1
$\begingroup$

Let $A$ be the $2\times2$ matrix, $A= \begin{bmatrix} 0 & -b \\ b & 0 \end{bmatrix}$, then $A$ is of the form $A=b \begin{bmatrix} 0 & -I_1 \\ I_1 & 0 \end{bmatrix}$. In particular, $SAS^{-1}=\begin{bmatrix} 0 & -I_1 \\ I_1 & 0 \end{bmatrix}$ where $S= I_2$

I want to show that if $A$ is the $(2k)\times(2k)$ block diagonal matrix,

$\begin{bmatrix} 0 & -b & \cdots & \cdots & \cdots & \cdots & 0 \\ b & 0 & \cdots & \ddots & \ddots & \ddots & \vdots\\ 0 & \cdots & 0 & -b \\ 0 & \cdots & b & 0\\ 0 & \cdots & \cdots & \cdots & \ddots\\ 0 & \cdots & \cdots & \cdots & \cdots & 0 & -b\\ 0 & \cdots & \cdots & \cdots & \cdots & b & 0\\ \end{bmatrix}$

Then by a permutation matrix $S$, $SAS^{-1}=b\begin{bmatrix} 0 & -I_k \\ I_k & 0 \end{bmatrix}$. I tried induction but did not get very far. And by brute force I have not seen any patterns.

$\endgroup$
1
  • $\begingroup$ The question is confused. $A$ is first $2\times2$ then $2k\times2k$. Also you are conjugating by the identity matrix $S$ in the first paragraph, which is curious. Please clarify. $\endgroup$ Commented Dec 14, 2011 at 9:18

2 Answers 2

3
$\begingroup$

What you seem to require here is the $(2k)\times(2k)$ perfect shuffle permutation,

$$\mathbf S=[\mathbf e_1,\mathbf e_{k+1},\mathbf e_2,\mathbf e_{k+2},\dots,\mathbf e_k,\mathbf e_{2k}]$$

where $\mathbf e_j$ is the $j$-th column of the identity matrix. (I previously talked about these permutation matrices here.) Your matrix is in fact very nearly a Golub-Kahan tridiagonal, except that it is skew-symmetric instead of being symmetric.


Try it out in Mathematica:

perfectShuffle[n_Integer?EvenQ] := 
 IdentityMatrix[n][[All, Flatten[Transpose[Partition[Range[n], n/2]]]]]

(* block diagonal *) With[{n = 10}, 
  SparseArray[{Band[{2, 1}] -> Riffle[ConstantArray[C, n/2], 0], 
    Band[{1, 2}] -> -Riffle[ConstantArray[C, n/2], 0]},
    {n, n}]] // MatrixForm

(* shuffled matrix *) With[{n = 10}, 
  perfectShuffle[n] .
  SparseArray[{Band[{2, 1}] -> Riffle[ConstantArray[C, n/2], 0], 
               Band[{1, 2}] -> -Riffle[ConstantArray[C, n/2], 0]},
              {n, n}].Transpose[perfectShuffle[n]]] // MatrixForm
$\endgroup$
0
$\begingroup$

You may take $S$ such that the entries

  • $(1,1), (3,2), (5,3), \ldots, (2k-1,k)$ and
  • $(2,k+1), (4,k+2), (6,k+3),\ldots, (2k,k+k)$

contain ones and other entries contain zeroes.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .