matricies

For a fixed positive integer n, E(r,s) (1 ≤ r ≤ n, 1 ≤ s ≤ n) will denote the basic n x n matrices.

E(r,s)ij = δriδsj
(i think this is the kroneeker delta stuff)

now suppose that n≥1000. Show by a formal matrix calculation (i.e using the definition) that

E(1,999)E(999,1000) = E(1,1000)

and also that E(1,999)E(1000,999) = 0

Reply 1

gap-year-mat

Reply 2

Jonny W

Matrix multiplication is defined by

(AB)_{ij} = (sum over k) A_{ik}B_{kj}

So

( E(1,999)E(999,1000) )_{ij}
= (sum over k) E(1,999)_{ik} E(999,1000)_{kj}
= (sum over k) [delta(1, i)delta(999, k)delta(999, k)delta(1000, j)]
= delta(1, i)delta(1000, j) (sum over k) delta(999, k)
= delta(1, i)delta(1000, j) * 1
= delta(1, i)delta(1000, j)
= E(1, 1000)_{ij}

So the (i, j)th entry of E(1,999)E(999,1000) is equal to the (i, j)th entry of E(1, 1000).

Since i and j can be anything it follows that E(1,999)E(999,1000) = E(1, 1000).

--

( E(1,999)E(1000,999) )_{ij}
= (sum over k) E(1,999)_{ik} E(1000,999)_{kj}
= (sum over k) [delta(1, i)delta(999, k)delta(1000, k)delta(999, j)]
= (sum over k) [delta(1, i) * 0 * delta(999, j)] . . . because k can't equal 999 and 1000 at the same time
= 0

So E(1,999)E(1000,999) = 0.