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Show that there is a scalar $c$ (depending on $n$) for which $cJ_n$ is a projection matrix. ($J_n$ is the $n×n$ matrix of all 1’s.) Then, for that same $c$, show that $R = 2cJ_n − I$ is a reflection matrix.

So far, all I have found is that a projection matrix is any $n×n$ matrix $P$ for which $P_2$ = $P$. A reflection matrix is any matrix $R$ with the property that $R_2$ = $I$, and I am unsure on how to incorporate that.

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  • $\begingroup$ Does $c = 1/n$ work? $\endgroup$ – amarney Apr 19 '17 at 0:30
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First we find what $c$ we need to have a projection. Notice that $$(c J_n) = \begin{bmatrix}c & ... & c \\ \vdots & & \vdots \\ c & .. & c \end{bmatrix},$$

so $$(cJ_n)^2 = \begin{bmatrix}c & ... & c \\ \vdots & & \vdots \\ c & .. & c \end{bmatrix} \begin{bmatrix}c & ... & c \\ \vdots & & \vdots \\ c & .. & c \end{bmatrix} = \begin{bmatrix} nc^2 & ... & nc^2 \\ \vdots & & \vdots \\ nc^2 & .. & nc^2 \end{bmatrix},$$ so you need $c = nc^2$, hence $c = 1/n$ works.

Now we have to check that we can form a reflection matrix. I will get you started. We have $$ R = 2(c J_n) - I = \begin{bmatrix}2/n - 1 & ... & 2/n \\ \vdots & & \vdots \\ 2/n & .. & 2/n-1 \end{bmatrix}, $$ so $$ R^2 = \begin{bmatrix}2/n - 1 & ... & 2/n \\ \vdots & & \vdots \\ 2/n & .. & 2/n-1 \end{bmatrix} \begin{bmatrix}2/n - 1 & ... & 2/n \\ \vdots & & \vdots \\ 2/n & .. & 2/n-1 \end{bmatrix} = \begin{bmatrix} & ... & \\ \vdots & & \vdots \\ & .. & \end{bmatrix}. $$

Now fill in the blanks.

EDIT: Big caveat, note that we excluded the $c = 0$ case by doing division. It turns out that choosing $c = 0$ gives you what you want immediately, but you probably don't want this uninteresting case.

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