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Robert Israel
Robert Israel
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In the case $n=4$ you could have $D = \pmatrix{0 & 0 & 0 & 1\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 0\cr 1 & 0 & 0 & 0}$, in which case $A_1 A_2 A_3 A_4 A_1 A_2 A_3 A_4 = \pmatrix{0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 1 & 1 & 0\cr 1 & 0 & 0 & 1\cr}$ has eigenvalues $0$ and $1$, both with multiplicity $2$.

EDIT: In fact, in this case for any permutation of $A_1 \ldots A_4 A_1 \ldots A_4$ you always get the eigenvalues $0,0,1,1$.

In the case $n=4$ you could have $D = \pmatrix{0 & 0 & 0 & 1\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 0\cr 1 & 0 & 0 & 0}$, in which case $A_1 A_2 A_3 A_4 A_1 A_2 A_3 A_4 = \pmatrix{0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 1 & 1 & 0\cr 1 & 0 & 0 & 1\cr}$ has eigenvalues $0$ and $1$, both with multiplicity $2$.

In the case $n=4$ you could have $D = \pmatrix{0 & 0 & 0 & 1\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 0\cr 1 & 0 & 0 & 0}$, in which case $A_1 A_2 A_3 A_4 A_1 A_2 A_3 A_4 = \pmatrix{0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 1 & 1 & 0\cr 1 & 0 & 0 & 1\cr}$ has eigenvalues $0$ and $1$, both with multiplicity $2$.

EDIT: In fact, in this case for any permutation of $A_1 \ldots A_4 A_1 \ldots A_4$ you always get the eigenvalues $0,0,1,1$.

Source Link
Robert Israel
Robert Israel
  • 54.2k
  • 1
  • 76
  • 152

In the case $n=4$ you could have $D = \pmatrix{0 & 0 & 0 & 1\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 0\cr 1 & 0 & 0 & 0}$, in which case $A_1 A_2 A_3 A_4 A_1 A_2 A_3 A_4 = \pmatrix{0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 1 & 1 & 0\cr 1 & 0 & 0 & 1\cr}$ has eigenvalues $0$ and $1$, both with multiplicity $2$.