Do column 'vectors' need a basis?

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etotheipi
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Consider the transformation of the components of a vector ##\vec{v}## from an orthonormal coordinate system with a basis ##\{\vec{e}_1, \vec{e}_2, \vec{e}_3 \}## to another with a basis ##\{\vec{e}'_1, \vec{e}'_2, \vec{e}'_3 \}##

The transformation equation for the components of ##\vec{v}## looks something like$$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} = \begin{bmatrix}
\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\
\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\
\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}
\
\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}
$$Now since ##\vec{v} = v_i \vec{e}_i = v'_i\vec{e}'_i##, and the matrix is not the identity, the structures ##\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix}## and ##\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}## in the above expression can't be ##\vec{v}##. Instead, they just appear to be 3-tuples of numbers with no apparent basis.

Similarly, in different contexts we sometimes put vectors in such a column structure, like ##\begin{pmatrix}\vec{e}_1\\\vec{e}_2\\\vec{e}_3\end{pmatrix}## e.g. when looking at how the basis transforms.

It appears to me then that something like ##\begin{pmatrix}a\\b\\c\end{pmatrix}## can represent the raw tuple ##(a,b,c)##, or an expression with certain basis vectors like ##a\vec{e}_1 + b\vec{e}_2 + c\vec{e}_3## , depending on the context. I wondered if someone could clarify whether this is along the right lines?
 
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  • #2
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Consider the transformation of the components of a vector ##\vec{v}## from an orthonormal coordinate system with a basis ##\{\vec{e}_1, \vec{e}_2, \vec{e}_3 \}## to another with a basis ##\{\vec{e}'_1, \vec{e}'_2, \vec{e}'_3 \}##

The transformation equation for the components of ##\vec{v}## looks something like$$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} = \begin{bmatrix}
\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\
\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\
\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}
\
\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}
$$Now since ##\vec{v} = v_i \vec{e}_i = v'_i\vec{e}'_i##, the structures ##\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix}## and ##\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}## in the above expression can't be ##\vec{v}##.
Sure they can. They are just different representations of ##\vec v## in different bases.
etotheipi said:
Instead, they just appear to be 3-tuples of numbers with no apparent basis.
No. Any time you have the coordinate representation of a vector, the entries are the coordinates of the underlying basis.

For a simpler example, consider the vector ##\vec v = \begin{pmatrix}1\\2 \end{pmatrix}##. Without any other information, we would naturally assume that the coordinates are those of the standard basis for ##\mathbb R^2##, ##\hat {e_1}## and ##\hat{e_2}##.
If we instead write ##\vec v## in terms of the basis ##\{\begin{pmatrix}1\\0 \end{pmatrix}, \begin{pmatrix}1\\1 \end{pmatrix}\}##, the representation of ##\vec v## in terms of that basis would be ##\begin{pmatrix} -1\\2 \end{pmatrix}##.
etotheipi said:
Similarly, in different contexts we sometimes put vectors in such a column structure, like ##\begin{pmatrix}\vec{e}_1\\\vec{e}_2\\\vec{e}_3\end{pmatrix}## e.g. when looking at how the basis transforms.

It appears to me then that something like ##\begin{pmatrix}a\\b\\c\end{pmatrix}## can represent the raw tuple ##(a,b,c)##
Again, there's really no such thing as a "raw tuple". The coordinates a, b, and c are the scalar multipliers of whatever vectors are in the basis you're using.
etotheipi said:
, or an expression with certain basis vectors like ##a\vec{e}_1 + b\vec{e}_2 + c\vec{e}_3## , depending on the context. I wondered if someone could clarify whether this is along the right lines?
 
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  • #3
etotheipi
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Sure they can. They are just different representations of ##\vec v## in different bases.
I don't agree. In the expression $$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} = \begin{bmatrix}
\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\
\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\
\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}
\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}
$$ The matrix is not the identity, so $$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} \neq \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$ It is instead ##\vec{v}## that is independent of the coordinate system, i.e. ##v'_i \vec{e}'_i = v_i \vec{e}_i = \vec{v}##.
 
  • #4
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I don't agree. In the expression $$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} = \begin{bmatrix}
\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\
\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\
\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}
\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}
$$ The matrix is not the identity, so $$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} \neq \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$ It is instead ##\vec{v}## that is independent of the coordinate system, i.e. ##v'_i \vec{e}'_i = v_i \vec{e}_i = \vec{v}##.
That's not the point. Of course the matrix is not the identity matrix. Also, I'm not saying that the coordinates of a vector in one basis will be pairwise equal to the coordinates of the same vector in another basis.

In the example I gave, we have two different representations of the same vector; namely (1, 2), in the standard basis for R^2, and (-1, 2), in another basis. Obviously the coordinates are different, but they nevertheless represent a single vector.
 
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  • #5
etotheipi
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That's not the point. Of course the matrix is not the identity matrix. Also, I'm not saying that the coordinates of a vector in one basis will be pairwise equal to the coordinates of the same vector in another basis.

In the example I gave, we have two different representations of the same vector; namely (1, 2), in the standard basis for R^2, and (-1, 2), in another basis. Obviously the coordinates are different, but they nevertheless represent a single vector.
Though my question is really about the specific structure $$
\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} = \begin{bmatrix}

\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\

\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\

\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}

\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$ The column structures in that expressions are evidently not equal to ##\vec{v}##. So I wonder what basis they are represented in.
 
  • #6
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I don't agree. In the expression $$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} = \begin{bmatrix}
\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\
\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\
\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}
\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}
$$ The matrix is not the identity, so $$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} \neq \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$ It is instead ##\vec{v}## that is independent of the coordinate system, i.e. ##v'_i \vec{e}'_i = v_i \vec{e}_i = \vec{v}##.
It depends what you mean by the ##=## sign. Some notations I use are:
$$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix}' = \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$
$$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix}_{b2} = \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}_{b1}$$
$$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} \leftrightarrow \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$
All these indicate that it's the same vector represented by a different column of numbers in different bases.
 
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  • #7
etotheipi
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But it is not the case that: $$

(v'_1 \vec{e}'_1 + v'_2 \vec{e}'_2 + v'_3 \vec{e}'_3) =

\begin{bmatrix}

\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\

\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\



\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}



(v_1 \vec{e}_1 + v_2 \vec{e}_2 + v_3 \vec{e}_3)

$$ I.e. my point is that it makes no sense to write: $$

\vec{v} =

\begin{bmatrix}

\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\

\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\



\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}



\vec{v}

$$since the matrix is not the identity. So in the original expression $$
\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} = \begin{bmatrix}

\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\

\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\

\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}

\

\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$those column vectors must not equal ##\vec{v}##! I then wondered what the basis was for each of those column vectors... :wink:
 
  • #8
Math_QED
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Of course, there is something like 'rawr' tuples. ##\mathbb{R}^3## is exactly the collections of formal tuples ##(a,b,c)## with ##a,b,c \in \mathbb{R}##. I don't need linear algebra, or coordinates, or bases to talk about this.

Given a vector ##(a,b,c) \in \mathbb{R}^3##, we can look at the coordinates of this vector w.r.t. to a basis ##E=(e_1, e_2, e_3)##. What are these coordinates? Well, we can write ##(a,b,c) = \lambda_1 e_1 + \lambda_2 e_2 + \lambda_3 e_3## for unique ##\lambda_1, \lambda_2, \lambda_3 \in \mathbb{R}##. The coordinates of ##(a,b,c)## w.r.t. the basis ##E## is then ##(\lambda_1, \lambda_2, \lambda_3)##. Let us introduce some notation. Let's write ##[(a,b,c)]_E= (\lambda_1, \lambda_2, \lambda_3)##.

Your question is basically: given bases ##E= (e_1, e_2, e_3)## and ##E' = (e_1', e_2', e_3')##, what is the relation between ##[(a,b,c)]_E## and ##[(a,b,c)]_{E'}##? The answer is that you can get from one to the other via a matrix multiplication.

I guess in your post ##(v_1, v_2, v_3)^T## is the coordinates of a vector w.r.t. ##E## and ##(v_1', v_2', v_3')^T## is the coordinates of the same vector w.r.t. ##E'##.
 
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  • #9
Stephen Tashi
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But it is not the case that: $$

(v'_1 \vec{e}'_1 + v'_2 \vec{e}'_2 + v'_3 \vec{e}'_3) =

\begin{bmatrix}

\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\

\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\



\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}



(v_1 \vec{e}_1 + v_2 \vec{e}_2 + v_3 \vec{e}_3) $$
When you say "is not the case", are you assuming the notation on the right hand side of the equal sign has a defined meaning?
 
  • #10
etotheipi
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Your question is basically: given bases ##E= (e_1, e_2, e_3)## and ##E' = (e_1', e_2', e_3')##, what is the relation between ##[(a,b,c)]_E## and ##[(a,b,c)]_{E'}##? The answer is that you can get from one to the other via a matrix multiplication.
Thank you, yes this was my suspicion. That the column vectors on the LHS and RHS were tuples of the components with no basis, transformed via matrix multiplication.

When you say "is not the case", are you assuming the notation on the right hand side of the equal sign has a defined meaning?
I suppose I took some liberties with that questionable notation to make the point, but essentially I meant that the column vectors on the RHS and LHS aren't even representations of the same vector in a different basis. But instead as @Math_QED mentioned they are just tuples of numbers.
 
  • #11
Math_QED
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I suppose I took some liberties with that questionable notation to make the point, but essentially I meant that the column vectors on the RHS and LHS aren't even representations of the same vector in a different basis. But instead as @Math_QED mentioned they are just tuples of numbers.
The column vectors on the RHS an LHS are representing the same vector. Say your fixed vector is ##(a,b,c) \in \mathbb{R}^3##. Then ##(v_1, v_2, v_3) = [(a,b,c)]_E## and ##(v_1', v_2', v_3')= [(a,b,c)]_{E'}##.
 
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  • #12
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But it is not the case that: $$

(v'_1 \vec{e}'_1 + v'_2 \vec{e}'_2 + v'_3 \vec{e}'_3) =

\begin{bmatrix}

\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\

\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\



\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}



(v_1 \vec{e}_1 + v_2 \vec{e}_2 + v_3 \vec{e}_3)

$$ I.e. my point is that it makes no sense to write: $$

\vec{v} =

\begin{bmatrix}

\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\

\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\



\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}



\vec{v}

$$since the matrix is not the identity. So in the original expression $$
\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} = \begin{bmatrix}

\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\

\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\

\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}

\

\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$those column vectors must not equal ##\vec{v}##! I then wondered what the basis was for each of those column vectors... :wink:
There is a duality in linear algebra. Let's start with a vector, ##u## and a linear transformation ##T##. We have:
$$v = Tu$$
Where ##v## is another vector. Now, if we choose any basis, this equation takes the form of a matrix/tuple equation:
$$
\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} = \begin{bmatrix}

T_{11} & T_{12} & T_{13} \\
T_{21} & T_{22} & T_{23} \\
T_{31} & T_{32} & T_{33} \end{bmatrix}

\

\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}$$
But, we can also interpret a matrix equation like this as a change of basis, where now the numbers ##v_1, v_2, v_3## represent the components of ##u## in a new basis. So, you need to be clear about what you are doing.
 
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  • #13
etotheipi
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There is a duality in linear algebra. Let's start with a vector, ##u## and a linear transformation ##T##. We have:
$$v = Tu$$
Where ##v## is another vector. Now, if we choose any basis, this equation takes the form of a matrix/tuple equation:
$$
\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} = \begin{bmatrix}

T_{11} & T_{12} & T_{13} \\
T_{21} & T_{22} & T_{23} \\
T_{31} & T_{32} & T_{33} \end{bmatrix}

\

\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}$$
But, we can also interpret a matrix equation like this as a change of basis, where now the numbers ##v_1, v_2, v_3## represent the components of ##u## in a new basis. So, you need to be clear about what you are doing.
Thanks, this sums it up nicely. With linear transformations you get out a new vector, but with component transformations you get out a representation of the same vector. So in the tuple equation you wrote up, ##\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} \not\equiv v_1 \hat{x} + v_2 \hat{y} + v_3 \hat{z}##, whilst this would be true for the linear transformation.

You can get other equations like

$$


\begin{pmatrix}\vec{e}'_1\\\vec{e}'_2\\\vec{e}'_3\end{pmatrix} = \begin{bmatrix}



T_{11} & T_{12} & T_{13} \\

T_{21} & T_{22} & T_{23} \\

T_{31} & T_{32} & T_{33} \end{bmatrix}



\



\begin{pmatrix}\vec{e}_1\\\vec{e}_2\\\vec{e}_3\end{pmatrix}$$ where again this is matrix multiplication and the objects in each row of the column aren't coefficients of some basis vector (that wouldn't even make semantic sense in this case!).
 
  • #14
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Thanks, this sums it up nicely. With linear transformations you get out a new vector, but with component transformations you get out a representation of the same vector.
The change-of-basis matrix is a linear transformation, so the distinction isn't as clear-cut as you seem to think.
etotheipi said:
So in the tuple equation you wrote up, ##\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} \not\equiv v_1 \hat{x} + v_2 \hat{y} + v_3 \hat{z}##, whilst this would be true for the linear transformation.
This is unclear. Is the vector ##\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}## in terms of the basis ##\{\hat x, \hat y, \hat z \}##? If so, then ##\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}## does equal ##v_1 \hat{x} + v_2 \hat{y} + v_3 \hat{z}##.
etotheipi said:
You can get other equations like

$$
\begin{pmatrix}\vec{e}'_1\\\vec{e}'_2\\\vec{e}'_3\end{pmatrix} = \begin{bmatrix}
T_{11} & T_{12} & T_{13} \\
T_{21} & T_{22} & T_{23} \\
T_{31} & T_{32} & T_{33} \end{bmatrix}
\



\begin{pmatrix}\vec{e}_1\\\vec{e}_2\\\vec{e}_3\end{pmatrix}$$ where again this is matrix multiplication and the objects in each row of the column aren't coefficients of some basis vector (that wouldn't even make semantic sense in this case!).
Which of the five columns do you mean?
 
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  • #15
Stephen Tashi
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There is a duality in linear algebra.
Consider the transformation of the components of a vector ##\vec{v}## from an orthonormal coordinate system with a basis ##\{\vec{e}_1, \vec{e}_2, \vec{e}_3 \}## to another with a basis ##\{\vec{e}'_1, \vec{e}'_2, \vec{e}'_3 \}##

The transformation equation for the components of ##\vec{v}## looks something like$$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix} = \begin{bmatrix}
\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\
\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\
\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}
\
\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}
$$
We could try to distinguish between the dual ideas by using the notation:

$$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix}_{b'} = \begin{bmatrix}
\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\
\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\
\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}
\
\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}_b
$$

when both sides of the equation represent the same vector.
We could even go as far as putting parenthesis around the entire right hand side of the above equation and subscripting it with a ##b'##. (Something I don't know how to do in LaTex!)


That would contrast to the notation for a linear transformation that maps a vector to a different vector. By analogy to expressions like ##y = 3x ## or ##x' = 3x## we could write

$$\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix}_{b} = \begin{bmatrix}
\vec{e}'_1 \cdot \vec{e}_1 & \vec{e}'_1 \cdot \vec{e}_2 & \vec{e}'_1 \cdot \vec{e}_3 \\
\vec{e}'_2 \cdot \vec{e}_1 & \vec{e}'_2 \cdot \vec{e}_2 & \vec{e}'_2 \cdot \vec{e}_3 \\
\vec{e}'_3 \cdot \vec{e}_1 & \vec{e}'_3 \cdot \vec{e}_2 & \vec{e}'_3 \cdot \vec{e}_3 \end{bmatrix}
\
\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}_b
$$

with the understanding that ##\begin{pmatrix}v'_1\\v'_2\\v'_3\end{pmatrix}_{b} ## is a notation for a variable that represents a vector different than ##\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}_{b}##.
 
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  • #16
etotheipi
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Actually I think I see what you guys were trying to say now. Really when we write $$\vec{v} = \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$this is shorthand for $$[\vec{v}]_{\beta} = \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$since the column structure is just a tuple, that can undergo matrix multiplication. I was previously under the (I think, now, erroneous) impression that the column structure "had basis vectors built in" but after this discussion I don't think that makes sense. It seems much more coherent for it to just be a tuple of numbers.

Sorry for the confusion, and thanks for helping out!
 
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Actually I think I see what you guys were trying to say now. Really when we write $$\vec{v} = \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$this is shorthand for $$[\vec{v}]_{\beta} = \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$$since the column structure is just a tuple, that can undergo matrix multiplication.
The numbers ##v_1, v_2,## and ##v_3## are the scalar multiples of the basis vectors, so in a column vector, the basis vectors are built in.
etotheipi said:
I was previously under the (I think, now, erroneous) impression that the column structure "had basis vectors built in" but after this discussion I don't think that makes sense. It seems much more coherent for it to just be a tuple of numbers.
As I understand what you wrote in this thread, your previous impression was that a column vector was just a "raw tuple," as you put it. That's the impression that was erroneous. This has nothing to do with linear transformations. If the basis is not explicitly shown, we usually assume that we're dealing with the standard basis (i.e., Euclidean basis) for whatever space is being considered.
 
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  • #19
etotheipi
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The numbers ##v_1, v_2,## and ##v_3## are the scalar multiples of the basis vectors, so in a column vector, the basis vectors are built in.
That's not what the Colorado notes seem to say. It appears that when we express the vector in column form (as a coordinate vector), we map the vector to a tuple in ##\mathbb{R}^n## (with entries as coefficients of a certain basis), and a tuple does not require any basis vectors (it's just a list).

I think this was what @Math_QED alluded to in #8.
 
  • #20
etotheipi
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The numbers ##v_1, v_2,## and ##v_3## are the scalar multiples of the basis vectors, so in a column vector, the basis vectors are built in.
As I understand what you wrote in this thread, your previous impression was that a column vector was just a "raw tuple," as you put it. That's the impression that was erroneous. This has nothing to do with linear transformations. If the basis is not explicitly shown, we usually assume that we're dealing with the standard basis (i.e., Euclidean basis) for whatever space is being considered.
For instance, ##\vec{v} = B[\vec{v}]_{\beta} = [\vec{e}_1, \vec{e}_2, \vec{e}_3]\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}##, where we manually have to matrix multiply by the basis in order to obtain the vector. So the basis vectors aren't built-in to the tuple, we need to put them in manually.
 
  • #21
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That's not what the Colorado notes seem to say. It appears that when we express the vector in column form (as a coordinate vector), we map the vector to a tuple in ##\mathbb{R}^n## (with entries as coefficients of a certain basis), and a tuple does not require any basis vectors (it's just a list).
A mapping onto ##\mathbb{R}^n## implies a basis, namely the vectors that get mapped to ##(1,0,0)## etc. A good idea to try to distinguish these things is to think of the polynomials as the vector space. Then you have a clear definition of your vectors, which aren't just tuples (like we have with ##\mathbb{R}^n## itself).
 
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  • #22
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A mapping onto ##\mathbb{R}^n## implies a basis, namely the vectors that get mapped to ##(1,0,0)## etc. A good idea to try to distinguish these things is to think of the polynomials as the vector space. Then you have a clear definition of your vectors, which aren't just tuples (like we have with ##\mathbb{R}^n## itself).
Yes, sorry that's what I had in mind. I was trying to refer to the fact that the basis vector objects themselves aren't built into the column structure (a tuple), but instead must be put in as in #20.
 
  • #23
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Final conceptual (though maybe actually notational :wink:) question; take the standard Cartesian basis ##\beta = \{\hat{x}, \hat{y}, \hat{z}\}##. If we were going to be pedantic, would we say: $$[\hat{x}]_{\beta} = (1,0,0)$$I ask because even the most thorough references, who make use of the notation ##[\vec{v}]_{\beta}##, still write ##\hat{x} = (1,0,0)##.

To confirm, this is just because it's considered obvious that ##\hat{x} \, \, (= 1\hat{x} + 0\hat{y} + 0\hat{z})## is expressed in the standard basis, right? There's no mathematical trickery going on that makes this any sort of exception?
 
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Final conceptual (though maybe actually notational :wink:) question; take the standard Cartesian basis ##\beta = \{\hat{x}, \hat{y}, \hat{z}\}##. If we were going to be pedantic, would we say: $$[\hat{x}]_{\beta} = (1,0,0)$$I ask because even the most thorough references, who make use of the notation ##[\vec{v}]_{\beta}##, still write ##\hat{x} = (1,0,0)##.

To confirm, this is just because it's considered obvious that ##\hat{x} \, \, (= 1\hat{x} + 0\hat{y} + 0\hat{z})## is expressed in the standard basis, right? There's no mathematical trickery going on that makes this any sort of exception?
This is too specfic. Until you define a norm, there is no concept of a "unit" vector. If the vector space is explicitly ##\mathbb{R}^n## then you can imagine things like ##\hat x##. But, if your vector space is the quadratic polynomials (and below), then what is ##\hat x##? What is the length of the vector ##x^2 + 3x + 1##? What is your "default Cartesian basis"?
 
  • #25
etotheipi
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This is too specfic. Until you define a norm, there is no concept of a "unit" vector. If the vector space is explicitly ##\mathbb{R}^n## then you can imagine things like ##\hat x##. But, if your vector space is the quadratic polynomials (and below), then what is ##\hat x##? What is the length of the vector ##x^2 + 3x + 1##? What is your "default Cartesian basis"?
I see your point. In pure maths, it it seems to makes perfect sense to have a basis consisting of tuples of elements of ##\mathbb{R}^n##. So your polynomial would have a coordinate vector of ##(1,3,1)## in the standard basis ##\beta = \{(1,0,0), (0,1,0), (0,0,1)\}## (Correction by @PeroK: it should of course be ##\beta = \{x^2, x, 1\}##). The basis vectors basically define themselves :wink:, I don't actually see any issue with the equality ##\vec{e}_1 = (1,0,0)## in pure maths.

But I have some trouble translating this to Physics. We have an affine base space, in which we choose an origin in addition to three (let's say orthogonal, not necessarily normalised since we have no metric yet) geometrical vectors (arrows). Now we can really just express any other arrow as a sum of multiples of the basis arrows.

The best we can do is something like ##\vec{e}_1 = 1\vec{e}_1 + 0\vec{e}_2 + 0\vec{e}_3##. To map that to the reals, I would have thought we need to write ##[\vec{e}_1]_{\beta} = (1,0,0)##. Maybe I don't know any better, but I'm slightly weary in this case about equating ##\vec{e}_1## and ##(1,0,0)##.

Perhaps this doesn't make much sense... I have been learning LA from Axler but it takes a very pure approach!
 
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