The Hadamard matrices H0, H1, H2, . . . are defined as...

The Hadamard matrices H0, H1, H2, . . . are defined as follows: • H0 is the 1 × 1 matrix 1 • For k > 0, Hk is the 2 k × 2 k matrix Hk = Hk?1 Hk?1 Hk?1 ?Hk?1 Show that if v is a column vector of length n = 2 k , then the matrix-vector product Hkv can be calculated using O(n log n) operations. Assume that all the numbers involved are small enough that basic arithmetic operations like addition and multiplication take unit time.

The Hadamard matrices H0, H1, H2, . . . are defined as follows:
• H0 is the 1 × 1 matrix
1

• For k > 0, Hk is the 2
k × 2
k matrix
Hk =

Hk?1 Hk?1
Hk?1 ?Hk?1

Show that if v is a column vector of length n = 2
k
, then the matrix-vector product Hkv can be
calculated using O(n log n) operations. Assume that all the numbers involved are small enough
that basic arithmetic operations like addition and multiplication take unit time.

Transcript

00:01 Okay, so for problem 12, we're looking at the distributive property, and for part a, we want to take a plus b and multiply it by c.

00:14 Okay, so this is going to look like a matrix form.

00:19 We would have a plus b being a11 plus b11, all the way to our amk plus bmk, and then we multiply that by our c matrix matrix.

00:39 And our c matrix being k by n.

00:48 Okay, and so when we take our, for example, our first row by the first column, so this would be in our product matrix, only the element in the first row, first column, that element is gonna look like this.

01:07 We're gonna add a11 plus b11 times c11.

01:13 I'm gonna add that to a12 plus b12 multiplied by c21.

01:21 And we're going to add that.

01:22 We'll go all the way down to a1k plus b1k multiplied by ck1.

01:37 And you can, of course, do this, write it more generally with notation, index notation, but the indexes are confusing enough in the discrete case.

01:46 So i'm hoping this will be more illuminating.

01:49 But in this one element, this first row, first column element, these values, we can use the distribution property, right? these are just regular all numbers and we know that that works.

02:02 Okay? so this is going to be equivalent to a11c11 plus b11 c11 all the way down to a1kk1, k1, k1.

02:27 Okay, and then we can sort of separate these, right? so we can take all of our elements with a, so a11, c111, all the way down to the a1k1, ck1, and add that to all of the elements with b, b11, c11, all that across to the b1kk1.

02:58 Okay, well this, this then is actually going to be the same as taking a, the product of a and c...

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