogonal?

Using the inner product of the new basis vectors, of course!

 

[Kashmir]

I think the really interesting question here is, given two vectors $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ in the new basis representation, how do you decide if they are orthogonal?

Here $x_i$ is the coefficient of the new basis vector $e'_i$?

 

[Kashmir]

Using the inner product of the new basis vectors, of course!

We can't use $x^T x$ as the definition of norm here I think.

 

[Azad Koshur]

yes but how is this norm presented in this new basis?

In general if $G$ is a matrix of the inner product then the orthogonality of a matrix $A$ means that $A^TGA=I$

Yes I was thinking the norm will still be given by the same way i did in the old basis. In this new basis I would convert the vectors back to standard basis by a matrix $P$, then the norm will be just $x^Tx$ (considering real elements only).

 

[Office_Shredder]

Here $x_i$ is the coefficient of the new basis vector $e'_i$?

That's correct

We can't use $x^T x$ as the definition of norm here I think.

This is also right. You have to transform the coordinates to a basis where the dot product is valid.

 

[WWGD]

As I understand, Ortho matrices do not just preserve the norm , but preserve the inner-product. Edit: meaning : <x,y>=<Tx, Ty> , for T orthogonal and <,> an inner product. Note that angles between vectors are also preserved.