# Need help with linear transformations (with projection and reflection)?

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The answer to the first question is yes. Non-zero vectors $x$ with both of those properties do exist.

The line $L$ passes through the origin. Let $x$ be any vector from the origin to any other point on line $L$. It is easy to see that both the reflection and the projection of that vector relative to the line is itself.

One possible vector is $(3,4)$.

So we do not need eigenvectors to solve this problem. The point seems to be to show without calculation that $1$ must be an eigenvalue of both transformations (and indeed of any reflection or projection transformation in a line through the origin).

Do you need more detail?

There are multiple ways to find the standard matrices for $S$ and $P$. One way is to find two linearly independent vectors and their images and solve for the transformation. There are several ways to do that solving: let's use matrices.

As for $S$, reflection about the line $L$, any vector perpendicular to the line will be mapped to its additive inverse. One vector perpendicular to the line is $(4, -3)$. (Do you see how I got that from a vector that we know is on the line?) So the transformation $S$ takes the vector $(4,-3)$ to its inverse $(-4,3)$, and we already know that it takes the vector $(3,4)$ to itself. Using those column vectors we get this matrix equation:

$$S\cdot \begin{bmatrix} 3 & 4 \\ 4 & -3 \\ \end{bmatrix} = \begin{bmatrix} 3 & -4 \\ 4 & 3 \\ \end{bmatrix}$$

I hope you can solve that equation for $S$.

As for $P$, projection onto the line $L$, any vector perpendicular to the line will be mapped to the zero vector, the origin. So the transformation $P$ takes the vector $(4,-3)$ to $(0,0)$, and we already know that it takes the vector $(3,4)$ to itself. Then using those column vectors we get this matrix equation:

$$P\cdot \begin{bmatrix} 3 & 4 \\ 4 & -3 \\ \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 4 & 0 \\ \end{bmatrix}$$

I hope you can solve that equation for $P$.

I hope you also see that we have already determined that the eigenvalues for $S$ are $-1$ and $1$, while the eigenvalues for $P$ are $0$ and $1$.

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### droman07

Updated on July 22, 2022

Let $L$ be the line given by the equation $4x − 3y = 0$. Let $S : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be reflection through that line, and let $P : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be projection onto that line.
Determine, geometrically without doing any computations, whether there exist non-zero vectors $x$ such that: $$a) \qquad S(x) = 1 · x$$ $$b)\qquad P(x) = 1 · x$$ Can someone tell me how to solve this? Does anyone know what kind of arithmatic this is? I wanna look it up on google but I don't know what it's called. Looks a bit like eigenvectors to me but I'm not entirely sure...