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In SVD, $A=U\Sigma V^{T}=(UV^{T})(V \Sigma V^{T})=QS$

Here is an example.
$A= \left(\begin{array}{cc} 1 & - 2\\ 3 & - 1 \end{array}\right) = \left(\begin{array}{cc} 0 & - 1\\ 1 & 0 \end{array}\right) \left(\begin{array}{cc} 3 & - 1\\ - 1 & 2 \end{array}\right) =QS$

I can't draw this answer.
Since $U$ contains columns which are eigenvectors of $AA^{T}$, and by calculation I get root value, so I think Q also has root value.
Can you help me to get that answer?

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    OMG! That's my huge mistake! I corrected the number in the first matrix!2012-12-07

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