Changing Basis: Matrix E->F vs. Matrix B->C

[LCSphysicist]

Homework Statement

I am not sure if i am interpreting it right, but i think yes.

Relevant Equations

All below

See this exercise: It ask for the matrix changing the basis E -> F

If you pay attention, it write F in terms of E and write the matrix.
Now see this another exercise:

It ask the matrix B -> C, writing B in terms of C

Which is correct? If it are essentially equal, where am i interpreting wrong?

 

[etotheipi]

I think they are different types of transformation matrices. The first transforms the basis vectors, i.e. expresses one set of basis vectors as linear combinations of the other set.

The second matrix $P_{C\leftarrow B}$ transforms the components of a vector written in the basis B to the components of a vector written in the basis C.

For instance, the components of the element of $\mathbb{R}^2$ $(1,2)$ written in the basis $C$ is $[-3, 5.5]^T$ whilst the components of that same element written in the basis $B$ is $[1,0]^T$. Now note that $$[-3, 5.5]^T = P_{C\leftarrow B} [1,0]^T$$ I was also confused by similar matters very recently, and one of the key take-aways was that it is important to know the exact context of the matrix equation. Is it transforming basis vectors? Is it showing how the vector changes during a linear transformation? Is it transforming components of the same vector between two different bases? They all look similar, but are distinct operations .

 

[Mark44]

I think they are different types of transformation matrices. The first transforms the basis vectors, i.e. expresses one set of basis vectors as linear combinations of the other set.
The second matrix $P_{C\leftarrow B}$ transforms the components of a vector written in the basis B to the components of a vector written in the basis C.

I disagree. Both examples are doing the same thing: converting from one basis to another.

The first example is a bit trickier to see, since the vectors aren't given in coordinate form. If we assume that the basis E is the Euclidean basis for $\mathbb R^3$, it's a lot easier to follow what's happening.
From the equations given for basis F in terms of E (and with my assumption that E is the standard basis), we have:
$\vec {f_1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix}$
$\vec {f_2} = \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$
$\vec {f_3} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}$
So the vectors in basis F are the first, second, and third columns of the matrix identified as M. Using the notation of the second example, $M = P_{F \leftarrow E}$, the matrix that transforms the basis E to the basis F.

Borrowing from your notation for the second example, and applying it to the first example, we have $[1, -1, 0]^T = P_{F \leftarrow E}[1, 0, 0]^T$.

 

[etotheipi]

Yes, my mistake! You are indeed right. The first is also a matrix which transforms the components.

 

[LCSphysicist]

I disagree. Both examples are doing the same thing: converting from one basis to another.

The first example is a bit trickier to see, since the vectors aren't given in coordinate form. If we assume that the basis E is the Euclidean basis for $\mathbb R^3$, it's a lot easier to follow what's happening.
From the equations given for basis F in terms of E (and with my assumption that E is the standard basis), we have:
$\vec {f_1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix}$
$\vec {f_2} = \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$
$\vec {f_3} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}$
So the vectors in basis F are the first, second, and third columns of the matrix identified as M. Using the notation of the second example, $M = P_{F \leftarrow E}$, the matrix that transforms the basis E to the basis F.

Borrowing from your notation for the second example, and applying it to the first example, we have $[1, -1, 0]^T = P_{F \leftarrow E}[1, 0, 0]^T$.

Yes, that is a thing that i noticed, when we are dealing with vectors like the first example, it seems a little more tricky, i saw an example:

So we will need to summarize all of this system in the matrix we want.
I am not sure if the logic will help in this case as it helps in the second, but since the maths agree, that's okay for a while

 

[Mark44]

The first is also a matrix which transforms the components.

I think the error here ties into the thread you started a few days ago. In that thread, you made the distinction between (1) a matrix that transforms a vector in one basis to its coordinates in another basis, and (2) a matrix that transforms a vector to some other vector.

In one sense, these operations are different: a linear transformation that converts from one basis to another is an operator -- a transformation from one space to itself. The matrix representation for such a transformation will always be square. OTOH, a generic linear transformation could carry a vector in one space to a vector in a different vector space, e.g., $T: \mathbb R^2 \rightarrow \mathbb R^3$. The matrix representation for this type of linear transformation doesn't have to be square. In this case, the matrix would be 3 X 2.

In the other sense, a transformation is a transformation -- as long as it's a linear transformation, a matrix representation for it can be found. Where things get interesting is in the diagonalization of a matrix. In this process you have a matrix that converts one basis to another, as well as another matrix that does the opposite conversion (i.e., the inverse of the first conversion matrix), together with a matrix that represents some arbitrary linear transformation.

 

[etotheipi]

I think that a change of basis is not a linear transformation of the vector space; i.e. where a linear transformation is a function $T: V\rightarrow W$.

For a change of basis we are just finding the components of the same vector in the same underlying vector space, in a new basis.

That is why I like to keep them separate in my mind. Even though we can represent both in terms of a matrix multiplication.

 

[Mark44]

I think that a change of basis is not a linear transformation of the vector space; i.e. where a linear transformation is a function $T: V\rightarrow W$.

You are mistaken. Possibly you are under the misconception that a linear transformation has to be a map between two different vector spaces. That is not true.
A change of basis is always a linear transformation from a vector space to itself; i.e., $T: V\rightarrow V$. Other kinds of linear transformations can map one vector space to another of different dimension.
The first example in post #1 shows the matrix M that represents this linear transformation.

For a change of basis we are just finding the components of the same vector in the same underlying vector space, in a new basis.

Which is a linear transformation, and for which there is a (square) matrix representation. Please read (or reread) what I wrote in post #6.

That is why I like to keep them separate in my mind. Even though we can represent both in terms of a matrix multiplication.

Any linear transformation can be represented as a matrix, and vice versa. From my earlier post:

​

In the other sense, a transformation is a transformation​

​

​

 

[etotheipi]

I think we must be careful here. Let us define the identity operator $I_v: V \rightarrow V$ as an automorphism.

Then we write $\vec{v} = I_v\vec{v}$, and transform this into the matrix equation $[\vec{v}]_{\beta} = [I_v]_{\beta'}^{\beta} [\vec{v}]_{\beta'}$.

I would say that the change of basis is specifically $[\vec{v}]_{\beta} = [I_v]_{\beta'}^{\beta} [\vec{v}]_{\beta'}$, i.e. just the relabelling of coordinates, and not the whole linear transformation $\vec{v} = I_v\vec{v}$.

So if I had to pick, I think a change of basis is at best a certain matrix representation of the identity transformation. Although I think we agree on the important mathematics, and are just debating the linguistics.

 

[Mark44]

I think we must be careful here. Let us define the identity operator $I_v: V \rightarrow V$ as an automorphism.

Then we write $\vec{v} = I_v\vec{v}$, and transform this into the matrix equation $[\vec{v}]_{\beta} = [I_v]_{\beta'}^{\beta} [\vec{v}]_{\beta'}$.

I would say that the change of basis is specifically $[\vec{v}]_{\beta} = [I_v]_{\beta'}^{\beta} [\vec{v}]_{\beta'}$, i.e. just the relabelling of coordinates, and not the whole linear transformation $\vec{v} = I_v\vec{v}$.

Relabelling the coordinates is a linear transformation. I don't understand why you are persisting with this false dichotomy that amounts to "some linear transformations are not linear transformations, but other linear transformations are linear transformations."

So if I had to pick, I think a change of basis is at best a certain matrix representation of the identity transformation. Although I think we agree on the important mathematics, and are just debating the linguistics.

No, we are not in agreement on the mathematics. An identity matrix does exactly what its name suggests: its output is identically equal to its input. I.e., $I\vec v = \vec v$.
A change-of-basis matrix (which I emphasize is a linear transformation) will never be the identity matrix if at least one basis vector is different between the two bases.

 

[etotheipi]

A change-of-basis matrix (which I emphasize is a linear transformation) will never be the identity matrix if at least one basis vector is different between the two bases.

I think this is incorrect. The identity linear transformation $I_v$ I defined above can be represented as a matrix given a pair of bases, but it is not a matrix. It is a very different object, and coordinate independent.

For a change of basis, the linear operator must be the identity otherwise we will have changed the vector itself (i.e. we must have $\vec{v} = I_v \vec{v}$). This doesn't mean that the matrix representing this operator necessarily has components $(I_v)_{ij} = \delta_{ij}$, though. E.g. if the two $\vec{v}$'s are decomposed to coordinate vectors w.r.t. different bases.

We must be careful to distinguish between not only vectors and coordinate vectors, but also linear transformations and matrices.

An identity matrix does exactly what its name suggests: its output is identically equal to its input. I.e., $I\vec v = \vec v$.

I would indeed hope that this be the case for a change of basis, as the vector is frame invariant!

Relabelling the coordinates is a linear transformation.

I think a linear transformation is a mapping $V \rightarrow W$ ($W$ can still be $V$). This is different to $[\vec{v}]_{\beta} = [I_v]_{\beta'}^{\beta} [\vec{v}]_{\beta'}$, which is just a matrix multiplication.

 

[Mark44]

I think this is incorrect. The identity linear transformation $I_v$ I defined above can be represented as a matrix given a pair of bases, but it is not a matrix. It is a very different object, and coordinate independent.

This makes no sense. "The identity linear transformation ... can be represented as a matrix given a pair of bases, but it is not a matrix."
You are conflating two different concepts -- the identity transformation (and its matrix), and a change-of-basis transformation (and its matrix).

1. The identity transformation $I(\vec x) = \vec x$. What you get out is exactly the same as its argument. The identity transformation is not involved in changing from one basis to another.
2. A change-of-basis transformation changes a vector from the coordinates in one basis to those in another basis. The two sets of coordinates do represent the same vector, but I don't see how you can consider two sets of different coordinates to be equal, which must happen if this kind of transformation is an identity.

For a change of basis, the linear operator must be the identity otherwise we will have changed the vector itself (i.e. we must have $\vec{v} = I_v \vec{v}$).

Or more to the point $vec v = I \vec v$ for an identity. The identity transformation gives the coordinates of the basis vectors in terms of themselves. It's not an identity transformation if there are two basses involved.

This doesn't mean that the matrix representing this operator necessarily has components $(I_v)_{ij} = \delta_{ij}$, though. E.g. if the two $\vec{v}$'s are decomposed to coordinate vectors w.r.t. different bases.

We must be careful to distinguish between not only vectors and coordinate vectors, but also linear transformations and matrices.

You are being too careful, trying to make distinctions where none exist.

I would indeed hope that this be the case for a change of basis, as the vector is frame invariant!

I think a linear transformation is a mapping $V \rightarrow W$ ($W$ can still be $V$). This is different to $[\vec{v}]_{\beta} = [I_v]_{\beta'}^{\beta} [\vec{v}]_{\beta'}$, which is just a matrix multiplication.

It is misleading to label and talk about $[I_v]_{\beta'}^{\beta}$ as some sort of identity.

 

[etotheipi]

You are conflating two different concepts -- the identity transformation (and its matrix), and a change-of-basis transformation (and its matrix)

2. A change-of-basis transformation changes a vector from the coordinates in one basis to those in another basis. The two sets of coordinates do represent the same vector, but I don't see how you can consider two sets of different coordinates to be equal, which must happen if this kind of transformation is an identity.

I argue that the change-of-basis transformation is necessarily the identity transformation.

What I am saying is that a linear transformation that is not the identity would change the actual vector; for a change of basis, we are interested in representing the same vector in a different basis (and a vector object $\vec{v}$ is invariant under changes of coordinate system). Hence the solution must involve only the trivial automorphism.

So suppose we forget about the coordinate representations for a minute, and take a coordinate-free approach. If the change of basis constitutes a linear transformation (something of the form $\vec{a} = T\vec{b}$), then it must involve only the identity operator $$\vec{v} = I_v \vec{v}$$ Any other operator would not make sense, because that would give us a different underlying vector object. The key is that now we convert this equation into a matrix equation, but by decomposing both $\vec{v}$'s into coordinate vectors w.r.t. different bases. We can do this because we can always find a matrix $[I_v]_{\beta}^{\beta'}$ such that $$[\vec{v}]_{\beta'} = [I_v]_{\beta}^{\beta'} [\vec{v}]_{\beta}$$ This is now a matrix equation, involving the coordinate representations of the same vector in two different bases linked by a matrix multiplication. $[I_v]_{\beta}^{\beta'}$ is the identity transformation decomposed w.r.t. $\beta$ and $\beta'$, and it is not the unit matrix unless $\beta = \beta'$.

I also add some references, to prove I am not just plucking this out of thin air:

1. These Carleton notes says the following

Matrix representation of the identity linear transformation

In fact, for any n-dimensional vector space $V$ and ordered bases $\Gamma$ and $\Omega$ of $V$, there is a unique matrix $A$ such that for every $u \in V$, ${\lbrack u \rbrack}_{\Omega} = A{\lbrack u \rbrack}_{\Gamma}$

To see this, let $\text{id}$ denote the identity linear transformation. That is, $\text{id}(u)=u$ for all $u \in V$. Then $[\text{id}]_{\Gamma}^{\Omega}$ is precisely the matrix we desire because $u$ is the output of $\text{id}(u)$, and the product $[\text{id}]_{\Gamma}^{\Omega} {\lbrack u \rbrack}_{\Gamma}$, as we saw in the previous segment, gives the tuple representation of the output with respect to the ordered basis $\Omega$.

We call $[\text{id}]_{\Gamma}^{\Omega}$ the change-of-basis matrix from $\Gamma$ to $\Omega$. Note that this matrix is necessarily invertible since $\text{id}$ is a bijection.

2. Proof Wiki says the same thing,

Let $A = \langle a_n \rangle$ and $B = \langle b_n \rangle$ be ordered bases of G... Let $I_G$ be the identity linear operator on $G$... Then $[I_G ; \langle a_n \rangle, \langle b_n \rangle ]$ is called the matrix corresponding to the change of basis from $\langle a_n \rangle$ to $\langle b_n \rangle$.

 

[Mark44]

The disagreement boils down to what it means for two vectors to be equal. For example, if $\vec u = (0, 0, 1)$ and $\vec v = (-1, 0, 1)$, in the absence of further information, no one would conclude that $\vec u = \vec v$, because the coordinates are different. By the same token, no one would claim that T was the identity transformation if $T(\vec u) = \vec v$, or equivalently, that $A\vec u = \vec v$ was the identity matrix, where A is the matrix representation of T.

On the other hand, if we agree that two different representations, in different bases, can refer to the same vector (your position), then what you say about a change-of-basis matrix does make sense. Although I took a year-long 400-level sequence in linear algebra while I was a grad student, the idea of a change-of-basis transformation/matrix being a sort of identity never came up. However, we did work with transformations and matrices using a notation similar to what you have been using. Of the several linear algebra textbooks I own, only one uses this notation: "Linear Algebra," by P.G. Kumpel and J.A. Thorpe.

In it they talk about the coordinates of a vector in terms of a basis B, using $M_B(\vec v) = \begin{bmatrix}c_1\\ \vdots \\ c_n \end{bmatrix}$.
For the change-of-basis matrix of a transformation T, relative to the bases B and B', the notation is $M^B_{~~B'}(T(\vec v))$. The columns of this matrix are $\left[ M_B(\vec{ v_1}) \dots M_B(\vec{ v_n} )\right]$. Finally, the book I mentioned has a theorem about the representation of $T(\vec v)$ relative to a different basis B': $M_{B'}(T(\vec v)) = M^B_{~~B'}(T) M_B (\vec v)$.

As an example, let $E = \{\vec{e_1}, \vec{e_2}, \vec{e_3} \}$ be the standard Euclidean basis for $\mathbb R^3$ and B be the basis $\{\vec{e_1}, \vec{e_2}, \vec{b_3} \}$, with $b_3 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$, and T be the identity transformation relative to the standard basis, or I.
Per the formula above, $M^E_{~~B}(I) = \begin{bmatrix} M_B(\vec{e_1}) M_B(\vec{e_2}) M_B(\vec{e_3}) (\end{bmatrix}$. Here each entry in the last matrix is a column vector.
This results in the matrix $\begin{bmatrix}1 & 0 & -1\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$.
This is the change-of-basis matrix, which changes a vector in E coordinates to its counterpart in B coordinates.

It's interesting to note that for this example $\left(M^E_{~~B}(I)\right) \left( M^E_{~~B}(I)\right) = M^E_{~~B}(I)$, a result we would expect when multiplying an identity matrix by itself.

 

[xaos]

$(1,2)=c_{1}v_{1}+c_{2}v_{2}=(c_{1},c_{2})[v_{1},v_{2}]=[(1,2)]_β[v_{1},v_{2}]$
$[(1,2)]_β = (c_{1},c_{2}) = (1,2)[v_{1},v_{2}]^{-1}$
$(1,2)=d_{1}w_{1}+d_{2}w_{2}=(d_{1},d_{2})[w_{1},w_{2}]=[(1,2)]_{β'}[w_{1},w_{2}]$
$[(1,2)]_{β'} = (d_{1},d_{2}) =(1,2)[w_{1},w_{2}]^{-1}$
$[(1,2)]_{β'} =[(1,2)]_β[[v_{1},v_{2}][w_{1},w_{2}]^{-1}]^{β}_{β'}$
(1,2) never changed, just how it is represented changed. Hense a 'change of basis' rather than a 'change of vector'

 

[LCSphysicist]

Just a funny fact to end the discussion (actually, already ended, but is interesting), i just learn (so it can be wrong, but probably not) a name for this
$$e_{\upsilon '} = a_{v'}^{v}e_{v} $$ the vectors quantities are named covariant
$$\alpha ^{\upsilon } = a_{v'}^{v}\alpha ^{v'} $$ are so named contravariant XD