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Call the matrix in the middle M Its specialized format a Vandermonde matrix gives it many remarkable properties, of which the following is particularly relevant to us
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If x0 , , xn 1 are distinct numbers, then M is invertible
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The existence of M 1 allows us to invert the preceding matrix equation so as to express coef cients in terms of values In brief,
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Evaluation is multiplication by M , while interpolation is multiplication by M 1
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This reformulation of our polynomial operations reveals their essential nature more clearly Among other things, it nally justi es an assumption we have been making throughout, that A(x) is uniquely characterized by its values at any n points in fact, we now have an explicit formula that will give us the coef cients of A(x) in this situation Vandermonde matrices also have the distinction of being quicker to invert than more general matrices, in O(n 2 ) time instead of O(n3 ) However, using this for interpolation would still not be fast enough for us, so once again we turn to our special choice of points the complex roots of unity Interpolation resolved In linear algebra terms, the FFT multiplies an arbitrary n-dimensional vector which we have been calling the coef cient representation by the n n matrix 1 1 1 1 row for 0 = 1 2 n 1 1 1 2 4 2(n 1) 2 Mn ( ) = j 2j (n 1)j 1 j (n 1) 2(n 1) (n 1)(n 1) n 1 1
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where is a complex nth root of unity, and n is a power of 2 Notice how simple this matrix is to describe: its (j, k)th entry (starting row- and column-count at zero) is jk Multiplication by M = Mn ( ) maps the kth coordinate axis (the vector with all zeros except for a 1 at position k) onto the kth column of M Now here s the crucial observation, which we ll prove shortly: the columns of M are orthogonal (at right angles) to each other Therefore they can be thought of as the axes of an alternative coordinate system, which is often called 68
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Figure 28 The FFT takes points in the standard coordinate system, whose axes are shown here as x1 , x2 , x3 , and rotates them into the Fourier basis, whose axes are the columns of Mn ( ), shown here as f1 , f2 , f3 For instance, points in direction x 1 get mapped into direction f1
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the Fourier basis The effect of multiplying a vector by M is to rotate it from the standard basis, with the usual set of axes, into the Fourier basis, which is de ned by the columns of M (Figure 28) The FFT is thus a change of basis, a rigid rotation The inverse of M is the opposite rotation, from the Fourier basis back into the standard basis When we write out the orthogonality condition precisely, we will be able to read off this inverse transformation with ease: Inversion formula Mn ( ) 1 =
1 1 n Mn ( )
But 1 is also an nth root of unity, and so interpolation or equivalently, multiplication by Mn ( ) 1 is itself just an FFT operation, but with replaced by 1 Now let s get into the details Take to be e 2 i/n for convenience, and think of the columns of M as vectors in Cn Recall that the angle between two vectors u = (u 0 , , un 1 ) and v = (v0 , , vn 1 ) in Cn is just a scaling factor times their inner product
u v = u0 v0 + u1 v1 + + un 1 vn 1 ,
where z denotes the complex conjugate4 of z This quantity is maximized when the vectors lie in the same direction and is zero when the vectors are orthogonal to each other The fundamental observation we need is the following Lemma The columns of matrix M are orthogonal to each other Proof Take the inner product of any columns j and k of matrix M , 1 + j k + 2(j k) + + (n 1)(j k)
This is a geometric series with rst term 1, last term (n 1)(j k) , and ratio (j k) Therefore it evaluates to (1 n(j k) )/(1 (j k) ), which is 0 except when j = k, in which case all terms are 1 and the sum is n
The complex conjugate of a complex number z = rei is z = re i The complex conjugate of a vector (or matrix) is obtained by taking the complex conjugates of all its entries
The orthogonality property can be summarized in the single equation M M = nI, since (M M )ij is the inner product of the ith and jth columns of M (do you see why ) This immediately implies M 1 = (1/n)M : we have an inversion formula! But is it the same formula we earlier claimed Let s see the (j, k)th entry of M is the complex conjugate of the corresponding entry of M , in other words jk Whereupon M = Mn ( 1 ), and we re done And now we can nally step back and view the whole affair geometrically The task we need to perform, polynomial multiplication, is a lot easier in the Fourier basis than in the standard basis Therefore, we rst rotate vectors into the Fourier basis (evaluation), then perform the task (multiplication), and nally rotate back (interpolation) The initial vectors are coef cient representations, while their rotated counterparts are value representations To ef ciently switch between these, back and forth, is the province of the FFT
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