A linear operator $L: \mathbb R^n \rightarrow \mathbb R^n$ is called a projection if $L^2 = L$. A projection $L$ is an orthogonal projection if $\ker L$ is orthogonal to $L(\mathbb R^n)$.

I've shown that the only invertible projection is the identity map $I_{\mathbb R^n}$ by using function composition on the identity $L^2(v) = L$.


Now suppose that $L$ is an orthogonal projection whose image $L(\mathbb R^n)$ has rank $1$. Show there exist a unit vector $v$ such that $L$ is defined by the matrix $v v^T$.

What I've tried:

Since $L(\mathbb R^n)$ has rank $1$ there exist a unit vector $v \in L(\mathbb R^n)$ such that $Span(v) = L(\mathbb R^n)$.

Let $u\in L(\mathbb R^n)$ such that $L(u) = v$. But $L(L(u)) = L(v) = v$ by the property of $L$, so $v$ is mapped to itself.

I can write every vector $t = P_{Span(v)}(t) + r$ uniquely as a direct sum, where $P_{Span(v)}(t)$ lies in $Span(v)$ (the orthogonal projection) and $r$ lies in the orthogonal complement to $Span(v)$. But then $L(t) = L(P_{Span(v)}(t)) + L(r) = L(cv)+L(r) = cv + L(r)$. If I could show $L(r) = 0$ then I'm done, since $L$ is then the orthogonal projection on $Span(v)$, but I haven't been able to do this ?


Since $L$ is an orthogonal projection, there exists an orthogonal matrix $P$ so that \begin{equation}P^TLP=\begin{bmatrix}1&\bf{0}\\\bf{0}&\bf{0}\end{bmatrix}.\end{equation}(Bold zeros indicate zero vector/matrix). So since $L$ is rank 1 and an orthogonal projection it is orthogonally similar to a matrix with a 1 in entry $(1,1)$ and zeros elsewhere.

So now let $v_1 =\begin{bmatrix}1 \\ \bf{0}\end{bmatrix}$ and let $v=Pv_1$.


You have the direct sum $\mathbb R^n = kern(L) \oplus image(L)$. Then $$ t = Lt + (t-Lt) $$ and $L(t-Lt) = Lt - L^2t=0$. Hence, the remainder $r$ is in $kern L$.


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