Let $v \in R^{k}$ ,with $v^{T} v \neq 0$ . Let $P=I-2 \frac{v v^{T}}{v^{T}v}$, where $I$ is the $k X k $ identity matrix.

Question

Let $v \in R^{k}$ ,with $v^{T} v \neq 0$ . Let $P=I-2 \frac{v v^{T}}{v^{T}v}$, where $I$ is the $k \times k $ identity matrix. Then prove that eigenvalue of $P$ are $1,-1$ and $P^{2}=I$

$P^{2}=(I-2 \frac{v v^{T}}{v^{T}v})(I-2 \frac{v v^{T}}{v^{T}v})$

=$(I-4 \frac{v v^{T}}{v^{T}v}+4\frac{v v^{T}}{v^{T}v} *\frac{v v^{T}}{v^{T}v})$

Don"t know how to proceed further !!

Answer 1

Hint: note that since $v^Tv$ is a scalar, $$I - 4 \frac { v v ^ { T } } { v ^ { T } v } + 4 \frac { v v ^ { T } } { v ^ { T } v } * \frac { v v ^ { T } } { v ^ { T } v } = I - 4 \frac { v v ^ { T } } { v ^ { T } v } + 4 \frac { (v ^ { T } v ) v v ^ { T } } { (v ^ { T } v)^2 }.$$ Can you take it from here?

Answer 2

Continuing from your work, note that $vv^\top vv^\top = v(v^\top v) v^\top = (v^\top v)(vv^\top)$ since $v^\top v$ is a scalar. So $$\frac{vv^\top}{v^\top v} \frac{vv^\top}{v^\top v} = \frac{vv^\top}{v^\top v}.$$

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Hint for the eigenvalue question: What is $Pv$? What is $Pw$ when $w$ is orthogonal to $v$? Can you construct a basis of $R^k$ consisting of eigenvectors?