J_{d}

shifts each standard basis vector one place toward the first coordinate. Therefore

(J_{d}^{n})_{ij} = {1, 0, j - i = n, otherwise .

Thus

J_{d}^{0} = I

, for

1 \leq n < d

the matrix has ones precisely on its

n

th superdiagonal, and

J_{d}^{n} = 0 for n \geq d .

ii

Solution

Since

λ I

commutes with

J_{d}

(λ I + J_{d})^{n} = k = 0 \sum m i n (n, d - 1) (k n) λ^{n - k} J_{d}^{k} .

Equivalently, its

k

th superdiagonal is constant with value

(k n) λ^{n - k}

for

0 \leq k < d

, with the convention that this value is zero when

k > n

There are no discussions about this article yet.

There are currently no matching articles.