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The single value decomposition produces orthonormal bases of v鈥檚 and u鈥檚 for the four fundamental sub-spaces.

Using those bases, A becomes a diagonal matrix\(\sum \)and\(A{v_i} = {\sigma _i}{u_i}\)\({\sigma _i}\)= singular value.

Let \(A = U\Sigma {V^T}\)be singular value decomposition of \(A\). Rewrite it as \(AV = U\Sigma \).

If we expand \(U\Sigma \), then

\(U\Sigma = \left[ {\begin{array}{*{20}{l}}{{\sigma _1}{{\dot u}_1}}&{{\sigma _2}{{\dot u}_2}}& \cdots &{{\sigma _m}{{\dot u}_m}}\end{array}} \right]\)

if \(n \ge m\), and

\(U\Sigma = \left[ {\begin{array}{*{20}{l}}{{\sigma _1}{{\dot u}_1}}&{{\sigma _2}{{\dot u}_2}}& \cdots &{{\sigma _m}{{\dot u}_m}}&0& \cdots &0\end{array}} \right]\)

if \(n < m\).

In any case, the columns of \(U\Sigma \) are orthogonal since \({\dot u_i}\) 's are orthogonal. That is, the columns of \(AV\) are orthogonal.

The given statement is TRUE.