00:01
So for this problem, we're given a matrix a, which is an n by p matrix, matrix d, which is a diagonal matrix with d1 up to dn along its diagonal.
00:10
And we are asked to show that d .a, the product of the two, is found by multiplying the i -th row vector of a by d -i, where one is less than or equal to i is less than or equal to n.
00:24
Now, i note here that for the diagonal matrix, the i -jth element is just going to to be d i times the chronicer delta so that is going to do the same thing as d i j here sorry i should put that outside of the square braces that's the same thing as d i j so now we can look at this from the point of view of looking at the index form for the matrix product so d a element i j is equal to the sum from k equals 1 up to n of d element i times a element k j we know what d element i k is going to be that is going to be one second here equals 1 up to n this di k is going to be d i k is going to be d i delta i k times the k jth element of a.
01:45
But now, because we have this chronic or delta here, this is going to make this sum be, or each term in the sum, be zero, whenever i isn't the same thing as k, or k isn't the same thing as i.
01:58
So because that kills off all the terms, we're just going to have the result.
02:02
It's going to be d -i times delta -i -i, which is just going to be 1 times a, element i j or that's going to be d i and element a i j so now if we consider what the i throw of this d a matrix is going to look like so we have d a is equal to you know we have one dot dot dot dot dot dot dot going along so we will have d -i -i -1, then we'll have d -i -2, and so on, up to d -i -a -i -p...