What is the difference between and projection and a reflection, in vector transformation?

In my text book I have the problems of finding the standard matrix of the given linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$;

1. Projection onto the line $y = -x$.

2. Reflection in the line $y = -x$.

Both of these give different answers. I was hoping for an explanation of the difference between a projection/reflection in a vector transformation.

• Aprojection $p$ satisfies the equality $p\circ p=p$, while a reflection $r$ staisfies $r\circ r=\operatorname{id}$. This is quite different. – Bernard Jun 3 '16 at 19:36

The (orthogonal) projection onto a line "compresses" every point in the plane onto the line. If you drop the perpendicular from the point to the line, the image of the point after projection is the intersection of the perpendicular with the line you are projecting onto.

The reflection across a line moves a point to its "mirror image" across the line. If you drop a the perpendicular from a point onto the line of reflection, then the mirror image is going to lie at the same distance from the line of reflection on this perpendicular, but it will be on the other side of the line of reflection from the original.

Of course, in both cases, a point already on the line of reflection or projection is going to stay where it started.

A huge difference between these two transformations is that reflections are always invertible (since $R^2=Id$), but projections are almost never invertible. An orthogonal projection of the plane onto a line is never invertible since every point on a perpendicular to the line of projection maps to the same point on the line you are projecting onto.

In terms of eigenvalues, the projection in this case would have eigenvalues $\{0,1\}$ whereas the reflection would have eigenvalues $\{-1,1\}$.

The answer of @rschwieb gives you the theoretical vision of the problem. I add a figure to support it in a geometrical intuitive way. The blue line is your $y=-x$. Given the points $P$ or $Q$, than $P'$ and $Q'$ are the orthogonal projection on this line and $P''$ and $Q''$ are the reflections. As you can see the prjections of the two points coincide, so the orthogonal projection is not invertible.

If you take $P=(1,0)^T$ (a standard basis vector) you can easely see that $P'=(\frac{1}{2},-\frac{1}{2})^T$ and $P''=(0,-1)^T$, and if you do the same for the other basis vector $N=(0,1)^T$, you find $N'=((-\frac{1}{2},\frac{1}{2})^T$ and $N''=(-1,0)^T$, so you have the matrices that represents the two transformations (in the standard basis):

$$\frac{1}{2} \begin{bmatrix} 1&-1\\-1&1 \end{bmatrix}$$ for the projection, and $$\begin{bmatrix} 0&-1\\-1&0 \end{bmatrix}$$

for the reflecion.

• I must beg to differ: my answer is almost purely geometric intuition. But what you've written is certainly a concrete and explicit analytical answer, which is good too :) – rschwieb Jun 4 '16 at 0:56