# Linear Algebra III by Bookboon.com

By Bookboon.com

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The above equals ) ( ( −1 ) 1 n/2 det S exp − trace (B) det (B) 2 ( ) Of course the thing to estimate is only found in B. Therefore, det S −1 can be discarded in trying to maximize things. Since B is symmetric, it is similar to a diagonal matrix D which has λ1 , · · · , λn down the diagonal. Thus it is desired to maximize ) )n/2 ( p ( p ∏ 1∑ λi λi exp − 2 i=1 i=1 Taking ln it follows that it suﬃces to maximize p p n∑ 1∑ ln λi − λi 2 i=1 2 i=1 Taking the derivative with respect to λi , n 1 1 − =0 2 λi 2 and so λi = n.

To begin with, here is a simple lemma. 1 Let A be an m × n matrix. Then A∗ A is self adjoint and all its eigenvalues are nonnegative. 2 Proof: It is obvious that A∗ A is self adjoint. Suppose A∗ Ax = λx. Then λ |x| = (λx, x) = (A∗ Ax, x) = (Ax,Ax) ≥ 0. 2 Let A be an m × n matrix. The singular values of A are the square roots of the positive eigenvalues of A∗ A. With this deﬁnition and lemma here is the main theorem on the singular value decomposition. In all that follows, I will write the following partitioned matrix ( ) σ 0 0 0 where σ denotes an r × r diagonal matrix of the form   σ1 0   ..

By the Cauchy Schwarz inequality again, ||x|| ≡ ≡ �� �� )1/2 ( n n n ��∑ �� ∑ ∑ �� �� 2 |xi | ||vi || ≤ |x| ||vi || xi vi �� ≤ �� �� �� i=1 i=1 i=1 δ −1 |x| . com 52 Linear Algebra III Advanced topics Norms This proves the ﬁrst half of the inequality. Suppose the second half of the inequality is not valid. Then there exists a sequence xk ∈ X such that � k� �� �� �x � > k ��xk �� , k = 1, 2, · · · . Then deﬁne yk ≡ It follows Letting yik � k� �y � = 1, k xk . |xk | � k� �� �� �y � > k ��yk �� . 4) be the components of y with respect to the given basis, it follows the vector ) ( k y1 , · · · , ynk is a unit vector in Fn .