Geometric Interpretation of Reversing the Components of a Vector

[Metronome]

Is there a nice geometric interpretation of reversing the components of a vector? So given some $\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}$, I want to visualize how it relates to $\begin{bmatrix}x_3 \\ x_2 \\ x_1\end{bmatrix}$. Is there a specific linear transformation that accomplishes this reversal? I have tried graphing a few vectors and their "algebraic inverses," and all I have noticed so far is that the length of the original vector is preserved.

Ultimately I am interested in a geometric interpretation of performing this action followed by taking a dot product with another vector (which in and of itself has a geometric interpretation of projection followed by scaling), as this is very close to a discrete convolution. Discrete convolution has its own geometric interpretation defined on functions of a discrete variable, but these are basically identical to vectors, so I was wondering what the operation would look like in geometric vector space.

 

[Dr.Peterson]

Is there a nice geometric interpretation of reversing the components of a vector? So given some $\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}$, I want to visualize how it relates to $\begin{bmatrix}x_3 \\ x_2 \\ x_1\end{bmatrix}$. Is there a specific linear transformation that accomplishes this reversal? I have tried graphing a few vectors and their "algebraic inverses," and all I have noticed so far is that the length of the original vector is preserved.

Ultimately I am interested in a geometric interpretation of performing this action followed by taking a dot product with another vector (which in and of itself has a geometric interpretation of projection followed by scaling), as this is very close to a discrete convolution. Discrete convolution has its own geometric interpretation defined on functions of a discrete variable, but these are basically identical to vectors, so I was wondering what the operation would look like in geometric vector space.

Try doing the equivalent in two dimensions, and you should see that it's a reflection across the line y=x.

What reflection is it in three dimensions?

 

[Metronome]

Try doing the equivalent in two dimensions, and you should see that it's a reflection across the line y=x.

What reflection is it in three dimensions?

It looks like this is a reflection around the span of the sum of the two vectors. $\begin{bmatrix}x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix}x_2 \\ x_1\end{bmatrix} = \begin{bmatrix}x_1 + x_2 \\ x_1 + x_2\end{bmatrix}$, whose span is $y = x$ for any initial choice of vector. $\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} + \begin{bmatrix}x_3 \\ x_2 \\ x_1\end{bmatrix} = \begin{bmatrix}x_1 + x_3 \\ 2 x_2 \\ x_1 + x_3\end{bmatrix}$, whose span is a different line depending on the initial choice of vector. So in three dimensions the transformation over the entire vector space is not a reflection?

 

[Dr.Peterson]

It looks like this is a reflection around the span of the sum of the two vectors. $\begin{bmatrix}x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix}x_2 \\ x_1\end{bmatrix} = \begin{bmatrix}x_1 + x_2 \\ x_1 + x_2\end{bmatrix}$, whose span is $y = x$ for any initial choice of vector. $\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} + \begin{bmatrix}x_3 \\ x_2 \\ x_1\end{bmatrix} = \begin{bmatrix}x_1 + x_3 \\ 2 x_2 \\ x_1 + x_3\end{bmatrix}$, whose span is a different line depending on the initial choice of vector. So in three dimensions the transformation over the entire vector space is not a reflection?

Not quite. Keep thinking ... but more simply. Don't bias yourself by starting in a way that depends on the given vector!

In three dimensions, you can reflect across a plane as well as a line. If you want to continue your line of thinking, consider what plane always contains the sum of the vectors?

 

[Steven G]

Might you be switching some of the axes???

 

[Metronome]

Not quite. Keep thinking ... but more simply. Don't bias yourself by starting in a way that depends on the given vector!

In three dimensions, you can reflect across a plane as well as a line. If you want to continue your line of thinking, consider what plane always contains the sum of the vectors?

Might you be switching some of the axes???

Think I got it, thanks! The action turns $\hat i$ into $\hat k$ and vice-versa, so the matrix is $\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}$. I found some manim code I could plug my matrix into and it looks like the plane of interest is $z = x$.