Question-50

Let \(\displaystyle A\) be a real invertible matrix whose SVD is given by:

\[\begin{equation*} A=U\Sigma V^{T} \end{equation*}\]

• \(\displaystyle A^{-1} =U\Sigma ^{-1} V^{T}\)
• \(\displaystyle A^{-1} =V\Sigma U^{T}\)
• \(\displaystyle A^{-1} =V\Sigma ^{-1} U^{T}\)
• \(\displaystyle A^{-1} =U^{-1} \Sigma ^{-1} V^{-1}\)

Answer

• \(\displaystyle A^{-1} =U\Sigma ^{-1} V^{T}\)
• \(\displaystyle A^{-1} =V\Sigma U^{T}\)
• \(\displaystyle A^{-1} =V\Sigma ^{-1} U^{T}\)
• \(\displaystyle A^{-1} =U^{-1} \Sigma ^{-1} V^{-1}\)

Solution

Since \(\displaystyle A\) is invertible, it has full rank. All three matrices, \(\displaystyle U,\Sigma ,V\), are square and invertible. \(\displaystyle \Sigma\) in particular is diagonal with non-zero entries on the diagonal. Recall that \(\displaystyle U\) and \(\displaystyle V\) are orthogonal. It follows that the transpose is the inverse for these two matrices. With all this the inverse becomes:

\[\begin{equation*} A^{-1} =V\Sigma ^{-1} U^{T} \end{equation*}\] We can check this:

\[\begin{equation*} AA^{-1} =\left( U\Sigma V^{T}\right)\left( V\Sigma ^{-1} U^{T}\right) =I \end{equation*}\]