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sunjin09
#13
Apr1-12, 08:51 PM
P: 312
Quote Quote by Stephen Tashi View Post
Are you saying that vector 'a' will be chosen so that the vector Ua will be 1 at the jth component iff the largest singular value occurs in S at location S[j][j] and the vector Ua will be zero elsewhere?


In these two examples, do we have the same matrix for A'D but different answers for the maximum angle? (My 4-D intuition isn't good, so I'm not sure.)

Example 1: [itex] A = \begin{pmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}}\\ \frac{2}{\sqrt{15}}\\ \frac{1}{\sqrt{15}} \end{pmatrix} ,\ D = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{pmatrix} [/itex]

Example 2: [itex] A = \begin{pmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{6}} \end{pmatrix} ,\ D = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{pmatrix} [/itex]
As I solved this example, since A'*D is the same, so are a and b, actually a=1 and b=[1/sqrt(2),1/sqrt(2)]. But x1=A1≠x2=A2. However the angle is the same, since <x,y>=(Ua)'*S*(Vb) is totally determined by A'*D. Seemingly logical.