Finding eigenvalues without doing calculations

Question

Find, without doing any calculations, the eigenvalues of the following linear transformations from $\mathbb{R}^2 \to \mathbb{R}^2$:

$A)\quad$ Projection on a straight line that contains $(0,0)$

$B)\quad$ Symmetry with respect to a straight line that contains $(0,0)$

$C)\quad$ Symmetry with respect to the origin

How can I, without doing any calculations, do that? I think it's related to the definition of eigenvalues and eigenvectors, but I can't manage to relate them to this excercise.

Answer 1

When you apply a transformation, the eigenvectors are the vectors that stay pointing in the same direction. They may change in length; the associated eigenvalue is the scale factor for the length change.

To invent an example, consider the transformation $\langle x, y\rangle \mapsto \langle 3x, y\rangle$. It triples the $x$ coordinate of every vector while keeping the $y$ coordinate the same. As a result of this transformation, most vectors will end up pointing in a different direction. There are two exceptions: vertical vectors $\langle 0, y\rangle$ remain completely unchanged; they are eigenvectors with eigenvalue 1. Horizontal vectors $\langle x, 0\rangle$ point in the same direction but become three times as long; they are eigenvectors with eigenvalue 3.

As a second example, consider rotating the plane around the origin by 45 degrees. With this rotation transformation, every (nonzero) vector will point in a new direction, so this transformation has no eigenvectors.

Answer 2

Consider the example of a projection: we have $Px = \lambda x$ but $P^2 = P$, hence $\lambda x = \lambda^2 x$. What are the possible values of $\lambda$? The other cases can be reasoned through in a similar way.

Answer 3

Eigenvectors of a transformation represent lines through the origin that are mapped to themselves by the transformation. The corresponding eigenvalue is a scale factor for this mapping: if the eigenvalue is $1$, the line is invariant—all of the points on this line are fixed points of the transformation; if negative, the “direction” of the line is reversed; if zero, the entire line is collapsed onto the origin.

A projection onto a straight line through the origin leaves points on this line fixed. From the above, that means one of its eigenvalues is $1$. Points not on this line are “collapsed” onto the line by the projection. The type of projection is unspecified in your question, but what’s often meant at the elementary level is orthogonal projection, in which each line that’s perpendicular to the line onto which we’re projecting is mapped to a point on the line. In particular, the perpendicular through the origin is projected onto the origin, so the other eigenvalue of this transformation is $0$.

Similar reasoning can be applied to the other two transformations. They are both isometries—distances don’t change, although orientation can. Since no scaling is involved, their eigenvalues are $\pm1$. Think about what these transformations do to various lines through the origin to determine what the signs must be.