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Suppose \begin{align} A=\begin{pmatrix} \frac{1}{\sqrt{n}}&\frac{1}{\sqrt{n}}&\frac{1}{\sqrt{n}}&\cdots&\frac{1}{\sqrt{n}} \\&&P \end{pmatrix} \end{align}

is a real orthogonal matrix of size $n\times n$, where the matrix $P$ is of size $(n-1)\times n$.

I have to show that $$P^\top P=I-\frac{1}{n}\mathbf1\mathbf1^{\top}\tag{1}$$

where $\mathbf1$ is an $n$ component column vector of all ones.

The choice of $P$ is certainly not unique. The most obvious choice to me is that $P$ for which $A$ is a Helmert matrix:

\begin{align} P=\begin{pmatrix} \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0&0&\cdots&0&0 \\\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{6}}&-\frac{2}{\sqrt{6}}&0&\cdots&0&0 \\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots \\\frac{1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n(n-1)}}&\cdots&\frac{1}{\sqrt{n(n-1)}}&\frac{-(n-1)}{\sqrt{n(n-1)}} \end{pmatrix} \end{align}

I could now verify that $(1)$ holds true for this $P$ but that does not prove anything.

How do I find a general form of the matrix $P$ so that $A$ is an orthogonal matrix?

Or is it possible to prove $(1)$ without explicitly finding $P$? Any hint would be great.

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    $\begingroup$ An aside: the DFT matrix is often a easier choice to work with than the Helmert matrix, if you're looking for unitary matrices whose first row is $\frac 1{\sqrt{n}}1^T$. $\endgroup$ – Omnomnomnom Sep 21 '18 at 18:16
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It is certainly possible to prove your result without finding $P$. One way to do this efficiently is to use block-matrix multiplication.

In your case, I would write $$ A = \pmatrix{\frac 1{\sqrt n} 1^T\\ P} $$ The matrices $A$ and $A^T$ are partitioned conformally. As such, we can compute $$ A^TA = \pmatrix{\frac 1{\sqrt{n}}1 & P^T}\pmatrix{\frac 1{\sqrt n} 1^T\\ P} = \frac 1n 11^T + P^TP $$ Now, $A$ is orthogonal if and only if $A^TA = I$.

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  • $\begingroup$ Sorry, what is conformal partitioning? $\endgroup$ – StubbornAtom Sep 21 '18 at 18:17
  • $\begingroup$ @StubbornAtom "conformal" in this context just means that the matrices can be multiplied (using block-matrix multiplication). For block matrices, $A$ and $B$ are conformally partitioned if and only if we can multiply $AB$ (i.e. the number of columns in $A$ matches the number of rows in $B$) and the columns of $A$ are paritioned in the same way that the rows of $B$ are partitioned. In this case, we have partitioned both the columns of $A^T$ and the rows of $A$ into a segment of length $1$, then a segment of length $n-1$. $\endgroup$ – Omnomnomnom Sep 21 '18 at 18:21
  • $\begingroup$ Can't believe I did not apply the definition of an orthogonal matrix ! $\endgroup$ – StubbornAtom Sep 21 '18 at 18:25
  • $\begingroup$ Is there a general form of $P$ by the way? $\endgroup$ – StubbornAtom Sep 21 '18 at 18:37
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    $\begingroup$ @StubbornAtom once you have a particular choice of $P$: for any orthogonal matrix $Q$, we have $(QP)^T(QP) = P^TP$. I'm pretty sure that every possible $P$ can be attained with some orthogonal choice of $Q$. $\endgroup$ – Omnomnomnom Sep 21 '18 at 18:41
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We can prove it without explicitly finding the $P$'s.

Hint: For block matrices of adequate sizes we have in particular $$\pmatrix{U&V} \pmatrix{C\\D} =UC+VD$$ Apply it to $A^TA=I$.

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