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I am stuck on the following problem: For the following linear mapping, $L(\bar{x})=\bar{x}-2 \frac{\bar{x} \centerdot \bar{n}}{||\bar{n}||^{2}} \bar{n}$ where $\bar{n} \in \mathbb{R}^{n}$, find the algebraic and geometric multiplicities of all eigenvalues of L.

I have gotten to the following point: Firstly, two eigenvectors exist, $\bar{y} \in \mathbb{R}^{n}$ where $\bar{y}$ is orthogonal to $\bar{n}$. This has eigenvalue = 1, and eigenspace = span{$\bar{y}$}, thus the geometric multiplicity is 1.

Secondly, any multiple of $\bar{n}$ is an eigenvector, which has eigenvalue -1 and eigenspace = span{$\bar{n}$}, hence geometric multiplicity of 1.

However, I can't figure out a way of determining the algebraic multiplicities of the eigevalues, as the standard way of doing so requires the matrix form of the transformation, which I have tried to work with but is far too messy to come up with a characteristic polynomial for.

Any suggestions would be greatly appreciated. Thanks.

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2 Answers 2

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You have to start with eigenvalues (not eigenvectors): All vectors orthogonal to $n$ are eigenvectors to the eigenvalue $1$ - this should give a hint about the geometric multiplicity of $1$.

The vector $n$ is an eigenvector (if $n\ne0$) to the eigenvalue $-1$.

Since both $U:=span(n)$ and $U^\perp$ are eigenspaces of $L$, there will be no other eigenvectors and eigenvalues.

Now you have found the eigenvalues with their geometric multiplicity. The algebraic multiplicity is $\ge$ the geometric multiplicity. Moreover, the sum of all algebraic multiplicities of eigenvalues is $\le n$. This should allow to conclude all algebraic multiplicities.

An alternative would be to construct the matrix of the mapping: $$ L(x) = (I_n - 2 \frac1{\|n\|^2} nn^T)x. $$ It is the so-called Householder matrix. Then calculate the characteristic polynomial.

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Of course you can do the problem directly, but here is a slightly more general method which may provide a bit more insight into the problem.

Notice that your $L(\mathbf{x})$ is given by $$L = I - 2P$$ where $I$ is the identity map and where $$P(\mathbf{x}) = \frac{\mathbf{x}\cdot\mathbf{n}}{\|\mathbf{n}\|^2}\mathbf{n}$$ is the projection of $\mathbf{x}$ onto $\mathbf{n}$. Notice in particular that $P$, like all projections, satisfy $P^2 = P$.

You can prove, and this will be your main goal, that all such mappings $P$ will be diagonalizable. It doesn't matter what actual form $P$ takes, any $P$ which satisfies the idempotent property $P^2 = P$ will be diagonalizable.

Once you prove that $P$ is diagonalizable, it follows that $L$ will also be diagonalizable given the form that it has.

An additional hint in spoilers below:

One way to prove that $P^2 =P$ implies diagonalizability is to show that an eigenbasis for $P$ is given by any basis for the range of $P$ combined with any basis for the kernel of $P$. In other words, show that $\mathrm{range}(P)$ and $\ker(P)$ are eigenspaces for $P$ and that their respective dimensions sum up to $n$.

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