Consider a matrix A with the special property that Aor me positive integer m. Su

Question

Consider a matrix A with the special property that Aor me positive integer m. Such a matrix is special, and as such we might expect it to have special eigenvalues Say v is an eigenvector of A with eigenvalue t. By left-multiplying m times by the matrix A we get Since v is an eigenvector with eigenvalue t this translates into Since eigenvectors are non-zero the only way this can be satisfied is if tmv = tv. trn = t, implying that the eigenvalues of A must be either zero, or (m-1)th roots of unity An example of such a matrix is B= 13-2 1 3 Here m5 and the eigenvalues are the set Recall: a set in Maple notation is something of the form (1,1+2*I,3-2*I) remember to use capital i (1) to represent /2 can be entered using the Maple syntax sqrt (2)

Explanation / Answer

For the matrix B, we have B5 = B. Hence the eigenvalues of B are either 0 or the 4th roots of unity i.e. 0,+i or,-i. To confirm this, we know that the eigenvalues of B are solutions to its characteristic equation det(B-I3)= 0 or, 3+ = 0 or (2+1) = 0 or, (+i)(-i)=0 . Thus, the eigenvalues of B are 1=0, 2= i and 3=-i.

Note:

i and –i are the square rooths of -1 which is one of the square roots of unity, i.e. i and –I are 4th roots of unity

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