Ytukyg

We were asked to find a symmetric idempotent matrix $H$ with rank $n-1$ such that if $X$ is a column vector with $n$ observations, then ${1over n}X^THX$ is the variance of observations in $X$.

I found the matrix (for $n$ obs) to be $H_n=I_n-{1over n}A_n$ where $I_n$ is identity matrix of dimension $ntimes n$, $A_n$ is again $ntimes n$ with all observations being $1$ and $H_n$ is the required matrix.

It was easy to show this is symmetric and idempotent but I'm facing difficulty with showing its rank is $n-1$.
However, it is easy to see $R_1+R_2+dots+R_n=0$ where $R_i$ is the $i^{th}$ row. So its rank is strictly less than $n$.

I also noticed $R_1+R_2+dots+R_n-R_ine0$ for any $i$.

How should I proceed?

Denoting the column vector of all $1$s by $mathbf1$, we have$$H=I_n-frac{1}{n}mathbf{11}^top$$

Indeed as you say, $H$ is an idempotent matrix. Then we know that $$mathrm{rank}(H)=mathrm{trace}(H)=mathrm{trace}(I_n)-mathrm{trace}left(frac{1}{n}mathbf{11}^topright)=n-1$$

We can also use some trivial rank inequalities although this is quite unnecessary to prove the result:

We know that for any two matrices $A$ and $B$ having the same order, $$mathrm{rank}(A-B+B)le mathrm{rank}(A-B)+mathrm{rank}(B)$$

Or, $$mathrm{rank}(A-B)ge |mathrm{rank}(A)-mathrm{rank}(B)|$$

Noting that $mathbf{11}^top$ is a rank $1$ matrix, applying this inequality on $H$ we get,

$$mathrm{rank}(H)ge n-1$$

Now we can show that $mathrm{rank}(H)$ is never $n$ (the only other possibility) from the fact that $$det(H)=1-frac{1}{n}mathbf1^topmathbf1=1-1=0$$

So it must be that $$mathrm{rank}(H)=n-1$$

To prove the rank of $H$ is $n-1$, we may look at the linear system $HX = 0$ and prove the dimensions of the space of solutions is $1$. The system $HX = 0$ may be written as

begin{align*}
(S)left{begin{matrix}
x_1 &+& x_2 &+& ldots &x_n &=& n x_1 \
x_1 &+& x_2 &+& ldots &x_n &=& n x_2 \
vdots && &&&vdots && vdots \
x_1 &+& x_2 &+& ldots &x_n &=& n x_n \
end{matrix}right.
end{align*}
now substract the first line to all the other lines :
begin{align*}
(S) &Longleftrightarrow &
left{begin{matrix}
x_1 &+& x_2 &+& ldots &x_n &=& n x_1 \
&&&&& 0 &=& n (x_2-x_1) \
&&&&&vdots&& vdots \
&&&&& 0 &=& n (x_n-x_1) \
end{matrix}right.\
\
& Longleftrightarrow &
left{begin{matrix}
x_1 &+& x_2 &+& ldots &x_n &=& n x_1 \
&&x_2&&& &=& x_1 \
&&&ddots&&&& vdots \
&&&&& x_n &=& x_1 \
end{matrix}right.
end{align*}
Next we subtract all lines $2, ldots, n$ to line $1$ to get
begin{align*}
(S) &Longleftrightarrow &
left{begin{matrix}
x_2&& &=& x_1 \
&ddots &&& vdots \
&&x_n &=& x_1 \
end{matrix}right.
end{align*}
whose solutions are the $1$-dimensional space generated by $begin{pmatrix}1 \ 1 \ vdots \ 1end{pmatrix}$.

Let $e_k = (1, omega^k, omega^{2k}, ldots, omega^{(n-1)k})$ where $omega=e^{2pi i/n}$. You have shown that $e_0$ is in the null-space of $H$. For $0<k<n$, $e_k$ is an eigenvector of $H$ of eigenvalue $1$. As the $(e_k)$ are independent (why?), this shows that the rank is $n-1$.