6
$\begingroup$

Is there an established term for the following type of square matrices?

$\begin{pmatrix} c & c & c & c & \cdots & c & c \\ c & a & b & b & \cdots & b & b \\ c & b & a & b & \cdots & b & b \\ c & b & b & a & & b & b \\ \vdots & \vdots & \vdots & & \ddots & & \vdots \\ c & b & b & b & & a & b \\ c & b & b & b & \cdots & b & a \\ \end{pmatrix}$

The matrix contains just 3 different items $a, b, c$:

  • The first row is $c$.
  • The first column is $c$.
  • The diagonal is $a$, except for the upper left corner.
  • The remaining items are $b$.

Background: $a, b, c$ can be chosen such that the matrix is orthogonal, but has a constant first row. If the dimension is a square (i.e. the matrix is a $r^2 \times r^2$ matrix) then it is possible to choose all entries to be integers - up to a common (usually irrational) factor in front of the matrix for normalization.

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ I've found these useful as well! Premultiplying by one of these matrices converts a vector of zero mean to a vector with empty initial coordinates -- which is useful for dealing with centred data in Procrustes analysis. $\endgroup$ Commented Feb 13, 2017 at 15:58

1 Answer 1

Reset to default
7
$\begingroup$

When $b = 0$, we have an $n \times n$ symmetric arrowhead matrix. When $b \neq 0$, we have

$$\begin{bmatrix} c & c & c & \cdots & c & c \\ c & a & b & \cdots & b & b \\ c & b & a & \cdots & b & b \\ c & b & b & \cdots & b & b \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ c & b & b & \cdots & a & b \\ c & b & b & \cdots & b & a \\ \end{bmatrix} = \begin{bmatrix} c-b & c-b & c-b & \cdots & c-b & c-b \\ c-b & a-b & 0 & \cdots & 0 & 0 \\ c-b & 0 & a-b & \cdots & 0 & 0 \\ c-b & 0 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ c-b & 0 & 0 & \cdots & a-b & 0 \\ c-b & 0 & 0 & \cdots & 0 & a-b \\ \end{bmatrix} + b \, 1_n 1_n^{\top}$$

which is the sum of a symmetric arrowhead matrix and a (nonzero) multiple of the all-ones matrix.

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.