# Showing $P^TP=I_n-\frac1n11^T$ if $\left(\begin{smallmatrix}\frac1{\sqrt n}&\cdots&\frac1{\sqrt n}\\&P\end{smallmatrix}\right)$ is orthogonal

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.

• 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 $\frac 1{\sqrt{n}}1^T$. – Omnomnomnom Sep 21 '18 at 18:16

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 $$A^TA = I$$.

• Sorry, what is conformal partitioning? – StubbornAtom Sep 21 '18 at 18:17
• @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 $n-1$. – Omnomnomnom Sep 21 '18 at 18:21
• Can't believe I did not apply the definition of an orthogonal matrix ! – StubbornAtom Sep 21 '18 at 18:25
• Is there a general form of $P$ by the way? – StubbornAtom Sep 21 '18 at 18:37
• @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 $Q$. – Omnomnomnom Sep 21 '18 at 18:41

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 $A^TA=I$.