vectors simultaneously orthogonal in two different bilinear forms with diagonal matrices

Question

I'm interested in the orthogonal similarity transformation of diagonal matrices, see also my previous questions properties of orthogonal similarity transformation and orthogonal similarity transformation of a diagonal matrix by a permutation matrix: reverse direction. Assume that $\mathbf{D}$ is a diagonal matrix with positive and pairwise different elements $d_1 > \ldots > d_n > 0$ and $\mathbf{Q}$ an orthogonal matrix. If the orthogonal similarity transformation $\mathbf{Q}^T \mathbf{D} \mathbf{Q}$ is a diagonal matrix, we have expressions $$\mathbf{q}_i^T \mathbf{D} \mathbf{q}_j = 0$$ for the off-diagonal elements ($i \neq j$). At the same time, we have $$\mathbf{q}_i^T \mathbf{q}_j = \mathbf{q}_i^T \mathbf{I} \mathbf{q}_j = 0$$ for these elements, and $\|\mathbf{q}_i\| = 1$.

My guess is that unit vectors $\mathbf{q}_i$ and $\mathbf{q}_j$ can only be orthogonal in both bilinear forms (i.e. both expressions above are zero) if they have an element $\pm 1$.

Does anyone have a counter-example or an idea for a proof? Can the proof address individual elements $(i,j)$ or would we have to consider all $(i,j)$? Would the conjectured property be lost with different assumptions on $\mathbf{D}$, i.e. if some elements of $\mathbf{D}$ are identical or if some are negative?

I'm grateful for any ideas.

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I made a bit of progress, but got stuck at a later point.

$ \newcommand{\matQ}{\mathbf{Q}} \newcommand{\matD}{\mathbf{D}} \newcommand{\matC}{\mathbf{C}} \newcommand{\vecz}{\mathbf{z}} \newcommand{\vecq}{\mathbf{q}} \newcommand{\vecp}{\mathbf{p}} \newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}} \newcommand{\vecnull}{\mathbf{0}} $

In the following I give up the sorting assumption of the diagonal elements.

We first look at the system of equations for two columns $\vecq$ and $\vecp$ of $\matQ$:

\begin{eqnarray} \vecq^T \matD \vecp &=& 0\\ \vecq^T \vecp &=& 0. \end{eqnarray}

We can transform this system into a homogeneous linear system

\begin{equation} \pmat{ d_1 & d_2 & d_3 & \ldots & d_n\\ 1 & 1 & 1 & \ldots & 1} \pmat{ z_1\\ \vdots\\ z_n} = \pmat{0\\0} \end{equation}

where $z_i = q_i p_i$. We transform the coefficient matrix

\begin{equation} \matC = \pmat{ d_1 & d_2 & d_3 & \ldots & d_n\\ 1 & 1 & 1 & \ldots & 1} \end{equation}

into reduced row echelon form: We normalize $d'_i = d_i / d_1$,

\begin{equation} \pmat{ 1 & d'_2 & d'_3 & \ldots & d'_n\\ 1 & 1 & 1 & \ldots & 1}, \end{equation}

subtract the first row from the second,

\begin{equation} \pmat{ 1 & d'_2 & d'_3 & \ldots & d'_n\\ 0 & 1-d'_2 & 1-d'_3 & \ldots & 1-d'_n}, \end{equation}

normalize the second row,

\begin{equation} \pmat{ 1 & d'_2 & d'_3 & \ldots & d'_n\\ 0 & 1 & \frac{1-d'_3}{1-d'_2} & \ldots & \frac{1-d'_n}{1-d'_2}}, \end{equation}

and subtract a scaled version of the second row from the first

\begin{equation} \pmat{ 1 & 0 & d'_3 - \frac{1-d'_3}{1-d'_2} d'_2 & \ldots & d'_n - \frac{1-d'_n}{1-d'_2} d'_2\\ 0 & 1 & \frac{1-d'_3}{1-d'_2} & \ldots & \frac{1-d'_n}{1-d'_2}}. \end{equation}

We introduce for $i \geq 3$

\begin{eqnarray} a_i &:=& d'_i - \frac{1-d'_i}{1-d'_2} d'_2 = \frac{d'_i (1-d'_2) - (1-d'_i) d'_2}{1-d'_2} = \frac{d'_i - d'_2}{1-d'_2}\\ % b_i &:=& \frac{1-d'_i}{1-d'_2} \end{eqnarray}

and obtain the reduced row echelon form

\begin{equation} \matC' = \pmat{ 1 & 0 & a_3 & \ldots & a_n\\ 0 & 1 & b_3 & \ldots & b_n}. \end{equation}

For $d_i \neq d_j\, \forall i \neq j$ we see that $1-d'_i \neq 0\, \forall i \geq 2$, thus the denominators are defined and $b_i \neq 0\, \forall i \geq 3$ and $a_i \neq 0\, \forall i \geq 3$.

(This paragraph considers violations of the assumption of pairwise different elements in $\matD$. It may be useful later. If $\exists i,j\,(i\neq j): d_i = d_j$, we have to distinguish two cases. If all $d_i$ are equal, then the $z_i$ can be chosen arbitrarily under the constraint $\sum_{i=1}^n z_i = 0$. If there is at least one pair of diagonal elements which differ from each other, we can select them w.l.o.g. to be $d_1$ and $d_2$, thus $d_1 \neq d_2$ and $1-d'_2 \neq 0$. In addition, we select the diagonal elements such that $\exists i\,(i\geq 3): d_2 = d_i$. In this case we have $a_i = 0$. If also $\exists j\,(j\geq 3): d_1 = d_j$, we also have $b_j = 0$. The case $a_i = b_i = 0$ is not possible.)

We now determine the null space of $\matC$. From $\matC' \vecz = \vecnull$ we obtain

\begin{equation} \begin{matrix} z_1 & & + z_3 a_3 & \ldots & + z_n a_n & = 0\\ & z_2 & + z_3 b_3 & \ldots & + z_n b_n & = 0 \end{matrix} \end{equation}

where $z_3,\ldots,z_n$ are free parameters. This leads to

\begin{equation} \vecz = \pmat{ -z_3 a_3 & -z_4 a_4 & \ldots & -z_n a_n\\ -z_3 b_3 & -z_4 b_4 & \ldots & -z_n b_n\\ z_3 & & & \\ & z_4 & & \\ & & \ddots & \\ & & & z_n } \end{equation}

(note that this is not a matrix, but a vector written such that each $z_i$ appears in a separate column) thus the null space is spanned by the columns of the $n \times (n-2)$ matrix

\begin{equation} \pmat{ -a_3 & -a_4 & \ldots & -a_n\\ -b_3 & -b_4 & \ldots & -b_n\\ 1 & & & \\ & 1 & & \\ & & \ddots & \\ & & & 1 }. \end{equation}

Clearly, we can find a $\vecz \neq \vecnull$ such that both bilinear forms become zero. Therefore, individual vectors $\vecq, \vecp$ fulfilling both equations are not necessarily vectors with a single non-zero element in different positions: Assume that $z_i \neq 0$, then $q_i \neq 0$ and $p_i \neq 0$.

Thus we obviously have to consider the entire matrix $\matQ$:

\begin{eqnarray} \vecq_i^T \matD \vecq_j &=& 0\quad \forall i \neq j\\ \vecq_i^T \vecq_j &=& 0 \quad \forall i \neq j. \end{eqnarray}

If we could show that only $\vecz = \vecnull$ allows to fulfill all these equations, we could demonstrate that all vectors $\vecq_i$ can only have exactly one non-zero element (which has to be $\pm 1$ since the vectors are unit vectors) which appears in different positions. The argument proceeds as follows: If $z_k = q_{i,k} q_{j,k} = 0$ for all $k$, then $\vecq_i$ has zero elements where $\vecq_j$ has non-zero elements and vice versa. Assume that one vector $\vecq_i$ has more than one non-zero element. Since the remaining $n-1$ vectors $\vecq_j$ ($j \neq i$) have at least one non-zero element (as they are unit vectors), it would not be possible to find $n-1$ other vectors $\vecq_j$ which have their non-zero element at the $n-2$ remaining zero positions of $\vecq_i$.

Any ideas how to show this? Thanks!

Answer 1

Let $A$ be an $n \times n$ matrix. Then $A$ is diagonal if and only if each of the standard basis vectors $e_1, \ldots, e_n$ is an eigenvector of $A$. If $Q$ is an invertible matrix then $Q^{-1} A Q$ is diagonal if and only if the columns of $Q$ are eigenvectors of $A$.

In your case, you have a diagonal matrix $D$ with distinct entries, meaning that each standard basis vector $e_i$ is an eigenvector. Because the eigenvalues are distinct, every eigenvector is a multiple of some $e_i$. This means that when you suppose that $Q^{-1} D Q$ is diagonal for some invertible matrix $Q$, you already know that the columns of $Q$ must contain multiples of the eigenvectors $e_i$. For example, a potential $Q$ could look like $$ Q = \begin{pmatrix} 0 & 6 & 0 \\ -\sqrt{2} & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$

Now apply the fact that your $Q$ is orthogonal, and so every entry must be $\pm 1$.

Answer 2

Thanks a lot to Joppy for the proof. I rewrote the proof in my notation, I hope I got all the steps right:

$$ \newcommand{\matD}{\mathbf{D}} \newcommand{\matQ}{\mathbf{Q}} \newcommand{\matXi}{{\mathbf\Xi}} \newcommand{\matP}{\mathbf{P}} \newcommand{\matI}{\mathbf{I}} \newcommand{\vece}{\mathbf{e}} \newcommand{\vecq}{\mathbf{q}} \newcommand{\vecv}{\mathbf{v}} \newcommand{\vecnull}{\mathbf{0}} \newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}} $$

What we ultimately prove is the following Lemma:

Let $\matD$ be a diagonal matrix with non-zero and pairwise different entries. Let $\matQ$ be an orthogonal matrix. If $\matQ^T \matD \matQ$ is diagonal then $\matQ = \matP'$ where $\matP' = \matXi \matP$ is a signed permutation matrix ($\matXi$ is a diagonal sign matrix with entries $\pm 1$, $\matP$ is a permutation matrix).

(I'm not sure whether the requirement "non-zero" is necessary.)

Proof:

Statement 1: The eigenvectors of the diagonal matrix $\matD$ with pairwise different, non-zero entries are the basis vectors $\pm\vece_i$, $i=1,\ldots,n$. Proof: The characteristic equation $\det\{\matD - \lambda \matI\} = 0$ leads to $\lambda_i = d_i$. The eigenvector equation $(\matD - \lambda_i \matI) \vecv_i = \vecnull$ becomes $(\matD - d_i \matI) \vecv_i = \vecnull$ which leads to the solution $(\vecv_i)_i = \pm 1$ and $(\vecv_i)_j = 0$, $\forall j\neq i$, and thus $\vecv_i = \pm\vece_i$ which concludes the proof of statement 1.

Statement 2: Assume that $\matQ$ is an invertible matrix ($\matQ^{-1} = \matQ^T$ is a special case). If $\matQ^{-1} \matD \matQ$ is diagonal, then the columns of $\matQ$ are the eigenvectors of $\matD$. Proof: We express $\matQ$ by its columns $\matQ = \pmat{\vecq_1 & \ldots & \vecq_n}$. Since $\matQ$ is invertible, the columns of $\matQ$ span the entire $\mathbb{R}^n$ (since the $n$ columns must be linearly independent). Thus any vector can be expressed as a superposition of the column vectors, also the vector $\matD \vecq_i$:

\begin{equation} \matD \vecq_i = \sum_{j=1}^n a_{ij} \vecq_j. \end{equation}

We look at column $i$ of $\matQ^{-1}\matD\matQ$:

\begin{eqnarray} (\matQ^{-1}\matD\matQ)\vece_i &=& \matQ^{-1}\matD\vecq_i\\ &=& \matQ^{-1}\sum_{j=1}^n a_{ij} \vecq_j\\ &=& \sum_{j=1}^n a_{ij} \vece_j. \end{eqnarray}

Since we assume that $\matQ^{-1}\matD\matQ$ is diagonal (i.e. equals a diagonal matrix $\matD'$ with elements $d'_i$), we know that

\begin{eqnarray} (\matQ^{-1}\matD\matQ)\vece_i &=& d'_i \vece_i\\ \sum_{j=1}^n a_{ij} \vece_j &=& d'_i \vece_i \end{eqnarray}

thus $a_{ii} = d'_i$ and $a_{ij} = 0$, $\forall j \neq i$. Therefore the superposition equation above turns into

\begin{equation} \matD \vecq_i = d'_i \vecq_i,\quad i = 1,\ldots,n. \end{equation}

We see that the vectors $\vecq_i$, $i=1,\ldots,n$, are the eigenvectors of $\matD$, which concludes the proof of statement 2.

According to statement 1, the eigenvectors of $\matD$ are the basis vectors $\pm \vece_j$, and according to statement 2, the eigenvectors of $\matD$ are the columns $\vecq_i$. The assignment of $j$ to $i$ can be any permutation $\pi$:

\begin{equation} \vecq_i = \pm\vece_{\pi(i)} \end{equation}

which concludes the proof of the Lemma.