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I have this matrix which has a line of invariant points $y=0.5x$ and a line of invariance $y=0.25x$. This matrix is:

\begin{bmatrix} -3&8\\ -1&3\\ \end{bmatrix}

I am further told that it is a combination of a Shear and a reflection in the line $y=0.5x$. My trouble arises when I am asked to give a geometrical description of the sheer which I calculated to be:

$0.5\begin{bmatrix} -13&36\\ -9&23 \end{bmatrix}$

The answer states that this is a sheer that is in the direction parallel to the line $y=0.5x$. I am confused as to how this sheer direction is determined? How does the line of invariant points relate to the direction of the shear in a matrix, can anyone convince me?

EDIT: I have made a realization with the aid of a drawing, I realized that given $y=0.5x$ is a line of invariant points, the position of the points after the matrix will lie along that line hence moving parallel to the line $y=0.5x$.

Lo-urc
  • 567
  • What’s the difference between “line of invariant points” and “line of invariance?” – amd May 22 '17 at 00:10
  • i) A line that points lie on that once transformed stay in the same position, ii) a line that points lie on once transformed that may move a scalar distance along. – Lo-urc May 22 '17 at 16:04

2 Answers2

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A shear in two dimensions leaves points along some line fixed and shifts other points in a direction parallel to this line proportionally to their distance from the fixed line. In this case, the fixed line is the “line of invariant points” $2x=y$, so the shear is indeed parallel to this line.

A pure shear matrix only has the one fixed line, but this transformation also maps the line $4x=y$ to itself, so there’s a bit more going on. This line can be represented parametrically as $(4t,t)$ and from $$\begin{bmatrix}-3&8\\-1&3\end{bmatrix}\begin{bmatrix}4t\\t\end{bmatrix}=\begin{bmatrix}-4t\\-t\end{bmatrix}$$ we see that the transformation maps points along this line to their reflections in the origin. I’d hesitate to describe this as a reflection relative to the line $2x=y$, though, because that description usually applies to a transformation that reverses the line perpendicular to $2x=y$. I would instead describe this transformation as decomposing a vector into components in the directions of $(2,1)$ and $(4,1)$ and then reversing the latter component.

Incidentally, the question of which lines are mapped to themselves by a linear transformation is one of the motivations for the idea of eigenvectors and eigenvalues.

amd
  • 53,693
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I have made a realization with the aid of a drawing, I realized that given $y=0.5x$ is a line of invariant points, the position of the points after the matrix will lie along that line hence moving parallel to the line $y=0.5x$.

Lo-urc
  • 567