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Linear Algebra, Theory And Applications, 2012a

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104 DETERMINANTS<br />

13. Let U be an open set in R n and let g :U → R n be such that all the first partial<br />

derivatives of all components of g exist and are continuous. Under these conditions<br />

form the matrix Dg (x) givenby<br />

Dg (x) ij<br />

≡ ∂g i (x)<br />

≡ g i,j (x)<br />

∂x j<br />

The best kept secret in calculus courses is that the linear transformation determined<br />

by this matrix Dg (x) is called the derivative of g and is the correct generalization<br />

of the concept of derivative of a function of one variable. Suppose the second partial<br />

derivatives also exist and are continuous. Then show that<br />

∑<br />

(cof (Dg)) ij,j<br />

=0.<br />

j<br />

Hint: First explain why ∑ i g i,k cof (Dg) ij<br />

= δ jk det (Dg) . Next differentiate with<br />

respect to x j and sum on j using the equality of mixed partial derivatives. Assume<br />

det (Dg) ≠ 0 to prove the identity in this special case. Then explain using Problem 10<br />

why there exists a sequence ε k → 0 such that for g εk (x) ≡ g (x)+ε k x, det (Dg εk ) ≠0<br />

and so the identity holds for g εk . Then take a limit to get the desired result in general.<br />

This is an extremely important identity which has surprising implications. One can<br />

build degree theory on it for example. It also leads to simple proofs of the Brouwer<br />

fixed point theorem from topology.<br />

14. A determinant of the form ∣∣∣∣∣∣∣∣∣∣∣∣∣ 1 1 ··· 1<br />

a 0 a 1 ··· a n<br />

a 2 0 a 2 1 ··· a 2 n<br />

.<br />

.<br />

.<br />

a n−1<br />

0 a n−1<br />

1 ··· a n−1<br />

n<br />

a n 0 a n 1 ··· a n ∣<br />

n<br />

is called a Vandermonde determinant. Show this determinant equals<br />

∏<br />

(a j − a i )<br />

0≤ii.<br />

Hint: Show it works if n = 1 so you are looking at<br />

∣ 1 1 ∣∣∣<br />

∣ a 0 a 1<br />

Then suppose it holds for n − 1 and consider the case n. Consider the polynomial in<br />

t, p (t) which is obtained from the above by replacing the last column with the column<br />

( ) 1 t ··· t<br />

n T<br />

.<br />

Explain why p (a j )=0fori =0, ··· ,n− 1. Explain why<br />

n−1<br />

∏<br />

p (t) =c (t − a i ) .<br />

i=0<br />

Of course c is the coefficient of t n . Find this coefficient from the above description of<br />

p (t) and the induction hypothesis. Then plug in t = a n and observe you have the<br />

formula valid for n.

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