Question

For any \(n \times m\) matrix \(A\) there exists an orthogonal \(m \times m\)matrix \(S\) such that the columns of matrix \(AS\) are orthogonal.

Short answer

The given statement is TRUE.

Step-by-step solution

Definition of a singular value decomposition

The single value decomposition produces orthonormal bases of v’s and u’s 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.

Step 2: To Find TRUE or FALSE

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.

Step .3: Final Answer

The given statement is TRUE.