mjhjmtu

I am suppose to figure out if this is false or true.

If $I$ is the $n$ x $n$ identity matrix, and $J$ is an $n$ x $n$ matrix consisting entirely of ones, then the matrix $A = I - fracJn+1$ is idempotent (ie $A^2$ = $A$)

I understand obviously what $I$ and $J$ are, my issue is with the $A = I - fracJn+1$

I searched my textbook and found no reference to it.

What does it mean?

It is not hard to manually obtain $J^2=nJ$. Now it's just a matter of some matrix algebra:

$$A^2=(I-frac1n+1J)^2=I^2-frac1n+1IJ-frac1n+1JI+frac1(n+1)^2J^2 = \
= I - frac2n+1J + fracn(n+1)^2J = I + fracn-2(n+1)(n+1)^2J = \
= I + frac-n-1(n+1)^2J = I - frac1n+1J = A$$

The expression $A = I - fracJn+1$ is matrix addition, where $fracJn+1$ represents scalar multiplication of the matrix $J$ by $frac1n+1$. Since the matrix $J$ consists completely of $1$s, each entry of the scaled matrix $fracJn+1$ is $frac1n+1$. Here is the matrix addition:

$$ I - fracJn+1 =
left[beginarraycccc
1 & 0 & cdots & 0 \
0 & 1 & cdots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & cdots & 1 \
endarrayright]
-
left[beginarraycccc
frac1n+1 & frac1n+1 & cdots & frac1n+1 \
frac1n+1 & frac1n+1 & cdots & frac1n+1 \
vdots & vdots & ddots & vdots \
frac1n+1 & frac1n+1 & cdots & frac1n+1 \
endarrayright]$$

$$ implies A = left[beginarraycccc
fracnn+1 & frac-1n+1 & cdots & frac-1n+1 \
frac-1n+1 & fracnn+1 & cdots & frac-1n+1 \
vdots & vdots & ddots & vdots \
frac-1n+1 & frac-1n+1 & cdots & fracnn+1 \
endarrayright] $$

An interesting matrix. You could describe it by $A_ij=fracnn+1$ if $i=j$ and $A_ij=frac-1n+1$ if $ineq j$.