# Find all eigenvalues for $L(\vec{x})=\vec{x}-2proj_\vec{n}(\vec{x})$ for some non-zero vector $\vec{n}$

given a linear transformation $L:\Bbb R^n \to \Bbb R^n$ defined by $L(\vec{x})=\vec{x}-2proj_\vec{n}(\vec{x})$ for some non-zero vector $\vec{n}$. How would you Find the algebraic and geomtric multiplicites of all eigenvalues of $L$

I know that $\vec{x}=\vec{n}$ then $(-1,\vec{n})$ is an eigenpair and also that if $\vec{x} \cdot \vec{n}=0$ then $(1,\vec{x})$ is also an eigenpair.

But are these the only ones and how could I find the rest

• Does $\text{proj}_{\vec n}(\vec x) = (\vec n \cdot \vec v) \vec n$? i.e., the profection of the vector $\vec x$ onto the one-dimensional subspace spanned by $\vec n$? Apr 5, 2018 at 22:03
• Is there any part of $\mathbb R^n$ left after you compute the sum of these eigenspaces?
– amd
Apr 5, 2018 at 22:33

Think about the problem geometrically. $L$ represents reflection through the hyperplane with normal $\vec{n}$. As you say, this reflection will leave any vector parallel to the hyperplane (an $n-1$-dimensional subspace of $\mathbb{R}^n$) untouched. It will reflect any vector normal to the plane (a 1-dimensional subspace). It should be immediate what the eigenvalues are, and with what multiplicities.

To formalize this intuition, extend $\vec{n}$ to an orthonormal basis $\{\vec{n}, \vec{b}_2, \vec{b}_3, \ldots\}$ of $\mathbb{R}^n$ using e.g. Gram-Schmidt. Compute $L\vec{b}_i$ for each $i$. What can you conclude?

We can look at this a little more generally to get an idea of what is going on.

Let $V$ be any finite dimensional vector space, of dimension $d$, and $P:V\rightarrow V$ any projection map. A projection map is defined to be any linear map so that $P^2=P$. Now set $L(v)=v-2P(v)$. This means that $$\begin{eqnarray*} L^2(v) &=& L(v) - 2P(L(v))\\ &=& v -2P(v) - 2P(v-2P(v))\\ &=& v-4P(v) +4P(v)\\ &=& v \end{eqnarray*}$$ That is, $L^2=\textrm{id}$. This means that the only eigenvalues of $L$ are $\pm 1$.

If $v$ is in the image of $P$, then we have $v=P(w)$ and so $P(v)=P^2(w)=P(w)=v$. Therefore, $L(v)=-v$. On the other hand, if $v$ is in the kernel of $P$, then we get that $L(v)=v$. This means that the geometric multiplicity of the eigenvalue $1$ will be at least $a=\dim(\ker(P))$, and the geometric multiplicity of the eigenvalue $-1$ will be at least $b=\dim(Im(P))$. But the Rank-Nullity Theorem says that $a+b=d$, which means that the above "at least"s are actually equality. In particular, the $1$-eigenspace is $\ker(P)$ and the $-1$-eigenspace will be $Im(P)$.

Since the geometric multiplicity cannot exceed the algebraic multiplicity, this means that they are equal in this case. Moreover, the characteristic polynomial will be $$c_L(x) = (x-1)^{a}(x+1)^b$$

In the case that we are projecting onto a single vector, we will have $a=d-1$ and $b=1$. Moreover, your list is complete.

• how can you conclude from $L^2=I$ that the eigenvalues are $1,-1$? Apr 17, 2018 at 20:37
• @Skrrrrrtttt The minimal polynomial of $L$ is $x^2-1$
– C.W
Apr 18, 2018 at 0:17