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Let $A\in M_n$ and $s_j$ be singular value of $A$ with associated left singular vectors $u_1,u_2,....u_n$ and associated right singular vectors $v_1,v_2,....v_n$

Suppose $T=[u_1,...,u_n][v_1,...,v_n]^*$

Why does $Tv_j=u_j$?

(for all $j$)

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1 Answer 1

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Hint: $T = UV^*$, and $v_j = Ve_j$ where $e_j$ denotes the $j$th standard basis vector

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    $\begingroup$ $V^*v_j = V^*(Ve_j) = (V^*V)e_j$ $\endgroup$
    – Ben Grossmann
    Mar 3, 2016 at 13:31

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