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Consider a subspace $V$ in $\mathbb R^n$ with $\dim(V) = m$.

  • Suppose a matrix $A$ in $\mathbb R^{n\times n}$ that represents the orthogonal projection over $V$. What can you say about the eigenvalues and their algebraic and geometric multiplicities?

  • Same requirements but for $B$ in $\mathbb R^{n\times n}$ that represents the reflection in $V$.


I know that the eigenvalues for an orthogonal projection are $0$ and $1$ while for a reflection they are $1$ and $-1$.

I also know what algebraic and geometric multiplicities mean and that the geometric multiplicity is lower or equal than the algebraic multiplicity.

However, although I've read somewhere that the multiplicity of the eigenvalue $1$ for a projection is $m$ and for the eigenvalue $0$ is $m-n$ I don't understand why. Also I don't know how to find the geometric multiplicities from this fact.

Thank you

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  • $\begingroup$ If $x\in V$ then $Ax=x$ and if $x$ is orthogonal to every member of $V$ then $Ax=0.$ That gets you a space of dimension $m$ that is the eigenspace corresponding to the eigenvalue $1$, and another space in which every vector is an eigenvector with eigenvalue $0. \qquad$ $\endgroup$ Commented Jun 14, 2017 at 16:06
  • $\begingroup$ Ok, thanks. What about the algebraic multiplicities then? Also, for the reflection case, Would the eigenvalue $1$ correspond to the space $V$ and $-1$ to a space, let's say $U$ of dimension %n-m$ ? $\endgroup$ Commented Jun 14, 2017 at 17:14

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In the first case, $A$ acts as the identity on $V$, and if we have chosen an orthogonal basis, it maps all other basis elements to zero. Hence it is a diagonalisable matrix (any orthogonal basis will do) - the diagonal will have $1$'s first (for the 'part' it acts as the identity on), and zeroes after that. So the geometric and algebraic multiplicities are equal.

A reflection has the basis elements in $V^{\top}$ (again, with an orthogonal basis) as eigenvectors with eigenvalue $-1$, and as such is diagonalisable as well.

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